Chem 1110 Chap. 4 wileyplus

Which statement below is not correct in describing a solution? a) A solution is a homogeneous mixture of (at least) two substances. b) The concentration of a solution is the sum of the moles of solute and solvent. c) The solute in a solution is the substance that dissolves in the solvent. d) The solvent in a solution is usually the component present in largest amount.

B

What is the molarity of a solution prepared by adding 16.0 g of sodium nitrate (NaNO3, molecular mass = 85.0 g/mol) to enough water to make 250.0 mL of solution? a) 0.753 M b) 0.188 M c) 5.44 M d) 0.0471 M

a) 0.753 M Why: Molarity = moles / volume To determine the number of moles of a particular solute requires the mass of the solute as well as its molecular mass. You realized that you needed these data and you used the data correctly. The answer: (16.0 g NaNO3) (1 mol / 85.0 g) = 0.188 mol NaNO3 and (0.188 mol NaNO3) / (0.250 L) = 0.753 M NaNO3

What is the molarity of a solution formed by mixing 120.0 mL of 0.820 M HCl with 80.0 mL of 1.40 M HCl? a) 1.05 M b) 1.11 M c) 2.22 M d) 1.17 M

a) 1.05 M Why: Molarity = moles / volume For a single solute the number of moles can be calculated from the mass and the molecular mass. When mixing solutions of known molarities you need to determine the total number of moles and the total volume before using the equation above. You did exactly that: (0.120 L) (0.820 mol/L) = 0.0984 mol HCl (0.0800 L) (1.40 mol/L) = 0.112 mol HCl Total moles = 0.210; Total volume = 0.200 L Molarity = 0.210 mol HCl / 0.200 L = 1.05 M HCl

If 125 mL of a 6.00 M KNO3 solution is diluted with 100.0 mL of water, the molarity of the resulting solution would be a) 3.33 M b) 0.0267 M c) 7.50 M d) 2.67 M

a) 3.33M Why: For a single solute the number of moles can be calculated from the mass and the molecular mass. When mixing a solution of known molarity with water, you need to determine the total number of moles. The number of moles is found by using the equation above (as all the solute comes from one source). Next, the total volume is needed. Lastly, the above equation is used once more to determine the final molarity. moles solute: (0.125 L) X (6.00 mol/L) = 0.750 moles Molarity: (0.750 mol) / (0.225 L) = 3.33 M Another way to solve this problem is to use the equation: Md x Vd = Mc x Vc Thus, Md x 225 mL = 6.00 mol/L x 125 mL; Solving gives Md = 3.33 mol/L.

Which equation correctly represents a dissociation reaction for an electrolyte? a) Na2SO4(s) → 2Na+(aq) + SO4 2-(aq) b) CaBr2(s) → Ca2+(aq) + Br2(l) c) KHCO3(s) → KH2+(aq) + CO3 2-(aq) d) MgSO3(s) → MgO(s) + SO2(g)

a) Na2SO4(s) → 2Na+(aq) + SO4 2-(aq)

For which acid or base reaction below are the products written wrong? a) NaHCO3(aq) + HCl(aq) → NaOCl(aq) + H2(g) + CO2(g) b) Fe2O3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2O(l) c) NaOH(aq) + NH4Cl(aq) → NaCl(aq) + NH3(g) + H2O(l) d) Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l)

a) NaHCO3(aq) + HCl(aq) → NaOCl(aq) + H2(g) + CO2(g) Why: Acids will react with carbonates and bicarbonates to produce carbon dioxide.

Which statement below is correct? a) Of the four aqueous hydrogen halides, only HF is a weak acid. b) All strong oxoacids have four oxygens. c) All four chlorine oxoacids are strong acids. d) All four aqueous hydrogen halides are strong acids.

a) Of the four aqueous hydrogen halides, only HF is a weak acid. why: There are six common strong acids. Among these are three aqueous hydrogen halides. These are HCl, HBr, and HI. Missing is HF as this acid is a weak acid.

