Solutions and Solubility Equilibria
A complex ion is an ion formed when electron pair donors and electron pair acceptors form coordinate covalent bonds.
How are complex ions formed?
If the entropy of an endothermic reaction is greater than the enthalpy, the change in Gibbs Free Energy will be negative, making the reaction spontaneous.
How can an endothermic reaction be spontaneous?
Multiply mOsm/L by 10^-3
How can mOsm/L be converted to mol/L?
The addition of a solute decreases the evaporation rate of a solvent without affecting the condensation rate, thus decreasing the vapor pressure of a solvent.
How does adding a solute to a solvent affect the vapor pressure of a solvent?
The solubility of solids can be increased by increasing the temperature.
How does temperature affect the solubility of solids?
Solubility of gases can be increased by decreasing temperature or increasing the partial pressure of the gas above the solvent.
How does temperature and partial pressure affect the solubility of gases?
m = moles solute / kg solvent
How molality is calculated: m = _________ solute/________ solvent
M = moles of solute/liters of solution
How molarity of a solute is calculated: M = _________ of solute/_________ of solution
X = moles of a specific species/total moles of all species
How mole fraction is calculated: X = #moles of _____________________/#moles of __________________
N = #equivalents * molarity
How normality is calculated: N = _______________ * ________________
%composition by mass = (mass of solute/mass of solution) * 100%
How percent composition by mass is calculated: %composition by mass = (mass of _________/mass of ___________) * 100%
MiVi=MfVf Mi = Initial concentration Mf = Final concentration Vi = Initial volume Vf = Final volume
How the concentration of a solution after dilution is calculated: MiVi = MfVf Mi = ____________ Mf = ____________ Vi = _____________ Vf = _____________
∏ = i * M * R * T ∏ = osmotic pressure i = van't Hoff factor M = Molarity R = Ideal Gas Constant (0.082 L*atm/molK) T = Temperature
How to calculate osmotic pressure: ∏ = i * M * R * T ∏ = ____________________ i = ______________________ M = ___________________________ R = _______________________ T = ____________________
ΔT(b) = i*K(b)*m ΔT(b) = increase in boiling point m = molality K(b) = proportionality constant of solvent i = van't Hoff factor
How to calculate the change in boiling point for a solvent when a solute is added: ΔT(b) = i*K(b)*m ΔT(b) = ____________________ m = ________________ K(b) = __________________________ i = ________________________
ΔT(f) = i * K(f) * m ΔT(f) = decrease in freezing point i = van't Hoff factor K(f) = proportionality constant of solvent m = molality
How to calculate the change in freezing point for a solvent when a solute is added: ΔT(f) = i * K(f) * m ΔT(f) = ___________________________ i = _______________________ K(f) = ________________________________ m = __________________
P = Vapor pressure of solvent with solute added X = Mole fraction of solvent P° = Vapor pressure of solvent without solute
How to calculate the change in vapor pressure of a solvent when a solute is added: P = X * P° P = __________________ X = _________________ P° = __________________
The addition of a complex ion results in a certain ion in a solution to be taken up by the it, essentially resulting in a decrease of that ion in the solution. The solubility of the solute added to the solution will increase with the introduction of the complex ion.
How would the addition of a complex ion affect the solubility of a solute.
Acceptable answers: -Group 1 metals -Ammonium (NH4+) -Nitrate (NO3-) -Acetate (CH3COO-)
Name two ions that form salts that are always soluble.
1. 2 - one particle of Na+ and one particle of Cl- 2. 4 - one particle of Ag(3+) and three particles of Cl- 3. 3 - one particle of Pb(2+) and two particles of Cl-
1. What is the van't Hoff factor of NaCl? 2. What is the van't Hoff factor of AgCl3? 3. What is the van't Hoff factor of PbCl2?
ΔT(f) = i * K(f) * m i = 4 since AgCl3 dissociates into 4 particles. m = moles solute/kg solvent 400 g AlCl3/133 g/mol = 3 mol 1.5 L water = 1500 mL * 1 g/mL = 1500 g = 1.5 kg water 3 mol AlCl3/1.5 kg water = 2 mol/kg ΔT(f) = 4 * 1.86 K*kg/mol * 2 mol/kg ΔT(f) = 14.9 K Freezing point of water is 273 K 273 K - 14.9 K = 258.1 K 258.1 K - 273 = -14.9°C
400 g AlCl3 is dissolved in 1.5 L of water at room temperature (K(f) = 1.86 K*kg/mol). What is the new freezing point of this solution?
ΔT(b) = i*K(b)*m i=4 since AlCl3 dissociates into 1 particle Al(3+) and 3 particles Cl-. m = mol solute/kg solvent 1.5 L water = 1500 mL water * 1 g/mL = 1500 g water = 1.5 kg water 400 g AlCl3/133 g/mol AlCl3 = 3 mol AlCl3 3 mol AlCl3/1.5 kg water = 2 mol/kg m = 2 mol/kg ΔT(b) = 4 * 0.512 K*kg/mol * 2 mol/kg ΔT(b) = 4 K
400 g AlCl3 is dissolved in 1.5 L of water at room temperature (Kb = 0.512 K*kg/mol). How much does the boiling point increase after adding thee AlCl3?
Dilute Concentrated
A ____________ solution is a solution in which the proportions of solute to solvent is considerably small. A __________________ solution is a solution in which the proportions of solute to solvent is considerably large.
Saturated Unsaturated Super-Saturated
A ______________ solution is a solution in which the concentration of solute is at its maximum value in the solution. A __________________ solution is a solution in which the concentration of solute is not at its maximum value in the solution. A __________________ solution is a solution in which the concentration of solute has exceeded its maximum value in the solution.
Electrolyte More Electrolyte Less Electrolyte
A _________________ is a substance that creates ions in a solution. The __________ soluble a molecule is, the stronger the ______________________. The ___________ soluble a molecule is, the weaker the _______________.
