Question

Determine the [H+], [OH-] and the pOH for the NH3 solution 0.80M (Kb= 1.8x10^-5)?

`NH_3 (ac) + H^2O (l) ⇌ NH_4 (ac) + OH^- (ac)`

Answer 1

Well....we solve for `[HO^-]`...

Explanation:

We address the equilibrium...

`NH_3(aq) + H_2O(l)rightleftharpoonsNH_4^+ + HO^-`

For which we have the equilibrium constant....

`1.80xx10^-5=([NH_4^(+)][HO^(-)])/([NH_3])`

Now initially...`[NH_3]=0.80*mol*L^-1`...and if `x*mol*L^-1` dissociates we got...

`1.80xx10^-5=x^2/(0.80-x)`...and so ....

`x=sqrt(1.80xx10^-5xx(0.80-x)`

And reasonably....if `0.80">>"x`...then....

`x~=sqrt(1.80xx10^-5xx0.80)`

`x_1=sqrt(1.44xx10^-5)=3.79xx10^-3`

...and we can plug the approx. in again to see how `x` evolves..

`x_2=sqrt(1.80xx10^-5xx(0.80-3.79xx10^-3))=3.79xx10^-3`

And so the approximation is sound...

And so `[HO^-]=3.79xx10^-3*mol*L^-1`...`pOH=-log_10(3.79xx10^-3)=2.42`...`pH=14-2.42=11.58`...and `[H_3O^+]=10^(-11.58)*mol*L^-1=2.63xx10^-12*mol*L^-1`