A 0.350 M solution of the salt NaA has pH=9.60. Calculate Ka for the acid HA
Could someone explain how to do this please?
Could someone explain how to do this please?
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9.6 = -log[H+]. [H+] equals 10^(-pH).
10^-9.6 = [H+] = 2.51188643x10^-10
Weak acids dissociate into equal amounts of H+ and their conjugate base
Ka = (2.5188643x10^-10)^2 / .35 = 1.8x10^-19
10^-9.6 = [H+] = 2.51188643x10^-10
Weak acids dissociate into equal amounts of H+ and their conjugate base
Ka = (2.5188643x10^-10)^2 / .35 = 1.8x10^-19