chemistry 1120 ch. 8, 9, 10

(v/v)% concentration problem: what is the concentration of a solution made by dissolving 37 mL of isopropyl alcohol in enough water to give a final volume of 53 mL?

(37 mL / 53 mL) x 100 = 69.8 % (v/v)

solute

-the substance that is present in lesser amount; the substance that is dissolved by the solvent -cannot be separated by filtration -are not visible but can give a color to the solution

dilutions

-volume increases -concentration decreases -mass of solute stays the same M1 V1 = M2 V2

avogadro's law - volume & molar amount

-volume is directly related to molar amount -P & T are constant V1 / n1 = V2 / n2

using molar volume - problem: 1) what is the volume occupied by 2.5 moles of N2 gas at STP? 2) how many gas of He are present in 8.0 L of gas at STP?

1) 2.5 mol --> ? L N2 2.5 mol N2 x 22.4 L N2 / 1 mol N2 = 56 L N2 gas 2) 8.0 L He --> ? g He 8.0 L He --> mol He --> g He (mmHe = 4.00 g/mol) 8.0 L He x 1 mol / 22.4 L He x 4.00 g He / 1 mol He = 1.4 g He

solubility problem: at 20 degrees celsius, the solubility of KCl is 34g / 100g of water. lets say you mic 75g of KCl with 200g of water at temperature of 20 degrees celsius. 1) how much of the KCl can dissolve? 2) is the solution saturated or unsaturated? 3) what is the mass (if any) of solid KCl on the bottom of the container?

1) 200g x 34 g KCl / 100g = 68g 2) saturated, only 68g is dissolved of the 75g 3) 75g - 68g = 7g KCl

identifying conjugate acids & bases: 1) H3PO4 + NH3 --> H2PO4- + NH4+ 2) HF + H2O -->H30+ + F-

1) H3PO4 = acid / its conj base = H2PO4- NH3 = base / its conj acid = NH4+ 2) HF = acid / conj base = F- H20 =base / conj acid = H30+

(m/v)% concentration - problem: a solution of heparin contains 14 g heparin dissolved water at a final volume of 12 mL. what is the concentration % (m/v) of the solution?

%(m/v)= (14 g / 12 mL) x 100 = 116.6 % (m/v)

ideal gas law problem: an aerosol spray can of deodorant with a volume of 350mL contains 3.2g of propane (mw= 44.10 g/mol). what is the pressure in atm, in the can at 20 degrees celsius?

350 mL x 1 L / 1000 mL = 0.350 L 3.2 g x 1 mol / 44.10g = 0.0726 mol 20 + 273.15 = 293.15 P= (0.0726 mol) (0.082 L*atm/mol*K) (293.15K) / 0.350 L = 4.99 atm

Dalton's law of Partial Pressures

P total = P (gas 1) + P (gas 2) + P (gas 3).....

combined gas law

P1 V1 / T1 = P2 V2 / T2

gay-lussac's law - problem: driving on a hot day causes tire temperature to rise. what is the pressure inside an automobile tire at 45 degrees celsius if the tire has a pressure of 30 psi at 15 degrees celsius ?

P1= 30 --> P2= ? T1= 30 --> T2= 45 T1 = 30 + 273.15= 288.15 T2= 45 + 273.15 = 318.15 30 psi/288.15 = (P2 / 318.15) x 318.15 = (0.104) x 318.15 =33.08 --> 33 psi

ideal gas law

PV = nRT (R is the gas constant) R= 0.0821 L*atm/mol*K R= 62.4 L*mmHg/mol*K the numerical value of R depends on the units used to measure the gas pressure

which acid is weaker? acetic acid, Ka= 1.8 x 10^-5 carbonic acid, Ka= 4.3 x 10^-7

carbonic acid because of the smaller value in Ka

strong electrolyte

completely ionized in water NaCl (s) --> Na+ (aq) + Cl- (aq)

solutions

homogenous mixtures where particles are so small they cannot be detected by eye

colloids

homogenous mixtures with slightly larger particles that can be seen as a cloudy appearance

concentration (v/v)%

for a solution of liquids dissolved in liquids (volume of solute, mL / volume of solution, mL) X 100%

conjugate acid

formed by the addition of H+ to a base

conjugate base

formed by the loss of H+ from an acid

boyle's law - pressure & volume

if volume decreases, the pressure increases P1 V1 = P2 V2

nonelectrolyte

not ionized in water (dissolves in water as molecule, doesn't produce ions in water) Glucose (s) <---> Glucose (aq)