Ideal solution
A _________________________ is a solution with an enthalpy of dissolution that is equal to zero.
a) Since we have PbCl2 precipitate, we know that this is a saturated solution of Pb(2+) and Cl-. Out of the 10 grams of PbCl2 added, only 0.22 grams gets dissolved into solution. solubility g/L = 0.22g/0.05L = 4.4 g/L To find molar solubility, we must find the number of moles of the solute dissolved. 0.22 g/278.1 g/mol = 7.9 x 10^-4 mol solubility mol/L = 7.9 x 10^-4 mol/0.05 L = 0.016 M Solubility (g/L) = 4.4 g/L Molar Solubility (mol/L) = 0.016 M b) Since the solution is saturated, the rate of reaction of dissolution and precipitation is in equilibrium. PbCl2 (s) <-----> Pb(2+) (aq) + Cl- (aq) Balance reaction: 1PbCl2 (s) <-----> 1Pb(2+) (aq) + 2Cl- (aq) From a), we found out the molarity of PbCl2 is 0.016 M. Use ICE to find molarity of products: *multiply molarities by coefficients 1PbCl2 (s) <-----> 1Pb(2+) (aq) + 2Cl- (aq) N/A 0 0 - 1*(0.016) + 1*(0.016) + 2*(0.016) N/A 0.016 0.032 Ksp = [Pb(2+)][Cl-] (PbCl2 is left out because we always leave out pure solids and liquids in equilibrium expressions) Raise concentrations to the power of coefficients: Ksp = [Pb(2+)]^1 * [Cl-]^2 Ksp = [0.016 M]^1 * [0.032 M]^2 Ksp = 0.016 M * 0.001 M Ksp = 1.6 x 10^-5 M (Solving for Ksp is like finding Ka or Kb in acid-base equilibria) c) Nothing will change since only 0.22 g of that 100 g will be the total amount that will dissolve, regardless of how much solute is added. Molar solubility will still remain the same as well as our solubility product constant Ksp.
A chemist adds 10.00 g of PbCl2 to 50.00 mL of water at 25°C and finds that only 0.22 g of the solid goes into solution. (Molar mass PbCl2 = 278.1 g/mol) a) Calculate the solubility (in g/L and mol/L) of PbCl2 in water at 25°C b) Calculate the solubility product constant Ksp (at 25°C) for PbCl2. c) How would the results change if we had tried to dissolve 100 g of PbCl2 in the same volume of water?
MiVi=MfVf 1.1 M * 0.3 L = 5.5 M * Vf 1.1 M * 0.3 L = Vf 5.5 M Vf = 0.06 L
A chemist wishes to prepare 300 mL of a 1.1 M NaOH solution from a 5.5 M NaOH stock solution. What volume of stock solution should be diluted with pure water to obtain the desired solution?
Spontaneous
A reaction that has a decrease in free energy (ΔG negative), or favors the formation of products is ____________________.
Nonspontaneous
A reaction that has an increase in free energy (ΔG positive), or favors the formation of reactants is ________________________.
Answer: D. Since both salts have a formula MX3 (one of one particle, and three of another), it is possible to directly compare the molar solubilities of each. When the solutions are mixed, [OH-] is above saturation levels for both the cobalt and the thallium in the solution. Since thallium (III) hydroxide has a smaller Ksp than that of cobalt(III) hydroxide, it will react first. The ion product of the mixed solution is higher than the Ksp for thallium(III) hydroxide, and the system will shift left to precipitate solid thallium(III) hydroxide. After the thallium(III) hydroxide precipitates, a small excess of OH- will remain, which gives an ion product slightly above the Ksp of cobalt(III) hydroxide. This will cause a small amount (1% - 3%) of cobalt(III) hydroxide to also precipitate.
A saturated solution of cobalt(III) hydroxide (Ksp = 1.6 x 10^-44) is added to a saturated solution of thallium(III) hydroxide (Ksp = 6.3 x 10^-46). What is likely to occur? A. Both cobalt(III) hydroxide and thallium(III) hydroxide will remain stable in solution. B. Cobalt(III) hydroxide precipitates and thallium(III) hydroxide remains stable in solution. C. Thallium(III) hydroxide precipitates and cobalt(III) hydroxide remains stable in solution. D. Both thallium(III) hydroxide and cobalt(III) hydroxide precipitate.
Mole fraction: 90 g NaCl/58.5 g/mol NaCl = 1.5 mol NaCl 10 L H2O = 10000 mL H2O * 1 g/mL = 10000 g H2O 10000 g H2O/18 g/mol H2O = 555.6 mol H2O 555.6 + 1.5 = 557.1 mol solution 1.5 mol NaCl/ 557.1 mol solution = 0.0027 mole fraction = 2.7 x 10^-3 Percent composition by mass: 10000 g H2O + 90 g NaCl = 10090 g solution 90 g NaCl/10090 g solution * 100% = 0.9% percent composition by mass = 0.9%
A stock solution for making typical IV saline bags contains 90.0 g of NaCl per 10 liters of water (density = 1 g/mL) What is the mole fraction and the percent composition by mass of NaCl in the saline solutions?
Answer: A. The equation ΔTb = iKbm can be used to solve this problem. The change in boiling point is 374 - 373 = 1 K. Then plug that into: ΔTb/iKb = m 1 K/(1 * 0.512 K*kg/mol) = 2 m The van't Hoff factor for this solute is 1 because the molecule does not dissociate into smaller components. Then convert to grams of solute using the definition of molality. mol solute/kg solvent = molality mol solute = 2 m * 0.1 kg = 0.2 mol The mass used in this equation is 0.1 kg because 100 mL of water has a mass of 0.1 kg. Then determine the molar mass: molar mass = 70 g/0.2 mol = 350 g/mol, which is closest to choice A.
An aqueous solution was prepared by mixing 70 g of an unknown nondissociating solute into 100 g of water. The solution has a boiling point of 101.0°C. What is the molar mass of the solute? (Note: Kb = 0.512 K*kg/mol). A. 358.4 g/mol B. 32.3 g/mol C. 123.2 g/mol D. 233.6 g/mol
Positive
An endothermic reaction absorbs heat from the surroundings as a reaction proceeds, making the enthalpy (ΔH) of the reaction _______________.
Reactants Products
An endothermic reaction favors the _______________ in a reaction. An exothermic reaction favors the ________________ in a reaction.
Negative
An exothermic reaction gives off heat into the surroundings as a reaction proceeds, making the enthalpy (ΔH) of the reaction ___________________
increases
At constant temperature entropy always ___________ upon dissolution.
