Answer to Question #265075 in General Chemistry for Jay

Question #265075

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


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Expert's answer
2021-11-13T01:33:22-0500

Moles of CH3COONa =2.2082.0=0.02683  mol= \frac{2.20}{82.0} = 0.02683 \;mol

Volume V = 78.0 mL = 0.078 L

[CH3COONa] =0.026830.078=0.3439  M= \frac{0.02683}{0.078} = 0.3439 \;M

[CH3COOH] = 0.25 M

pH=log(Ka)+log([CH3COONa][CH3COOH])pH=log(1.75×105)+log(0.34390.25)pH=4.89pH = -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


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