Question

A 0.240 M solution of the salt NaA has pH = 8.40. Calculate Ka for the acid HA.

Answer 1

Lets see the reaction

conjugate base A^- reacts with water as

A^- + H₂O $\rightleftharpoons$ HA + OH^-  

We have

pH = 8.40

And we know that pH + pOH = 14

So pOH = 14 - pH = 14 - 8.40 = 5.6

pOH = -log[OH^-]

then

[OH^-] = 10^-pOH = 10^-5.6 = 2.5  $\times$ 10^-6 M

then

[HA] = 2.5  $\times$ 10^-6 M

For reaction

A^- + H₂O $\rightleftharpoons$ HA + OH^-  

equilibrium constant Kb will be

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

= (2.5  $\times$ 10^-6 )(2.5  $\times$ 10^-6 ) / 0.24 M

= 2.6 $\times$ 10^-11

we know that

Ka $\times$ Kb = 1.0 $\times$ 10^-14  

then

Ka = 1.0 $\times$ 10^-14 / Kb = 1.0 $\times$ 10^-14 / 2.6 $\times$ 10^-11

= 3.8 $\times$ 10^-4   

So our answer is 3.8 $\times$ 10^-4