Molar solubility = molarity Ni(OH)2 (s) <---> Ni(2+) (aq) + 2OH- I 0 0 C -x +x +2x E x 2x Ksp = [Ni(2+)] * [2OH-]^2 Ksp = [5.2 x 10^-6] * [2 * (5.2 x 10^-6)]^2 Ksp = 5.2 x 10^-6 * 1.1 x 10^-10 Ksp = 5.7 x 10^-16
Calculate the Ksp of Ni(OH)2 in water, given that its molar solubility is 5.2 x 10^-6 M.
a) 1AgCl (s) <-----> 1Ag+ (aq) + 1Cl- (aq) I N/A 0 0 C -x +x +x E x x Ksp = [Ag+]^1 * [Cl-]^1 1.8 x 10^-10 = x^2 sqrt(1.8 x 10^-10) = sqrt(x^2) x = 1.3 x 10^-5 [Ag+] = x = [AgCl] Molar Solubility = 1.3 x 10^-5 M b) The addition of NH3 results in the formation of Ag(NH3)2 complexes Ag+ (aq) + 2NH3 (aq) <-----> Ag(NH3)2+ Because Kf is high, the reaction favors the products, meaning that the reaction is exothermic. Thus, Ag+ will be taken by the NH3 and essentially removed from the solution, which leads to a decrease in Ag+ ions. This all leads to an increase in solubility for AgCl. Form a net reaction between the dissolution and formation reactions: AgCl (s) + Ag+ (aq) + 2NH3 (aq) <-----> Ag+ (aq) + Cl- (aq) + Ag(NH3)2+ (aq) AgCl (s) + 2NH3 (aq) <-----> Ag(NH3)2+ (aq) + Cl- (aq) Multiply the two equilibrium constants of the two reactions: 1.8 x 10^-10 * 1.6 x 10^7 2.9 x 10^-3 K = 2.9 x 10^-3 Use ICE for net reaction: AgCl (s) + 2NH3 (aq) <-----> Ag(NH3)2+ (aq) + Cl- (aq) I N/A 3.0 M 0 0 C -x -2x +x +x E N/A 3.0 - 2x x x K = ([Ag(NH3)2+]^1 * [Cl-]^1)/([NH3]^2) 2.9 x 10^-3 = x^2/(3.0 - 2x)^2 Since K > 1 x 10^-4, we cannot assume x<<1 2.9 x 10^-3 = (x/(3.0 - 2x))^2 sqrt(2.9 x 10^-3) = sqrt((x/(3.0 - 2x))^2) 0.054 = x/(3.0 - 2x) 0.054 * (3.0 - 2x) = x 0.016 - 0.11x = x 0.016 = 1.11x 0.016/1.11 = 1.11x/1.11 x = 0.14 M Molar Solubility in 3.0 M NH3 = 0.14 M
Calculate the molar solubility of AgCl (Ksp = 1.8 x 10^-10 at 25°C) in... a) pure water b) 3.0 M NH3 (Kf = 1.6 x 10^7 at 25°C)
KCl shares a common ion with PbCl2 PbCl2 (s) <-----> Pb(2+) (aq) + Cl- (aq) Balance reaction: 1PbCl2 (s) <-----> 1Pb(2+) (aq) + 2Cl- (aq) Use ICE: 1PbCl2 (s) <-----> 1Pb(2+) (aq) + 2Cl- (aq) I N/A 0 0.1 C -x + 1x + 2x E N/A x 2x + 0.1 Ksp = [Pb(2+)]^1 * [Cl-]^2 1.6 x 10^-5 = x * (2x + 0.1)^2 Since Ksp < 10^-4, to make things easier, we assume x<<1 1.6 x 10^-5 = x * (0.1)^2 1.6 x 10^-5 = x * 0.01 x = 0.0016 M x = [Pb(2+)] = [PbCl2] Molar solubility = 0.0016 M in KCl
Calculate the molar solubility of PbCl2 (Ksp = 1.6 x 10^-5 at 25°C) in a 0.100 M solution of KCl.
Find solubility in both mol/L and g/L Write and balance the reaction: Cu(OH)2 (s) <-----> Cu(2+) (aq) + OH- (aq) 1Cu(OH)2 (s) <-----> 1Cu(2+) (aq) + 2OH- (aq) I N/A 0 0 C - 1x + 1x + 2x E x 2x Ksp = [Cu(2+)]^1 * [OH-]^2 2.2 x 10^-20 = x^1 * (2x)^2 2.2 x 10^-20 = x^1 * 4x^2 2.2 x 10^-20 = 4x^3 2.2 x 10^-20 = 4x^3 4 4 5.5 x 10^-21 = x^3 cbrt(5.5 x 10^-21) = cbrt(x^3) x = 1.8 x 10^-7 M [Cu(2+)] = x [Cu(OH)2] = x so... [Cu(2+)] = [Cu(OH)2] Therefore: Molar solubility of Cu(OH)2 = 1.8 x 10^-7 M Now find solubility in g/L 1.8 x 10^-7 M = 1.8 x 10^-7 mol/L Multiply by molar mass: 1.8 x 10^-7 mol/L * 97.57 g/mol = 1.8 x 10^-5 g/L Solubility = 1.8 x 10^-7 mol/L & 1.8 x 10^-5 g/L
Calculate the solubility of copper (II) hydroxide. (Ksp = 2.2 x 10^-20 at 25°C).
ΔT(f) = i * K(f) * m i = 2 since NaCl dissociates into 2 particles m = moles solute/kg solvent 1800 g water = 1.8 kg water 58.5 g NaCl/58.5 g/mol NaCl = 1 mol NaCl 1 mol NaCl/1.8 kg of water = 0.56 mol/kg ΔT(f) = 2 * 1.86 K*kg/mol * 0.56 mol/kg = 2.08 K
Determine the freezing point depression of a solution containing 58.5 g of NaCl in 1800 g of water at room temperature. (K(f) = 1.86 K*kg/mol).
ΔT(b) = i * K(b) * m i = 3 since MgCl2 dissociates into 3 particles m = moles solute/kg solvent 190 g MgCl2/95 g/mol MgCl2 = 2 mol 1500 g water = 1.5 kg water 2 mol/1.5 kg = 1.333 mol/kg ΔT(b) = 3 * 0.512 K*kg/mol * 1.333 mol/kg = 2 K 373 K is the boiling point of water 373 K + 2 K = 375 K
Determine the new boiling point of a solution containing 190 g MgCl2 in 1500 g water at room temperature (K(b) = 0.512 K*kg/mol).
