Calculate the pH of a solution prepared by dissolving 1.60 g of sodium acetate, CH3COONa, in 50.0 mL of a 0.10 M acetic acid, CH3COOH (aq). Assume the volume change upon dissolving the sodium acetate is negligible. Ka of CH3COOH is 1.75 × 10^-5.​

pH = 5.35

Explanation:

Given 1.60 grams sodium acetate (NaOAc(aq))*** added to 50ml of 0.10M acetic acid (HOAc(aq)) solution.

Applying common ion effect keeping in mind that the addition of NaOAc provides the common-ion (OAc⁻).

     HOAc(aq) ⇄          H⁺(aq)            +                 OAc⁻(aq)

I      0.10m          1.32 x 10⁻³M ≈ ∅M*     (1.6g/82.03g/mol) / 0.050L = 0.39M

C       -x                         +x                                0.39M + x ≈ 0.39M**

E    0.10M - x                  x                                    0.39M

      ≈ 0.10M

Ka = [H⁺][OAC⁻]/[HOAC] => [H⁺] = Ka·[HOAc] / [OAc⁻]

[H⁺] = (1.75 X 10⁻⁵)(0.10) / (0.39) = 4.5 x 10⁻⁶M

∴ pH = -log[H⁺] = -log(4.5 x 10⁻⁶) = -(-5.35) = 5.35

* [H⁺] before adding NaOAc = SqrRt(Ka · [HOAc]) = SqrRt(1.75 x 10⁻⁵· 0.10) = 1.32 x 10⁻³M. Since this concentration value is so small, the initial [H⁺] is assumed to be zero molar (∅M).

** The added [H⁺] is negligible and dropped in the ICE table. That is, adding ~[H⁺] in the order of 10⁻³M does not change the H⁺ ion concentration sufficiently to affect problem outcome and is therefore dropped in the ICE table.

*** Acetic Acid and Sodium Acetate are frequently written HOAc and NaOAc where the OAc⁻ anion is the acetate ion (CH₃COO⁻) for brevity.

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