Answer:
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
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* [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.