P = XP° 540 g H2O/18 g/mol H2O = 30 mol H2O 190 g MgCl2/95 g/mol MgCl2 = 2 mol MgCl2 X(H2O) = 30 mol H2O/32 mol solution = 0.94 P = 0.94 * 3.2 kPa = 3 kPa
Determine the vapor pressure of a solution containing 190 g MgCl2 in 540 g water at room temperature. (Note: The vapor pressure of pure water at 25°C is 3.2 kPa).
High temperatures Low temperatures
Endothermic dissolutions are favored at ____________ temperatures Exothermic dissolutions are favored at ______________ temperatures.
Calculate moles of NaOH: 10 g NaOH/40 g/mol NaOH = 0.25 mol Find molality: 0.25 mol NaOH/0.5 kg H2O = 0.5 m
If 10 g NaOH are dissolved in 500 g of water, what is the molality of the solution?
Find moles: 184 g/92 g/mol = 2 mol C3H8O3 180 g/18 g/mol = 10 mol H2O Combine the moles to get moles of solution: 2 mol (C3H8O3) + 10 mol (H2O) = 12 mol Determine mole fractions of each: C3H8O3: 2 mol/12 mol = 1/6 = 0.17 H2O: 10 mol/12 mol = 5/6 = 0.83
If 184 g glycerol (C3H8O3) is mixed with 180 g water, what will be the mole fractions of the two components? (Note: Molar mass of H2O = 18 g; molar mass C3H8O3 = 92 g/mol)
Unsaturated dissolve super-saturated precipitation saturated not dissolve
If IP < Ksp, the solution is ____________, and the solute will ______________ If IP > Ksp, the solution is ________________, and the solute will _______________. If IP = Ksp, the solution is _______________, and the solute will ______________________________
10^-4
If K is less than or equal to __________, you assume x is negligible in an ICE chart?
Find moles: 11 g CaCl2/ 111.1 g/mol CaCl2 = 0.1 mol CaCl2 Find molarity: 0.1 mol CaCl2/ 0.1 L solution = 1 M
If enough water is added to 11 g of CaCl2 to make 100 mL of solution, what is the molarity of the solution?
Endothermic.
If intermolecular interactions prior to dissolution are stronger than intermolecular interactions formed by dissolution, the reaction will be __________________.
Exothermic.
If intermolecular interactions prior to dissolution are weaker than intermolecular interactions formed by dissolution, is the reaction will be _________________.
Not affect
If the anion in a solution is a conjugate of a strong acid, would the addition of an acid increase, decrease, or not affect the solubility of the solute
Increase
If the anion in a solution is a conjugate of a weak acid, would the addition of an acid increase, decrease, or not affect the solubility of the solute?
180 g/60 g/mol = 3 mol carbonate 250 mL * 1 g/mL = 250 g = 0.25 kg water 3 mol carbonate/0.25 kg water = 12 m molality = 12 m
If you mix 180 g of carbonate in 250 mL of water (density = 1 g/mL) what is its concentration in molality? (Note: Molar mass of carbonate is 60 g/mol)
Molality: Find kg water: 250 mL * 1 g/mL = 250 g = 0.25 kg H2O Find moles of glucose: 180 g/ 180 g/mol = 1 mol Find molality: 1 mol (glucose)/ 0.25 kg (water) = 4 m molality = 4 m
If you mix 180 g of glucose in 250 mL of water (density = 1 g/mL) what is its concentration in molality? (Note: Molar mass of glucose is 180 g/mol)
Answer: B. 200 ppb of Pb(2+) is equivalent to 200 grams of Pb(2+) in 10^9 grams solution; given the extremely low concentration of lead, the mass of the water can be assumed to be approximately 10^9 grams, as well. To solve, set up a dimensional analysis question. The units needed at the end are moles/L (molarity), so convert from grams of lead to moles of lead and grams of water to liters of water: g (Pb(2+))/10^9 g (H2O) * density (H2O)/molar mass (Pb(2+)) 200 g Pb(2+) * 10^3 g H2O = 1 mol Pb(2+) = 200 x 10^3 mol 10^9 g H2O..................1L....................207.2 g Pb(2+)......207.2 x 10^9 L = 9.65 x 10^-7 mol/L Note that the denominator was rounded to a smaller number, meaning the estimated answer is slightly larger than the actual value.
Lead is a toxic element that can cause many symptoms, including mental retardation in children. If a body of water is polluted with lead ions at 200 ppb (parts per billion), what is the concentration of lead expressed in molarity? (Note: the density of water is 1 g/mL, and ppb = grams per 10^9 grams of solution) A. 9.7 x 10^-10 M Pb(2+) B. 9.7 x 10^-7 M Pb(2+) C. 6.2 x 10^-7 M Pb(2+) D. 6.2 x 10^-6 M Pb(2+)
Hypobromite
Name the common polyatomic ion: BrO-
Bromite
Name the common polyatomic ion: BrO2-
Bromate
Name the common polyatomic ion: BrO3-
Perbromate
Name the common polyatomic ion: BrO4-
Oxalate
Name the common polyatomic ion: C2O4(2-)
Acetate
Name the common polyatomic ion: CH3COO-
Cyanide
Name the common polyatomic ion: CN-
Carbonate
Name the common polyatomic ion: CO3(2-)
Hypochlorite
Name the common polyatomic ion: ClO-
Chlorite
Name the common polyatomic ion: ClO2-
Chlorate
Name the common polyatomic ion: ClO3-
Perchlorate
Name the common polyatomic ion: ClO4-
Dichromate
Name the common polyatomic ion: Cr2O7(2-)
Chromate
Name the common polyatomic ion: CrO4(2-)
Dihydrogen phosphate
Name the common polyatomic ion: H2PO4-
Bicarbonate (hydrogen carbonate)
Name the common polyatomic ion: HCO3-
Biphosphate (hydrogen phosphate)
Name the common polyatomic ion: HPO4(2-)
Bisulfate (hydrogen sulfate)
Name the common polyatomic ion: HSO4-
Permanganate
Name the common polyatomic ion: MnO4-
Ammonium
Name the common polyatomic ion: NH4+
Nitrite
Name the common polyatomic ion: NO2-
Peroxide
Name the common polyatomic ion: O2(2-)
Hydroxide
Name the common polyatomic ion: OH-
Phosphate
Name the common polyatomic ion: PO4(3-)
Thiosulfate
Name the common polyatomic ion: S2O3(2-)
Thiocyanate
Name the common polyatomic ion: SCN-
Sulfite
Name the common polyatomic ion: SO3(2-)
Sulfate
Name the common polyatomic ion: SO4(2-)
Nitrate
Name the following polyatomic ion: NO3-
Answer: B. The mass percent of a solute equals the mass of the solute divided by the mass of the total solution times 100%. Plug in the values given for sucrose, the volume of water and the density of water to determine the %mass of sucrose. %mass = mass solute/total mass of solution * 100% %mass sucrose = 100 g sucrose/(292.5 g water + 100 g sucrose) * 100% = 25.5%. Keep in mind that in rounding while calculating, the denominator was estimated to be larger than the actual value, thus giving an answer that is slightly lower than the actual value. Thus the correct answer is (B.), 25.5%, (A.) results if rounding error is not taken into account. While these answers are very close, the mass of water must be slightly less than 300 g, given the density value, so percent composition of sucrose must be slightly higher than 25%. If the solute's mass is not added to the solvent's, the calculated value is 34.2%, which is (D.). (C.) neglects both the addition step and rounding error.