Two solutions are mixed. The first contains 7.00 g of calcium nitrate (formula mass = 164 g/mol) and the second contains 4.00 g of potassium fluoride (formula mass = 58.0 g/mol). The reaction is Ca(NO3)2 + 2KF → 2KNO3 + CaF2 The reaction proceeds with CaF2 precipitating from the solution. The mass of CaF2 formed is a) 11.0 g b) 2.69 g c) 3.33 g d) 5.38 g

b) 2.69 g Why: You realized that this is a limiting reagent problem. The equation tells us that 2 moles of potassium fluoride are required for each mole of calcium nitrate. Since we have the mass and molar masses of both of these reagents, we can determine the one that is used up. From the moles of this reagent we can find the mass of calcium fluoride that forms. Here is how I solved this problem: (7.00 g Ca(NO3)2) (1 mol Ca(NO3)2/164 g Ca(NO3)2) = 0.0427 mol Ca(NO3)2. This will require twice as many moles of KF or 0.0854 mol KF(4.00 g KF) (1 mol KF/58.0 g KF) = 0.0690 mol KF. This is less than the 0.0854 mol KF needed for the Ca(NO3)2 above. Thus, KF is the limiting reagent. (0.0690 mol KF) (1 mol CaF2/2 mol KF) (78.0 g CaF2/1 mol CaF2) = 2.69 g CaF2 product forms.

The volume (in mL) of 0.241 M HCl solution required to exactly neutralize 20.0 mL of a 0.325 M NaOH solution can be calculated to be a) 6.50 mL b) 27.0 mL c) 14.8 mL d) 37.1 mL

b) 27.0 mL Why: Once a chemical equation is balanced, there are three steps to follow when you are given the mass of one substance and asked for the mass of a second substance in the reaction. The three steps are (1) convert grams of what you have to moles (of what you have). For this problem, this step does not involve mass; rather it involves a volume and a molarity. You were able get moles from these two pieces of information: (20.0 mL) (1L / 1000 mL) (0.325 mol NaOH / L) = 0.00650 mol NaOH (2) convert moles of what you have to moles of what you want. (0.00650 mol NaOH) (1 mol HCl / 1 mol NaOH) = 0.00650 mol HCl (3) convert moles of what you want to grams of what you want. For this problem, this step does not involve mass; rather it asks for a volume of a given molarity of HCl. You were able get to do this: (0.00650 mol HCl) (1 L / 0.241 mol HCl) (1000 mL / 1 L) = 27.0 mL There is another way to solve this problem. In acid-base neutralization reactions, the moles of H+ = moles OH- . You can write this as Ma x Va x na(H+) = Mb x Vb x nb(OH-) (a = acid, b = base) For this problem: (0.241M) (Va ) (1) = (0.325 M) (20.0 mL) (1) Solving for Va gives an answer of 27.0 with the units the same as those of NaOH.

Which match below is not correct? a) H3PO4 - a triprotic acid b) CaO - an acidic anhydride c) HCHO2 - a monoprotic acid d) CH3NH2 - a base

b) CaO - an acidic anhydride why: CaO is a basic anhydride. These compounds react with water to form the hydroxide ion in solution. CaO (s) + H2O Ca2+(aq) + 2OH-(aq

A mixture of NaCl and NaClO3 weighed 7.50 g. The sample was dissolved in water and AgNO3 was added until the precipitation of AgCl was complete. The AgCl was filtered, dried and weighed. A total of 8.83 g AgCl was obtained. From these data, the percentage of NaCl in the mixture a) can be calculated to be 61.6%. b) cannot be calculated because you cannot have a larger mass of product than reactant. c) can be calculated to be 48.0%. d) can be calculated to be 29.1%. e) can be calculated to be 84.9%.

