Answer to Question #265075 in General Chemistry for Jay

Question

Calculate the pH of a solution prepared by dissolving 2.20 got sodium acetate, CH3COONa, in 78.0 mL of 0.25 M acetic acid, CH3COOH(aq). Assume the volume change upon dissolving the sodium acetate is negligible. Ka of CH3COOH is 1.75 x 10^-5

Answer 1

Moles of CH₃COONa $= \frac{2.20}{82.0} = 0.02683 \;mol$

Volume V = 78.0 mL = 0.078 L

[CH₃COONa] $= \frac{0.02683}{0.078} = 0.3439 \;M$

[CH₃COOH] = 0.25 M

$pH = -log(K_a) + log(\frac{[CH_3COONa]}{[CH_3COOH]}) \\ pH = -log(1.75 \times 10^{-5}) + log(\frac{0.3439}{0.25}) \\ pH = 4.89$