One hundred grams of sucrose are dissolved in a cup of hot water at 80°C. The cup of water contains 300.00 mL of water What is the percent composition by mass of sugar in the resulting solution? (Note: Sucrose = C12H22O11, density of water at 80°C = 0.975 g/mL) A. 25.0% B. 25.5% C. 33.3% D. 34.2%
Answer: B. Osmotic pressure is given by the formula ∏=iMRT. Entering the values from the question stem gives: ∏ = (1000 x 10^-3 mOsm/L)(0.082 atm/molK)(298 K) = 24.2 atm Notice that the concentration of seawater is given for all solutes, which represents i x M. It is also given in mOsm/L which is converted into moles/L by multiplying 10^-3. Also, the question asks for the minimum pressure required, which means that the correct answer choice must be slightly above the calculated pressure in order for reverse osmosis to proceed.
Reverse osmosis is a process that allows fresh water to be obtained by using pressure to force an impure water source through a semipermeable membrane that only allows water molecules to pass. What is the minimum pressure that would be required to purify seawater at 25°C that has a total osmolarity of 1000 mOsm/L? A. 23.5 atm B. 24.5 atm C. 24000 atm D. 24500 atm
AgI (s) <---> Ag+ (aq) + I- (aq) I 1 x 10^-5 0 C -x +x +x E 1 x 10^-5 + x x Ksp = [Ag+] * [I-] 8.5 x 10^-17 = (1 x 10^-5 + x) * x since Ksp < 1 x 10^-4, assume x<<1 8.5 x 10^-17 = 1 x 10^-5 * x 1 x 10^-5 1 x 10^-5 8.5 x 10^-12 = x x = [I-] [I-] = 8.5 x 10^-12 M
The Ksp of AgI in aqueous solution is 8.5 x 10^-17. If a 1 x 10^-5 M of AgNO3 is saturated with AgI, what will be the final concentration of the iodide ion?
Ba(OH)2 <---> Ba(2+) + 2OH- I 0 0 C -x +x +2x E x 2x IP = [Ba(2+)] * [2OH-]^2 IP = [x] * [2x]^2 Ba(2+) = 0.05 M IP = 0.05 * [2x]^2 If Ba(2+) = x, 2OH- = 2x, and Ba(2+) = 0.05, then 2OH- = 2(0.05) IP = 0.05 * [2 * 0.05]^2 IP = 0.05 * [0.1]^2 IP = 0.05 * 0.01 IP = 5 x 10^-4 M Ksp = 5.0 x 10^-3 IP < Ksp, means dissolution.
The Ksp of Ba(OH)2 is 5.0 x 10^-3. Assuming that barium hydroxide is the only salt added to form a solution, calculate the ion product of a solution with a concentration of Ba(2+) of 0.05 M. Then predict the behavior of the given solution (dissolution, equilibrium, precipitation).
Ba(OH)2 <---> Ba(2+) + 2OH- I 0 0 C -x +x +2x E x 2x IP = [Ba(2+)] * [2OH-]^2 IP = [x] * [2x]^2 Ba(2+) = 0.1 M IP = 0.1 * [2x]^2 If Ba(2+) = x, 2OH- = 2x, and Ba(2+) = 0.1, then 2OH- = 2(0.1) IP = 0.1 * [2 * 0.1]^2 IP = 0.1 * [0.2]^2 IP = 0.1 * 0.04 IP = 0.004 M Ksp = 5.0 x 10^-3 IP < Ksp, means dissolution.
The Ksp of Ba(OH)2 is 5.0 x 10^-3. Assuming that barium hydroxide is the only salt added to form a solution, calculate the ion product of a solution with a concentration of Ba(2+) of 0.1 M. Then predict the behavior of the given solution (dissolution, equilibrium, precipitation).
Ba(OH)2 <---> Ba(2+) + 2OH- I 0 0 C -x +x +2x E x 2x IP = [Ba(2+)] * [2OH-]^2 IP = [x] * [2x]^2 Ba(2+) = 0.5 M IP = 0.5 * [2x]^2 If Ba(2+) = x, 2OH- = 2x, and Ba(2+) = 0.5, then 2OH- = 2(0.5) IP = 0.5 * [2 * 0.5]^2 IP = 0.5 * [1]^2 IP = 0.5 M Ksp = 5.0 x 10^-3 IP > Ksp, means precipitate.
The Ksp of Ba(OH)2 is 5.0 x 10^-3. Assuming that barium hydroxide is the only salt added to form a solution, calculate the ion product of a solution with a concentration of Ba(2+) of 0.5 M. Then predict the behavior of the given solution (dissolution, equilibrium, precipitation).
solubility precipitate saturation
The _____________ of a solute is a measure of the amount of solute that can be dissolved in a solvent at a certain temperature. A ____________ is an insoluble solid that separates from a solution; generally the result of mixing two or more solutions or of a temperature change. The _____________ of a solution is the maximum amount of solute that can be dissolved in a particular solvent at a given temperature.
van't Hoff factor
The ______________________ is the number of particles into which a compound dissociates in a solution.
Solubility Product Constant (Ksp) Ion Product (IP)
The _________________________________ is the product of concentration of ions in a saturated solution. The ________________ is the product of concentration of ions in a solution that has not reached saturation.
concentration chemical identity
The colligative properties of a solution are the physical properties of a solution that depend on the __________________ of dissolved particles, but not on the _______________ _____________ of the particles.
