Determine the [OH-] and the [H+], and the pH of the following solutions

Question

Determine the [OH-] and the [H+], and the pH of the following solutions.

A. 1.0 M KCl

B. 1.0 M KC2H3O2

I know the answer for A is [OH-] and [H+] is 1.0x10^-7 because KCl is neutral due to coming from a strong acid and a strong base.

I don't know how to work B.

I'm not given any Ka's or Kb's for this problem.

The answer for B is: [OH-] 2.4x10^-5, [H+] 4.2x10^-10, pH is 9.38

please list steps for B so I can follow along.

Answer 1

You are correct about A. Now, for B, we proceed as follows:

KC₂H₃O₂ is the salt of a weak acid (HC₂H₃O₂) and a strong base (KOH). The salt will hydrolyze as follows:

C₂H₃O₂^- + H₂O ==> HC₂H₃O₂ + OH^-

And to find the [OH-], and thus the pOH and then the pH, we need either the Kb for C₂H₃O₂- or the Ka for HC₂H₃O₂ (acetic acid). Looking up the Ka for acetic acid we find it to be 1.8x10^-5

Since Ka x Kb = Kw, we can find Kb for C₂H₃O₂^-

Kb = 1x10^-14/1.8x10^-5 = 5.56x10^-10

Kb = 5.56x10^-10 = [HC₂H₃O₂] [ OH^-]/[C₂H₃O₂^-]

5.56x10^-10 = (x)(x)/1-x and assuming x is small relative to 1 we can ignore it in the denominator

x² = 5.56x10^-10

x = 2.36x10^-5 M = [OH^-] = 2.4x10^-5 M

[H⁺] = 1x10^-14/2.4x10^-5 = 4.17x10^-10 = 4.2x10^-10 = [H⁺]

pH = -log 4.2x10^-10 = 9.38 = pH