What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, NaC7H5O2, to 1.00 L of 0.0100 M benzoic acid, HC7H5O2? Assume that there is no change in volume. The K a for benzoic acid is 6.3 × 10–5.
pKa = -log Ka = -log(6.3*10^-5) = 4.2
moles of sodium benzoate = 3.96 gm/144.11g/mol = 0.0275 mol
moles of benzoic acid = 1L * 0.01M = 0.01 moles
pH = pKa + Log[benzoate/benzoic acid]
= 4.2 + log [0.0275/0.01] = 4.64
ph of the solution is 4.64
What is the pH of a buffer that was prepared by adding 3.96 g of sodium...
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