Answer: C. CaS will cause the most negative contribution to ΔS°(soln) through hydration effects because the Ca(2+) and S(2-) ions have the highest charge density compared to the other ions. All of the other ions have charges of +1 or -1, whereas Ca(2+) and S(2-) each have charges with a magnitude of 2.
The entropy change when a solution forms can be expressed by the term ΔS°(soln). When water molecules become ordered around an ion as it dissolves, the ordering would be expected to make a negative contribution to ΔS°(soln). An ion that has more charge density will have a greater hydration effect, or ordering of water molecules. Based on this information, which of the following compounds will have the most negative contribution to ΔS°(soln). A. KCl B. LiF C. CaS D. NaCl
Answer: D. The solubility of AgBr can be determined using the Ksp value given in the equation. Some amount of AgBr will dissolve; this is the molar solubility x for these conditions. When AgBr dissociates, there will be x amount of silver(I) formed and x amount of bromide - which is added to the 0.0010 M Br- already present from NaBr. AgBr <---> Ag+ + Br- Ksp = [Ag+][Br-] 5.35 x 10^-13 = (x)(0.0010 M + x) Given the Ksp of 5.4 x 10^-13, x will be negligible compared to 0.0010 M. Thus, the math can be simplified to: 5.35 x 10^-13 = (x)(0.0010) x = 5.35 x 10^-13 = 5.35 x 10^-10 0.0010 Therefore, x, the molar solubility, is 5.35 x 10^-10, which looks like (C.). However, the units of the answer choices are g/L, not mol/L, and the result must be multiplied by the molar mass (187.8 g/mol): 5.35 x 10^-10 mol/L x 187.8 g/mol = 1 x 10^-7 which is close to (D.). Note that a very accurate approximation was reached by rounding down the first number and rounding up the second, balancing the error.
The following equilibrium exists when AgBr (Ksp = 5.35 x 10^-13) is in solution: AgBr (s) <---> Ag+ (aq) + Br- (aq) What is the solubility of AgBr in a solution of 0.0010 M NaBr? A. 5.35 x 10^-13 g/L B. 1.04 x 10^-12 g/L C. 5.35 x 10^-10 g/L D. 1.04 x 10^-7 g/L
molar solubility = molarity molarity = 4 x 10^-10 M Fe(OH)3 (s) <-----> Fe(3+) (aq) + 3OH- (aq) I 0 0 C -x +x +3x E x 3x x = 4 x 10^-10 M Ksp = [Fe(3+)]^1 * [OH-]^3 Ksp = x * (3x)^3 Ksp = (4 x 10^-10 M) * (3 * (4 x 10^-10 M))^3 Ksp = (4 x 10^-10 M) * (1.7 x 10^-27 M) Ksp = 6.8 x 10^-37
The molar solubility of Fe(OH)3 in an aqueous solution was determined to be 4 x 10^-10 mol/L. What is the value of the Ksp for Fe(OH)3 at the same temperature and pressure?
Answer: D. The first step will most likely be endothermic because energy is required to break molecules apart. The second step is also endothermic because the intermolecular forces in the solvent must be overcome to allow incorporation of solute particles. The third step will most likely be exothermic because polar water molecules will interact with the dissolved ions, creating a stable solution and releasing energy.
The process of formation of a salt solution can be better understood by breaking the process into three steps: 1. Breaking the solute into its individual components 2. Making room for the solute in the solvent by overcoming intermolecular forces in the solvent 3. Allowing solute-solvent interactions to occur to form the solution Which of the following correctly lists the enthalpy of changes for these three steps, respectively? A. Endothermic, exothermic, endothermic B. Exothermic, endothermic, endothermic C. Exothermic, exothermic, endothermic D. Endothermic, endothermic, exothermic
Answer: A. Dissolution is governed by enthalpy and entropy, which are related by the equation ΔG°(soln) = ΔH°(soln) - TΔS°(soln). The cooling of the solution indicates that heat is used up in this bond breaking reaction. In other words, dissolution is endothermic, and ΔH is positive. The reaction is occurring spontaneously, so ΔG must be negative. The only way that a positive ΔH can result in a negative ΔG is if entropy, ΔS, is a large, positive value as in (A.). Conceptually, that means that the only way the solid can dissolve is if the increase in entropy is great enough to overcome the increase in enthalpy. (B.) is incorrect because it is clearly stated in the question stem that KCl dissolves; further, all salts of Group 1 metals are soluble. (C.) is incorrect because ΔS°(soln) must be positive in order for KCl to dissolve. Finally, (D.) is incorrect because solute dissolution would cause the boiling point to elevate, not depress. It is also not a piece of evidence that could be found by simply observing the beaker's temperature change.
The salt KCl is dissolved in a beaker. To an observer holding the beaker, the solution begins to feel colder as the KCl dissolves. From this observation, one could conclude that: A. ΔS°(soln) is large enough to overcome the unfavorable ΔH°(soln) B. KCl is mostly insoluble in water C. ΔS°(soln) must be negative when KCl dissolves D. boiling point depression will occur in this solution
ΔG = Gibbs Free Energy Change ΔH = Enthalpy Change ΔS = Entropy Change T = Temperature (Kelvin)
The spontaneity of dissolution is dependent on the change in Gibbs Free Energy, which is determined by: ΔG = ΔH - TΔS ΔG = ____________________ ΔH = _________________ ΔS = _______________ T = _____________
Answer: B. Benzene and toluene are both organic liquids and have very similar properties. They are both nonpolar and are almost exactly the same size. Raoult's law states that ideal solution behavior is observed when solute-solute, solvent-solvent, and solute-solvent interactions are all very similar. Therefore, benzene and toluene in the solution will be predicted to behave as a nearly identical solution.
Two organic liquids, pictured in the figure, are combined to form a solution. Based on their structures, will the solution closely obey Raoult's Law? A. Yes; the liquids differ due to the additional methyl group on toluene and, therefore, will not deviate from Raoult's Law. B. Yes, the liquids are very similar and, therefore, will not deviate from Raoult's Law. C. No, the liquids differ due to the additional methyl group on toluene and, therefore, will deviate from Raoult's Law. D. No, the liquids both contain benzene rings, which will interact with each other and cause deviation from Raoult's Law.
