What would be the pH of a 0.503 M solution of ammonia (NH3) at room temperature? The #K_b# of ammonia is #1.78 x 10^-5#

1 Answer
Apr 15, 2017

Answer:

#pH~=11.5#

Explanation:

We interrogate the equilibrium...............

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

#K_"eq"=([NH_4^+][HO^-])/([NH_3(aq)])=1.78xx10^-5#

Now we know that #[NH_3(aq)]=0.503*mol*L^-1# initially, and call the amount of association it undergoes #x#.

So at equilibrium, #[NH_4^+]=[HO^-]=x#; and #[NH_3(aq)]=(0.503-x)*mol*L^-1#.

#K_"eq"=1.78xx10^-5=x^2/(0.503-x)#

This is quadratic in #x#, which we could solve exactly, however, we make the approximation that #(0.503-x)~=0.503*mol*L^-1#.

Thus #x_1=sqrt(1.78xx10^-5xx0.503)=0.00299*mol*L^-1#

Now that we an approximation for #x#, we can we substitute this value back into the equation.

And thus..........................................................

#x_2=sqrt(1.78xx10^-5xx(0.503-0.00299))=0.00298*mol*L^-1#

Since the values have converged, we can take this value as the true value.

And so #[HO^-]=[NH_4^+]=2.98xx10^-3*mol*L^-1#, and thus #pOH=-log_(10)[HO^-]=-log_10(2.98xx10^-3)=2.53#.

But we need #pOH#, however, we know (how?) that #pH+pOH=14#.

And so #pH=14-2.53=??#