- #1

Jzhang27143

- 38

- 1

this type of question should go into homework section

You prepare 0.5 liters of a solution by adding 0.75 moles of a weak acid HA to water. For HA, Ka = 10^-1 Finally you dilute this solution to a final volume of 2.0 liters. What is the pH of the diluted solution?

My solution is this. [HA] = .75 mol /.5 L = 1.5 M. HA + H2O -> H3O+ + A-. If x is the equilibrium concentration of H3O+, x^2 /(1.5 - x) = 10^-1 so solving for x gives x = 0.34 M = [H3O+].

Moles of H3O+ must be 0.34M * 0.5 L = 0.17 mol. When the new volume is 2.0 L, [H3O+] = 0.085M so the pH = 1.1. However, the correct answer to this is 0.8. Where did I go wrong?

My solution is this. [HA] = .75 mol /.5 L = 1.5 M. HA + H2O -> H3O+ + A-. If x is the equilibrium concentration of H3O+, x^2 /(1.5 - x) = 10^-1 so solving for x gives x = 0.34 M = [H3O+].

Moles of H3O+ must be 0.34M * 0.5 L = 0.17 mol. When the new volume is 2.0 L, [H3O+] = 0.085M so the pH = 1.1. However, the correct answer to this is 0.8. Where did I go wrong?