calculate the pH of a solution prepared by dissolving 2.05 g of 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
"Moles~ of~ CH3COONa= ~\\cfrac{2.05}{82.0} = ~0.025 ~mol"
"Volume~ V = 78.0 ~mL = 0.078~ L"
"[CH_3COONa]=~ \\cfrac{0.025}{0.078} = ~0.32~M"
"[CH_3COOH] = 0.25~ M"
"pH =~ -log(K_a) +~ log\\bigg(\\cfrac{[CH_3COONa]}{[CH_3COOH]}\\bigg)"
"pH =~ -log(1.75 *~10^{-5}) + ~log\\bigg(\\cfrac{0.32}{0.25}\\bigg) \\\\"
"pH= 4.757 + 0.079 = 4.836"
Answer: pH = 4,836
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