What is the final [H⁺] if 1.00 L of 0.0550 M HCl is mixed with 1.00 L of 12.0 M acetic acid (Ka for CH₃COOH is 1.80 × 10⁻⁵)?

Answer 1

You are mixing 2 acids. A strong acid (HCl) and a weak acid (CH3COOH). To find the pH of the solution, we want to find the total [H⁺] and then take the negative log of that value.

Final volume of solution = 2.00 L

[HCl] = 0.0275 M

[CH3COOH] = 6.0 M

Finding [H⁺] from CH3COOH:

CH3COOH <==> CH3COO^- + H⁺

Ka = [CH3COO^-][H⁺] / [CH3COOH]

1.80x10^-5 = (x)(x) / 6.0

x² = 1.08x10^-4

x = [H⁺] = 0.01039 M

Finding [H⁺] from HCl

HCl ==> H⁺ + Cl^-

[H⁺] = 0.0275 M

Total [H⁺] = 0.01039 M + 0.0275 M = 0.03789 M

pH = -log 0.03789

pH = 1.421

(be sure to check all of the math)