A 0.40 M solution of a salt NaA has pH=8.95. What is the value of Ka...

A 0.40 M solution of a salt NaA has pH=8.95. What is the value of Ka for the parent acid HA?

Expert Solution

NaA(aq) -----------> Na^+(aq) + A^-(aq)

PH = 8.95

POH = 14-PH

= 14-8.95 = 5.05

POH = 5.05

-log[OH-] = 5.05

[OH-] = 10^-5.05 = 8.9*10^-6 M

A^- + H2O --------------> HA + OH^-

[HA] = [OH-] at equilibrium

Kb = [OH^-][HA]/[A^-]

= 8.9*10^-6*8.9*10^-6/0.4

= 1.98*10^-10

Ka = Kw/Kb

= 1*10^-14/1.98*10^-10

= 5.05*10^-5 >>>>answer

queen_honey_blossom answered 2 years ago

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