Chapter 4: Reactions in Aqueous Solutions

Answer Key

4.89 The reaction between KHP (KHC8H4O4) and KOH is:

KHC8H4O4(aq)  KOH(aq)  H2O(l)  K2C8H4O4(aq)

We know the volume of the KOH solution, and we want to calculate the molarity of the KOH solution.

need to find want to calculate

mol KOH M of KOH = L of KOH soln

given If we can determine the moles of KOH in the solution, we can then calculate the molarity of the solution. From the mass of KHP and its molar mass, we can calculate moles of KHP. Then, using the mole ratio from the balanced equation, we can calculate moles of KOH.

1 mol KHP 1 mol KOH ? mol KOH= 0.4218 g KHP创 = 2.0654 10-3 mol KOH 204.22 g KHP 1 mol KHP

From the moles and volume of KOH, we calculate the molarity of the KOH solution.

mol KOH 2.0654 10-3 mol KOH Mof KOH= = = 0.1106 M L of KOH soln 18.68 10-3 L soln

4.91 (a) In order to have the correct mole ratio to solve the problem, you must start with a balanced chemical equation.

HCl(aq)  NaOH(aq) 揪 NaCl(aq)  H2O(l)

From the molarity and volume of the HCl solution, you can calculate moles of HCl. Then, using the mole ratio from the balanced equation above, you can calculate moles of NaOH. 93 CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

2.430 mol HCl 1 mol NaOH ? mol NaOH= 25.00 mL创 = 6.075 10-2 mol NaOH 1000 mL soln 1 mol HCl

Solving for the volume of NaOH:

moles of solute liters of solution = M

6.075 10-2 mol NaOH volume of NaOH= = 4.278� 10-2 L 42.78 mL 1.420 mol/L

(b) This problem is similar to part (a). The difference is that the mole ratio between base and acid is 2:1.

H2SO4(aq)  2NaOH(aq) 揪 Na2SO4(aq)  H2O(l)

4.500 mol H SO 2 mol NaOH ? mol NaOH= 25.00 mL创 2 4 = 0.2250 mol NaOH 1000 mL soln 1 mol H2 SO 4

0.2250 mol NaOH volume of NaOH= =0.1585 L = 158.5 mL 1.420 mol/L

(c) This problem is similar to parts (a) and (b). The difference is that the mole ratio between base and acid is 3:1.

H3PO4(aq)  3NaOH(aq) 揪 Na3PO4(aq)  3H2O(l)

1.500 mol H PO 3 mol NaOH ? mol NaOH= 25.00 mL创 3 4 = 0.1125 mol NaOH 1000 mL soln 1 mol H3 PO 4

0.1125 mol NaOH volume of NaOH= =0.07923 L = 79.23 mL 1.420 mol/L

4.95 The balanced equation is given in the problem. The mole ratio between Fe2 2- and Cr2 O 7 is 6:1.

2 2- First, calculate the moles of Fe that react with Cr2 O 7 . CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS 94

0.0250 mol Cr O2- 6 mol Fe2+ 26.00 mL soln创 2 7 = 3.90 10-3 mol Fe 2 + 2- 1000 mL soln 1 mol Cr2 O 7

The molar concentration of Fe2 is:

3.90 10-3 mol Fe 2 + M= = 0.156 M 25.0 10-3 L soln

4.97 The balanced equation is given in Problem 4.95. The mole ratio between Fe2 2- and Cr2 O 7 is 6:1.

2- First, calculate the moles of Cr2 O 7 that reacted.

0.0194 mol Cr O2- 23.30 mL soln� 2 7 4.52 10-4 mol Cr O 2 - 1000 mL soln 2 7

Use the mole ratio from the balanced equation to calculate the mass of iron that reacted.

6 mol Fe2+ 55.85 g Fe 2 + (4.52创 10-4 mol Cr O 2 - )� 0.1515 g Fe 2 + 2 7 2- 2 + 1 mol Cr2 O 7 1 mol Fe The percent by mass of iron in the ore is:

0.1515 g �100% 54.3% 0.2792 g

4.99 First, calculate the moles of KMnO4 in 24.0 mL of solution.

0.0100 mol KMnO 4 �24.0 mL 2.40 10-4 mol KMnO 1000 mL soln 4

Next, calculate the mass of oxalic acid needed to react with 2.40  104 mol

KMnO4. Use the mole ratio from the balanced equation.

-4 5 mol H2 C 2 O 4 90.036 g H 2 C 2 O 4 (2.40创 10 mol KMnO4 )� 0.05402 g H 2 C 2 O 4 2 mol KMnO4 1 mol H 2 C 2 O 4 95 CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

The original sample had a mass of 1.00 g. The mass percent of H2C2O4 in the sample is:

0.05402 g mass %= �100% 5.40% H C O 1.00 g 2 2 4

4.101 The balanced equation shows that 2 moles of electrons are lost for each mole 2- - of SO3 that reacts. The electrons are gained by IO3 . We need to find the moles - of electrons gained for each mole of IO3 that reacts. Then, we can calculate the final oxidation state of iodine.

2- The number of moles of electrons lost by SO3 is:

0.500 mol SO2- 2 mol e- 32.5 mL创 3 = 0.0325 mole- lost 2- 1000 mL soln 1 mol SO3

The number of moles of iodate, that react is: IO- , 3

- 1 mol KIO3 1 mol IO 3 -3 - 1.390 g KIO3创 = 6.4953 10 mol IO 3 214.0 g KIO3 1 mol KIO 3

3 - 6.4953  10 mole of IO3 gain 0.0325 mole of electrons. The number of moles - of electrons gained per mole of IO3 is: 0.0325 mol e- = 5.00 mole- /mol IO - -3 - 3 6.4953 10 mol IO3

- The oxidation number of iodine in IO3 is 5. Since 5 moles of electrons are - gained per mole of IO3 , the final oxidation state of iodine is 5  5  0. The

iodine containing product of the reaction is most likely elemental iodine, I2.