Question

calculate [H3O] in a solution that is 0.0525M HCl and 0.0768 M NaC2H3O2 . Ka of HC2H3O2= 1.8E-5

Answer 1

Reaction between HCl and NaC₂H₃O₂
HCl(aq) + NaC₂H₃O₂(aq) ====> HC₂H₃O₂(aq) + NaCl(aq)
HCl is the limiting reactant (completely reacts).
After reaction :
[HC₂H₃O₂] = 0.0525 M
[C₂H₃O₂⁻] = [NaC₂H₃O₂] = (0.0768 - 0.0525) M = 0.0243 M

Consider the dissociation of HC₂H₃O₂ :
HC₂H₃O₂(aq) + H₂O(l) ⇌ C₂H₃O₂⁻(aq) + H₃O⁺(aq) …. Kₐ

we can write Ka as
Kₐ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

[H₃O⁺] = Kₐ [HC₂H₃O₂] / [C₂H₃O₂⁻] = (1.8 × 10⁻⁵) × 0.0525 / 0.0243 M = 3.9 × 10⁻⁵ M