The pH is determined by the hydrolysis of the salt.
..........A^- + HOH ==> HA + OH^-
initial..0.160M..........0.....0
change...-x...............x.....x
equil...0.160-x...........x.....x
Kb for A^- = (Kw/Ka for HA) = (x)(x)/(0.160-x
Solve for x = (OH^-)and convert to pH.
Calculate the pH when 39.0mL of 0.321 of HA is mixed with 39.0mL of 0.321M NaOH where
HA is a monoprotic weak acid with Ka= 7.5x10^-5 M ?
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