c) can be calculated to be 48.0%. Why: The steps needed to solve this problem are obtained from the roadmap g AgCl → mol AgCl → mol NaCl → g NaCl. Once g NaCl is determined, the percentage in the mixture can be found. The solution is: (1) (8.83 g AgCl) (1 mol AgCl / 143.3 g AgCl) = 0.0616 mol AgCl (2) (0.0616 mol AgCl) (1 mol NaCl / 1 mol AgCl) = 0.0616 mol NaCl (3) (0.0616 mol NaCl) (58.45 g NaCl / 1 mol NaCl) = 3.60 g NaCl (4) (3.60 g NaCl / 7.50 g sample) x 100 = 48.0 %

Ca2+ and F- form a precipitate. The reaction is Ca2+(aq) + 2F-(aq) → CaF2(s). Suppose that 40.0 mL of 0.250 M Ca(NO3)2 is added to 32.0 mL of 0.519 M NaF. Which grid below accurately tells what happens in this reaction? a) Limiting Reagent: Ca(NO3)2 Moles of CaF2 formed: 0.0100 M of product after reaction: 0.0917M F- b) Limiting Reagent: NaF Moles of CaF2 formed: 0.0083 M of product after reaction: 0.0236M Ca2+ c) Limiting Reagent: Ca(NO3)2 Moles of CaF2 formed: 0.0200 M of product after reaction: 0.0917M F- d) Limiting Reagent: NaF Moles of CaF2 formed: 0.0166 M of product after reaction: 0.0917M Ca2+ e) Limiting Reagent: NaF Moles of CaF2 formed: 0.0332 M of product after reaction: 0M Ca2+

b) Limiting Reagent: NaF Moles of CaF2 formed: 0.0083 M of product after reaction: 0.0236M Ca2+ Why: First you must determine the limiting reagent: (1) The reaction is Ca2+(aq) + 2F-(aq) → CaF2(s) (2) moles Ca2+ = (0.0400 L) x (0.250 mol L-1) = 0.0100 mol moles F- = (0.0320 L) x (0.520 mol L-1) = 0.0166 mol (3) (0.0100 mol Ca2+) x (2 mol F-/ 1 mol Ca2+) = 0.0200 mol F- needed There is only 0.0166 mol F-. This is not enough for all the Ca2+ and, thus, the NaF solution (or F-) is the limiting reagent. Now you need to determine the amount of CaF2 that forms (4) (0.0166 mol F-) x (1 mol CaF2/ 2 mol F-) = 0.00830 mol CaF2 Lastly, you calculate the concentration of the excess Ca2+ (5) Ca2+ left over = 0.0100 - 0.0083 = 0.0017 and 0.0017 mol Ca2+/0.072 L solution = 0.024 M Ca2+ (2 significant figures)

Which salt is unlikely to form a gas in a metathesis reaction with HCl? a) Na2SO3 b) Na2HPO4 c) Li2S d) KCN

b) Na2HPO4 why: Nitrates, sulfates and phosphates do not react with acids unless (except for nitrates) you are dealing with acid salts. However, even here, no gaseous products form. The salt Na2HPO4 will react with HCl. The reaction is Na2HPO4(aq) + HCl(aq) → NaH2PO4(aq) + NaCl(aq). If more HCl is added, a second reaction occurs to produce H3PO4. Unlike H2SO3 or H2CO3, H3PO4 does not decompose into water and a gas

In an experiment 20.6 g of potassium iodide (KI, molar mass = 166 g/mol) was added to 212 mL of water. The volume of the resulting solution was 237 mL. Which of the following is not correct? a) molarity of solution = 0.524 M b) density of solution = 0.907 g/mL c) moles of KI = 0.124 d) all the above are correct e) none of the above are correct

b) density of solution = 0.907 g/mL Why: The density of a solution is defined as the ratio of the mass of the solution to the volume of the solution. The value above is the ratio of the mass of the solute to the volume of the solution. The density of this solution is about 0.981 g/mL

