# Acid - Base Reactions

Please show all work.

Do NOT use the Henderson-Hasselbalch equation in solving any of these questions.

1. Ksp = 2 x 10^-19 for LaF3 in water at 25 degrees celcius. How many grams of LaF3 are present in 500 mL of saturated LaF3(aq) at 25 degrees celcius ? The atomic weight of La is 139.

2. Find [H+] in 0.25 M NaC2H3O2 (aq) at 25 degrees celcius given that Ka = 1.8 x 10^-5 for HC2H3O2 (aq) at 25 degrees celcius. You MUST begin by writing the chemical reaction that is at equilibrium.

3. Find [H+] in a solution prepared by dissolving 1.50 g of Ca(OH)2 in 600 mL of water and diluting the solution to a final volume of 800 mL.

4. Find the pH when 45.0 mL of 0.250 M NaOH (aq) is added to 35.0 mL of 0.200M HCl (aq).

5. Find [H+] in a solution prepared by adding 0.40 moles of NaCN to 500 mL of a 0.25M HCN(aq) solution. Assume that there is no volume change when the NaCN is added. For HCN(aq), Ka = 4.9 x 10^-10.

#### Solution Preview

Acid - Base Reactions

Please show all work.

Do NOT use the Henderson-Hasselbalch equation in solving any of these questions.

1. Ksp = 2 x 10^-19 for LaF3 in water at 25 degrees celcius. How many grams of LaF3 are present in 500 mL of saturated LaF3(aq) at 25 degrees celcius ? The atomic weight of La is 139.

SOLUTION:

Consider the dissociation of LaF3:

LaF3 La+3(aq) + 3F-(aq)

s s 3s

Where: s = solubility of LaF3 which is also equal to the concentration of a saturated solution of LaF3

Solving for s from the Ksp of LaF3:

Ksp = [La+3] [F-]3

Ksp = 2 x 10-19 = (s) (3s)3

2 x 10-19 = 27 s4

s4 = 2 x 10-19 / 27

S4 = 7.41 x 10-21

s = 9.28 x 10-6 M This is the concentration of a saturated solution of LaF3

Solving for the mass of LaF3 in 500 mL saturated solution:

Molarity = M = mol solute / liter solution

9.28 x 10-6 mol/L = x / 0.5 L

x = 4.64 x 10-6 ...

#### Solution Summary

The expert examines the base reaction for Henderson-Hasselbalch equations in acid.