Endothermic Exothermic
Unlike most dissolutions, dissolutions of gases into liquids are not ________________, they are ________________.
When a solute is added to a solvent, the boiling point of the solution will be higher than the pure solvent. It is higher since the addition of a solute requires more energy to get the vapor pressure to equal the ambient pressure.
When a solute is added to a solvent, does the solution have a higher or lower boiling point. Explain why.
CuBr (s) <-----> Cu+ (aq) + Br- (aq) I 0 0 C -x +x +x E x x Ksp = [Cu+] * [Br-] 6.27 x 10^-9 = x^2 sqrt(6.27 x 10^-9) = sqrt(x^2) x = 7.91 x 10^-5 M x = [Cu+], [Br-], and [CuBr] Molar solubility of [Cu+], [Br-], and [CuBr] = 7.91 x 10^-5 M 3g CuBr/143.4 g/mol CuBr = 0.02 moles 0.02 moles/1 L = 0.02 M 0.02 M = Molar Solubility from IP 7.91 x 10^-5 M = Molar Solubility from Ksp 0.02 M > 7.91 x 10^-5 M IP > Ksp The solution is super-saturated
What are the concentrations of each of the ions in a saturated solution of CuBr, given that the Ksp of CuBr is 6.27 x 10^-9 at 25°C? If 3 g CuBr is dissolved in water to make 1 L of solution at 25°C, would the solution be saturated, unsaturated, or supersaturated?
All halides are water soluble except for fluorides and halides formed with Ag+ (silver ion), Pb(2+) (lead ion), or Hg2(2+) (mercury ion)
What do the rules of solubility say in regards to halides?
All metal oxides are insoluble except for those formed with alkali metals, NH4+ (ammonium), Ca(2+) (calcium ion), Sr(2+) (strontium ion), or Ba(2+) (barium ion).
What do the rules of solubility say in regards to metal oxides?
All salts containing NH4+ (ammonium) cations or alkali metal cations are water soluble.
What do the rules of solubility say in regards to salts containing ammonium (NH4+) and alkali metal?
All salts containing nitrate (NO3-) anions and acetate (CH3COO-) anions are water soluble.
What do the rules of solubility say in regards to salts containing nitrate (NO3-) and acetate (CH3COO-)?
All salts containing SO4(2-) are water soluble except for those formed with Ca(2+) (calcium ion), Sr(2+) (strontium ion), Ba(2+) (barium ion), Pb(2+) (lead ion).
What do the rules of solubility say in regards to sulfates?
The common ion will lead to an increase in our products, causing our equilibrium to shift in favor of the reactants, which will cause us to have more of our solute at equilibrium, and decrease the solubility of our solute.
What happens when a solute is placed in a solvent that shares a common ion?
The ionic attraction between Cl- and Ag+ will be greater than the ion-dipole interactions between the water molecules and the ions, which causes the Cl- and Ag+ ions to form an ionic bond with one another, resulting in the formation of AgCl precipitate.
What happens when we add an aqueous solution of AgNO3 to an aqueous solution of NaCl?
P = X * P° Density of water at 100°C is close to 1 g/mL, and vapor pressure of 1 atm at the same temperature. 180g C3H6O3/90 g/mol C3H6O3 = 2 mol C3H6O3 0.18 L H2O = 180 mL H2O * 1 g/mL = 180 g H2O 180 g H2O/18 g/mol H2O = 10 mol H2O 10 mol + 2 mol = 12 mol solution 10 mol H2O/12 mol solution = 0.83 P = (0.83) * (1 atm H2O) = 0.83 atm 1 atm - 0.83 atm = 0.17 atm
What is the change in vapor pressure when 180 g of glyceraldehyde (C3H6O3) are added to 0.18 L of water at 100°C?
1000 g/L
What is the density of water in g/L?
Zn(OH)2 (s) <---> Zn(2+) (aq) + 2OH- (aq) I 0 0.1 C -x +x +2x E x 0.1 + 2x Ksp = [Zn(2+)] * [2OH-]^2 4.1 x 10^-17 = [x] * [01. + 2x]^2 Since Ksp < 10^-4, we assume x<<1 4.1 x 10^-17 = x * [0.1]^2 4.1 x 10^-17 = x * 0.01 0.01 0.01 4.1 x 10^-15 = x [Zn(2+)] = x = [Zn(OH)2] Molar Solubility Zn(OH)2 = 4.1 x 10^-15 M
What is the molar solubility of Zn(OH)2 (Ksp = 4.1 x 10^-17) in a 0.1 M solution of NaOH?
NaCl (aq) + AgNO3 (aq) ---> AgCl (s) + NaNO3 (aq) Na+ + Cl- + Ag+ + NO3- ---> AgCl + Na+ + NO3- Ag+ (aq) + Cl- (aq) ---> AgCl (s)
What is the net ionic equation of the combination of an aqueous solution of NaCl and an aqueous solution of AgNO3?
Mass of solute = 20 g NaCl (s) Mass of solution = 100 g NaCl (aq) 20 g/100 g * 100% = 20% percent composition by mass = 20% NaCl solution
What is the percent composition by mass of a salt water solution if 100 g of the solution contains 20 g of NaCl?
8.314 J/molK or 0.082 atm/molK
What is the value of the ideal gas constant (R)?
P = X * P° 58 g butane /58 g/mol butane = 1 mol butane 172 g hexane/86 g/mol hexane = 2 mol hexane 1 mol butane/3 mol solution = 1/3 butane 2 mol hexane/3 mol solution = 2/3 hexane P(butane) = 1/3 * 172 kPa = 57.3 kPa P(hexane) = 2/3 * 17.6 kPa = 11.7 kPa 57.3 kPa + 11.7 kPa = 69 kPa
What is the vapor pressure at room temperature of a mixture containing 58 g butane (C4H10) and 172 g hexane (C6H14)? (Note: The vapor pressures of pure butane and pure hexane are 172 kPa and 17.6 kPa, respectively, at 25°C).
All CO3(2-) (carbonates) are insoluble except those formed with alkali metals and ammonium.
What to the rules of solubility say in regards to carbonates (CO3(2-))?
All hydroxides are insoluble except for those formed with alkali metals, NH4+ (ammonium), Ca(2+) (calcium ion), Sr(2+) (strontium ion), or Ba(2+) (barium ion)
What to the rules of solubility say in regards to hydroxides?