How many moles of sodium hydroxide (NaOH, molecular mass = 40.0 g/mol) are there in 300.0 mL of a 0.400 M NaOH solution? a) 120 mol b) 4.80 mol c) 0.120 mol d) 1.33 mol

c) 0.120 mol Why: Molarity = moles / volume (300. mL) (1 L/1000 mL) (0.400 mol NaOH/L) = 0.120 mol NaOH

The concentration of ammonium ion in the solution of the salt called ferrous ammonium sulfate, formula [Fe(NH4)]2(SO4)3, is 0.360 M. The concentration of sulfate ions in this solution is a) 0.240 M b)0.360 M c) 0.540 M d) 1.08 M

c) 0.540 M Why: This is a compound you may not have seen before, but you treat it the same way as simple ionic compounds. However, you can treat this the same for CaCl2 or Pb(NO3)2. Ferrous ammonium sulfate yields 7 ions: 2Fe2+, 2NH4+, and 3SO42-. Thus (0.360 mol NH4+ L-1) x (3 mol SO42-/2 mol NH4+) = 0.540 mol SO42- L-1.

A sample of 30.0 mL of an H2SO4 solution was titrated with a solution of 0.720 M KOH. It was found that 46.90 mL of the base was required to exactly neutralize all of the acid. From these results the concentration of the H2SO4 solution can be calculated to be a) 2.25 M b) 0.230 M c) 0.563 M d) 1.13 M

c) 0.563 M Why: Once a chemical equation is balanced, there are three steps to follow when you are given the mass of one substance and asked for the mass of a second substance in the reaction. The three steps are (1) convert grams of what you have to moles (of what you have). For this problem, this step does not involve mass; rather it involves a volume and a molarity. You were able get moles from these two pieces of information: (46.90 mL) (1L / 1000mL) (0.720 mol KOH / L) = 0.0338 mol KOH (2) convert moles of what you have to moles of what you want. (0.0338 mol KOH) (1 mol H2SO4/ 2 mol KOH) = 0.0169 mol H2SO4 (3) convert moles of what you want to grams of what you want. For this problem, this step does not involve mass; rather it asks for the molarity of H2SO4. You were able get to do this: (0.0169 mol H2SO4 ) / 0.0300 L = 0.563 M There is another way to solve this problem. In acid-base neutralization reactions, the moles of H+ = moles OH-. You can write this as Ma x Va x na(H+) = Mb x Vb x nb(OH-) (a = acid, b = base) For this problem: (Ma) ( 30.0 mL) (2) = (0.720 M) (46.90 mL) (1) Solving for Ma gives an answer of 0.563 M H2SO4.

The mass of acetic acid (CH3COOH, molecular mass = 60.06 g/mol) needed to prepare 125 mL of a 0.450 M solution is a) 0.937 g b) 56.3 g c) 3.38 g d) 1.07 g

c) 3.38 g Why: Molarity = moles / volume To determine the number of grams of a particular solute requires the moles of the solute as well as its molecular mass. The number of moles can be determined using the above equation. I did it as follows (aa = acetic acid): (0.125 L) (0.450 mol aa/L) (60.06 g aa/mol aa) = 3.38 g acetic acid

What volume of a 4.50 M NH3 solution do you need to dilute to prepare 250.0 mL of 0.750 M NH3? a) 844 mL b) 1.50 mL c) 41.7 mL d) 0.188 mL

c) 41.7 mL Why: (0.250 L) x (0.750 mol/L) = 0.188 mol NH3 In your next step you divided this value by the molarity of the ammonia to arrive at the final answer. (0.188 mol NH3) x (1 L solution/4.50 mol NH3) = 0.0417 L or 41.7 mL Another way to solve this problem is to use the equation: Md x Vd = Mc x Vc Thus, 0.750 x 250.0 mL = 4.50 x Vc ; Solving gives Vc = 41.7 mL.