"likes dissolve likes"

polar molecules dissolve in polar solvents, non-polar molecules dissolve in non- polar solvents

properties of solutions

vapor pressure boiling point freezing point osmosis & osmotic pressure

effect of temperature on solubility

solid solutes: increase with increasing temperature gas solutes: decrease with increasing temperature

effect of pressure on solubility - Henry's law

solubility of a gas is directly proportional to its partial pressure C1 / P1 = C2 / P2

charles's law - volume & temperature

temperature of a gas increases, its volume increases V1/T1 = V2/T2

pressure conversion problem: A pressure of 640 torr is equivalent to how many atmospheres? (1 atm = 760)

torr --> ? atm 640 torr x 1 atm / 760 torr = 640/760 = 0.84 atm

[H3O+] & [OH-] concentrations

1.00 x 10^-7 M

rearranging charles's law

-temp to kelvin (k) -solve for T2 (consider cross multiplying) V1T2/V1 = V2T1/V1 ----> T2 = V2T1/V1

henry's law - problem: at 20 degrees celsius and a partial pressure of 760 mmHg the solubility of CO2 in water is 0.169 g/100 mL at this temperature. what is the solubility of CO2 at 2.5 x 10^4 mmHg?

(0.169 g / 100 mL) / (760 mmHg) = x / 2.5 x 10^4 mmHg x = 5.56 g/100 mL

boyle's law problem: the volume of a balloon is 2.6L at 1.00atm. if the volume expands to 5.3 L, will the new pressure be higher or lower?

(1.00)(2.6) = (P2)(5.3) V=2.6L --> 5.3L (increase) P2= 0.49 atm P=1.00 atm --> ? (decrease)

dilution problem: what is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution?

(1.80) (15.0) / (0.300) (V2) = 27 / 0.300 V2 = 90 mL

isotonic solution

(equal) maintains similar osmotic pressure; water flows freely across semipermeable membrane

hypertonic solution

(higher than) higher osmotic pressure ; higher concentration ; water flow out of RBC >results in crenation (shrink)

hypotonic solution

(lower than) lower osmotic pressure; lower concentration than RBC (RBC has higher concentration); water flow into the RBC >results in hemolysis (burst)

molar volume

-At STP, 1 mole of any gas (6.02 x 10^23 particles) occupies a volume of 22.4 L - (1 mol gas = 22.4 L)

concentration (m/v)%

-for solid solute in a liquid solvent (mass solute, g / mL of solution) x 100

solubility

-grams of solute in 100 mL (g/mL) of solvent grams solute / mL solvent OR grams solute / 100 g solvent

Gay-Lussac's law - pressure & temperature

-pressure from a gas increases, temperature in Kelvins increases -V & n are constant P1/T1 = P2/T2

concentration & volume dilution - problem: what is the final volume (V2) in milliliters when 0.943 L of an 80.7% (m/v) solution is diluted to 36.4% (m/v)?

0.943 L x 1000 mL / 1 L = 943 mL (80.7) (943) = (36.4) (V2) V2 = 2091 mL

pressure units

1 atm = 760 mmHg = 14.7 psi = 101,325 Pa 1 mmHg = 1 torr = 133.32 Pa

calculating concentrations of [H3O+] & [OH-] using Kw - problem: 1) what is the [H30+] of a solution is [OH-] is 1.0 x 10^-8 M? 2) if an acid is added to water, bringing the [H3O+] to 7.11 x 10^-2 M, what is [OH-]?

1) [H3O+] [1 x 10^-8] = 1 x 10^-14 M = 1 x 10^-6 M 2) [7.11 x 10^-2] [OH-] = 1x 10^-14 M [OH-] = 1.41 x 10^-13 M

properties of gases

1) volume (V) - space occupied 2) temperature (T) - avg. KE; MUST be in Kelvin 3) amount (n) - quantity of gas present 4) pressure (p) - force exerted by gas particles

(m/v)% concentration - conversion factor problem: how many grams of caffeine are required to prepare 355 mL of a 1.9 %(m/v) solution in water?

1.9 g caffeine / 100 mL solution 355 mL solution x 1.9 g caffeine / 100 mL solution =6.475 g caffeine

concentration molarity - conversion factor problem: the chlorine in swimming pools is actually HOCl (mm= 52.446 g/mol). if the concentration of HOCl in a pool is 0.0000095 M, how many grams of HOCl are in 3.5 L?

3.5 L --> ? g HOCl 0.0000095 mol HOCl / 1 L solution 3.5 L x 0.0000095 mol / 1 L x 59.4 g / 1 mol = 0.0017 g HOCl

concentration molarity - problem: how many milliliters of a 6.0 M HCl solution will provide 4.5 moles of HCl?

6.0 M HCl = 4.5 mol HCl / X L 6.0 X = 4.5 X= 0.75 L x 1000 mL / 1 L = 750 mL

the pKa of acetic acid is 4.74 if a buffer contains 0.0072 M acetic acid and 0.010 M, what is its pH?

