A buffer is created by adding 0.40 mol of sodium benzoate to 1.00 L of a 0.40 mol/L benzoic acid solution. The Ka of benzoic acid is 6.3 x 10-5. What is the pH of the buffer?
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C6H5COOH <--> H+ + C6H5COO-
Ka = {H+} {C6H5COO-} / {C6H5COOH}
Ka = (x) x (x) / (4.00x10^-1 -x) but x is small , so leave it out as it does not affect the final result.
x^2 = 6.3x 10^-5 x 4.0 x10 ^-1
" = 25.2 x10^-6
x = 5.02x10^-3
pH = 2..30
Ka = {H+} {C6H5COO-} / {C6H5COOH}
Ka = (x) x (x) / (4.00x10^-1 -x) but x is small , so leave it out as it does not affect the final result.
x^2 = 6.3x 10^-5 x 4.0 x10 ^-1
" = 25.2 x10^-6
x = 5.02x10^-3
pH = 2..30