Pyridine, C5H5N, is an important organic base. The equation for its ionization is a) C5H5N(aq) + H2O --> C5H4N+(aq) + H3O+(aq) b) C5H5N(aq) + H3O+(aq) --> C5H5NH+(aq) + H2O c) C5H5N(aq) + H2O --> C5H5NH+(aq) + OH-(aq) d) C5H5N(aq) + H2O --> C5H4NH2+(aq) + O-(aq)

c) C5H5N(aq) + H2O --> C5H5NH+(aq) + OH-(aq) why: According to the Arrhenius model, a base is a substance the produces OH- in aqueous solution. The water loses an H+ to the base. The general equation is: base + H2O baseH+ + OH-.

What is the molarity of a solution of hydrogen fluoride (HF, molecular mass = 20.0 g/mol) that contains 0.425 mol HF in 400.0 mL of solution? a) 0.940 M b) 0.0212 M c) 0.0531 M d) 1.06 M

d) 1.06 M why: Molarity = Mol/liters

A 0.100 M Pb(NO3)2 solution was added to 30.0 mL of 0.250 M MgI2. The volume of Pb(NO3)2 needed to completely react with the MgI2 solution is a) 37.5 mL b) 150. mL c) 300. mL d) 75.0 mL

d) 75.0 mL Why: The first step is to determine the stoichiometry of the reaction. It is Pb2+(aq) + 2I-(aq) → PbI2(s). Next we calculate the concentrations of the two reacting ions: a 0.100 M Pb(NO3)2 solution contains 0.100 M Pb2+ while 0.250M MgI2 contains 2(0.250) = 0.500 M I-. Lastly, we go through the steps to go from the volume of what we know (30.0 mL of the I- solution) to the volume of what we want (the Pb2+ solution). (1) (0.0300 L I- solution) x (0.500 mol I-/1 L I- solution) = 0.0150 mol I-. (2) (0.0150 mol I-) x (1 mol Pb2+/2 mol I-) = 0.0075 mol Pb2+ (3) (0.0075 mol Pb2+) x (1 L Pb2+ solution/0.0100 mol Pb2+) = 0.0750 L Pb2+ solution Thus, (0.0750 L Pb2+ solution) x (1000 mL / L) = 75.0 mL Pb2+ solution

When solutions of sodium carbonate and silver nitrate are mixed, a precipitate of silver carbonate forms while the compound sodium nitrate remains in solution. The net ionic equation for this reaction is a) CO3-(aq) + Ag+(aq) → AgCO3(s) b) NaCO3(s) + Ag+(aq) → AgCO3(s) + Na+(aq) c) Na2CO3(aq) + 2AgNO3(aq) → Ag2CO3(s) + 2NaNO3(aq) d) CO32-(aq) + 2Ag+(aq) → Ag2CO3(s) e) Na+(aq) + NO3-(aq) → NaNO3(s)

d) CO32-(aq) + 2Ag+(aq) → Ag2CO3(s)

In the net ionic equation for the reaction Cr2(SO4)3 + 3Ba(NO3)2 --> 3BaSO4 + 2Cr(NO3)3 the spectator ions are? a) Cr3+ and Ba2+ b) SO42- and NO3- c) Ba2+ and SO42- d) Cr3+ and NO3-

d) Cr3+ and NO3- why: all salts containing SO4 are soluble except those of Pb, Ca, Sr, Ba, and Hg2.

A net ionic equation can be written for a reaction when: a) a precipitate forms from a solution of soluble reactants. b) water forms in the reaction of an acid and a base. c) a weak electrolyte forms from a solution of strong electrolytes. d) two of the choices (a) - (c) are met, but not the third. e) any of the choices (a) - (c) are met.

why: A precipitate is formed from a mixture of soluble reactants. An acid and a base react together to form a salt and usually water. A weak electrolyte is formed from a mixture of strong electrolytes. A gas is formed from a mixture of reactants.