All PO4(3-) (phosphates) are insoluble except those formed with alkali metals and ammonium.
What to the rules of solubility say in regards to phosphates (PO4(3-))?
All S(2-) (sulfides) are insoluble except those formed with alkali metals and ammonium.
What to the rules of solubility say in regards to sulfides (S(2-))?
All SO3(2-) (sulfites) are insoluble except those formed with alkali metals and ammonium.
What to the rules of solubility say in regards to sulfites (SO3(2-))?
When a solute is added to a solvent, the freezing point of the solution will be lower than the pure solvent. It is lower since the solute particles interferes with the formation of lattice arrangement of solvent molecules in their solid state, meaning energy must be removed from the solution to solidify the solution.
When a solute is added to a solvent, does the solution have a higher or lower freezing point. Explain why.
At equilibrium
When a solution is saturated, the reaction between the solute and solvent is ________________________.
Answer: B. Formation of complex ions between silver ions and ammonia will cause more molecules of solid AgCl to dissociate. The equilibrium is driven toward dissociation because the Ag+ ions are exxentially being removed from solution when they complex with ammonia. This rationale is based upon Le Chatelier's principle, stating that when a chemical equilibrium experiences a change in concentration, the system will shift to counteract that change.
When ammonia, NH3, is used as a solvent, it can form complex ions. For example, dissolving AgCl in NH3 will result in the complex ion [Ag(NH3)](2+). What effect would the formation of complex ions have on solubility of a compound like AgCl in NH3? A. The solubility of AgCl will increase because complex ion formation will cause more ions to exist in solutions, which interact with AgCl to cause it to dissociate. B. The solubility of AgCl will increase because complex ion formation will consume Ag+ ions and cause the equilibrium to shift away fro, solid AgCl. C. The solubility of AgCl will decrease because Ag+ ions are in complexes, and the Ag+ ions that are not complexed will associate with Cl- to form solid AgCl. D. The solubility of AgCl will decrease because complex ion formation will consume Ag+ ions and cause the equilibrium to shift toward the solid AgCl.
Answer: A. Mixtures have a higher vapor pressure than predicted by Raoult's Law have stronger solvent-solvent and solute-solute interactions than solute-solvent interactions. Therefore, particles do not want to stay in solution and more readily evaporate, creating a higher vapor pressure than an ideal solution. Two liquids that have different properties, like hexane (hydrophobic) and ethanol (hydrophilic, small) in (A.), would not have many interactions with each other and would cause positive deviation; i.e. higher vapor pressure. (B.) and (C.) are composed of liquids that are similar to one another and would not show significant deviation from Raoult's Law. (D.) contains two liquids that would interact very well with each other, which would actually cause a negative deviation from Raoult's Law - when attracted to one another, solutes and solvents prefer to stay in liquid form and have a lower vapor pressure than predicted by Raoult's Law.
Which of the following combinations of liquids would be expected to have a vapor pressure higher than the vapor pressure that would be predicted by Raoult's Law? A. Ethanol and hexane B. Acetone and water C. Isopropanol and methanol D. Nitric acid and water
Answer: C. Melting point depresses upon solute addition, making (A.) and (B.) incorrect. Solute particles interfere with lattice formation, the highly organized state in which solid molecules align themselves. Colder-than-normal conditions are necessary to create the solid structure.
Which of the following explanations best describes the mechanism by which solute particles affect the melting point of ice? A. Melting point is elevated because the kinetic energy of the substance increases. B. Melting point is elevated because the kinetic energy of the substance decreases. C. Melting point is depressed because solute particles interfere with lattice formation. D. Melting point is depressed because solute particles enhance lattice formation.
Answer: B. The equation to determine the change in boiling point of a solution is as follows: ΔT = iKbm. m is the molarity of the solution, and Kb is the boiling point elevation constant. In this case, the solvent is always water, so Kb will be the same for each solution. What is needed is the number of dissociated particles from each of the original species. This is referred to as the van't Hoff factor (i) and is multiplied by molality to give normality (the concentration of the species of interest - in this case, all particles). The normality values determine which species causes the greatest change in boiling point. Species #moles #particles dissolved i x m (normality) CaSO4 0.4 2 0.8 Fe(NO3)3 0.5 4 2.0 CH3COOH 1.0 Either 1 or 2 Either 1.0 or 2.0 C12H22O11 1.0 1 1.0 The choice is between iron(III) nitrate and acetic acid. The fact that acetic acid is a weak acid indicates that only a few particles will dissociate into H+ and acetate. Therefore, the normality of the acetic acid will be much closer to 1.0 than 2.0.
Which of the following will cause the greatest increase in the boiling point of water when it is dissolved in 1.00 kg H2O? A. 0.4 mol calcium sulfate B. 0.5 mol iron(III) nitrate C. 1.0 mol acetic acid D. 1.0 mol sucrose
Answer: D. All three choices can make a solution as long as the two components create a mixture that is homogeneous (uniform in appearance). Hydrogen in platinum is an example of a gas in a solid. Brass and steel are examples of homogeneous mixtures of solids. The air we breathe is an example of a homogeneous mixture of gases; while these are more commonly simply referred to as mixtures, they still fit the criteria of a solution.
Which phases of solvent and solute can form a solution? I. Solid solvent, gaseous solute II. Solid solvent, solid solute III. Gaseous solvent, gaseous solute A. I and II only B. I and III only C. II and III only D. I, II, and III
M1V1=M2V2 100 ppm * V1 = 25 ppm * 100 mL V1 = 25 ppm * 100 mL 100 ppm V1 = 25 mL To get the final solution, start with 25 mL of the 100 ppm stock solution, and then add 75 mL of pure water to get 100 mL solution with 25 ppm of chlorine.
You are working in a sewage treatment facility and are assaying chlorine in a water sample. You need to dilute the water sample from 100 ppm stock to 25 ppm and create 100 mL of solution. Calculate the amount of stock solution needed and determine how you would create the final solution.
Solubility Saturation
_______________ is the amount of solute contained in a solvent. _______________ refers to maximum solubility of a compound at a given temperature.
Osmotic pressure Molarity
______________________________ is the pressure needed to be applied to a solution to prevent the transfer of water through a semipermeable membrane from an area of less solute concentration to an area of greater solute concentration. It is dependent on the ________________ of a solution.