=4.74 + log [0.010 / 0.0072] =4.74 + 0.1426675 pH = 4.88

6 strong acids******

> perchloric acid (HClO4) > sulfuric acid (H2SO4) > hydriodic acid (HI) > hydrobromic acid (HBr) > hydrochloric acid (HCl) > nitric acid (HNO3)

bronsted-lowry (B-L) base

>can accept H+ from another substance (aka proton acceptor)

bronsted-lowry (B-L) acid

>can donate H+ to another substance (aka proton donor)

conjugate acid-base pairs

>differ in a formula by a single H+ & their electrical charges are 1 unit different >one pair occurs in the forward direction and one pair occurs in the reverse direction

osmosis

>movement of water into or out of cells >flow of low solute concentration to high solute concentration through a semipermeable membrane

osmolarity

>particle concentration > osmol / L >the sum of the molarities of all dissolved particle in a solution >equal to the number of moles of dissolved particles per L of solution (ex: its the molarity & the # of particles that would arise from solute dissociation 0.2 M glucose --> 0.2 osmol/L 0.2 M NaCl --> 0.2 x 2 = 0.4 osmol/L

acid

>produces hydrogen ions, H+ (neutralize bases) HCl --> H+ + Cl-

base

>produces hydroxide ions, OH- (neutralize acids) NaOH --> Na+ + OH-

concentration & volume dilutions

C1 V1 = C2 V2

acid dissociation constant (Ka) HA + H2O <--> H3O+ + A-

Ka = [H3O+] [A-] / [HA] (the stronger acid, the bigger the Ka (LIQUIDS WILL NEVER GO IN THE EQUATION)

Kw

Kw = [H3O+] [OH-] = 1.00 x 10^-14 M

calculating [H30+] from pH

[H3O+] = 10^-pH

combined gas law problem: a gas has a volume of 675 mL at 35 degrees celsius and 0.850 atm pressure. what is the volume of the gas at -95 degrees celsius and a pressure of 802 mmHg? (1 atm = 760 mmHg)

T1= 35 + 273.15 = 308.15 T2= -95 + 273.15 = 178.15 P2= 802 mmHg x 1 atm / 760 mmHg = 1.055 atm (0.850) (675) / (308.15) = (1.055) (V2) / (178.15) (0.850) (675) (178.15) = (1.055) (308.15) (V2) V2= 314.4 mL

charles's law problem: a sample of oxygen gas has a volume of 420 mL at a temperature of 18 degrees celsius. At what temperature, in celsius, will the volume of the oxygen be 640 mL (P & n are constant)?

V1= 420 mL V2= 640 mL T1= 18 --> 18+273.15 = 291.15 K T2= (640)(291.15) / (420) = 443.66 K - 273.15 K = 171 degrees celsius

concentration set up

amount of solute / amount of solution

heterogenous mixture

concrete, gasoline

boiling point elevation in solutions

delta T (boiling)= (0.51 degrees celsius kg H2O / mol particles) (mol particles / kg water)

freezing point depression

delta T (freezing) = (-1.86 degrees celsius kg H2O / mol particles) (mol particles / kg water)

boiling point elevation - problem: what is the boiling point of a solution made using 600 g of sucrose in 0.300 kg of water? (mw sucrose = 342 g/mol)

mol sucrose = 600 g x 1 mol / 342 g = 1.75 mol sucrose / 0.300 kg H2O =(5.85 (mol sucrose/kg H2O) x 0.51 =2.975 degrees celsius 100 + 2.975 = 102.98 degrees

concentration (molarity)

mol/L moles of solute / liters of solution ***ALWAYS mol/L NOT mL*** (ex: 6 M solution means there's 6 moles of solute per 1 L solution)

acidic pH = pH < 7 neutral pH = pH = 7 basic pH = pH > 7

pH < 7 pH = 7 pH > 7

calculating pH of a buffer solution

pH = pKa + log [conj base] / [weak acid]

calculating pH - problem: what is the pH of stomach acid which has [H3O+] = 0.0316 M?

pH= -log [0.0316] =1.5

pH

pH= -log [H30+]

weak electrolyte

partially ionized in water (forms a solution of a few ions & un-dissociated molecules) CH3CO2H (l) <---> CH3CO2- (aq) + H+ (aq)

STP

standard temperature (0 degrees celsius or 273.15 K) & pressure (1 atm or 760 mmHg)

solvent

the substance that is typically present in larger amount; the substance in which the solute is dissolved

Dalton's Law - problem air 78% N2, 21% O2, & 1% Ar. what are the partial pressures of each gas at 760 torr total pressure?

tip: the partial pressure of each gas is the percent (decimal equivalent) times total pressure. Pn2 = 0.78 x 760 = 592.8 torr Po2= 0.21 x 760 = 159.6 torr Par=0.01 x 760 = 7.6 torr 592.8 + 159.6 + 7.6 = 760 torr