Question

Be sure to answer all parts. A solution is made by mixing exactly 500 mL of 0.117 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below. Ka of CH3COOH is 1.8 × 10^−5

[H+] M Enter your answer in scientific notation.

[OH−] M

[CH3COOH] M Enter your answer in scientific notation.

[Na+] M

[CH3COO−] M

Answer 1

Moles of NaOH = 500 x 0.117 = 58.5 mmol of NaOH

Moles of CH₃COOH = 500 x 0.100 = 50.0 mmol of CH₃COOH

50.0 mmol of CH₃COOH will react with 50.0 mmol of NaOH to form 50.0 mmol of CH₃COONa.

Unreacted moles of NaOH = (58.5 - 50.0) = 8.5 mmol of NaOH

Total volume of the solution after the reaction = (500 + 500) = 1000 mL

Molarity of unreacted NaOH = 8.5/1000 = 0.00850 M

Hence,

[OH^-] = 0.0085 M

Now,

K_w = [H⁺][OH^-]

or, 10^-14 = [H⁺] x 0.0085

or, [H⁺] = 10^-14/0.0085 = 1.18 x 10^-12 M

CH₃COOH is completely used up in the reaction

Hence, [CH₃COOH] = 0 M

Moles of Na⁺ = 58.5 mmol

Hence, [Na⁺] = 58.5/1000 = 0.0585 M

Moles of CH₃COO⁻ = 50.0

Hence, [CH₃COO⁻] = 50.0/1000 = 0.0500 M