Question

The weak base ammonia (NH3) accepts a proton (H+) from water to form a basic solution. What is the concentration of OH– ions in a 0.75 mol L–1 NH3 solution?

Answer 1

In the absence of `K_b` we have to supply one....

Explanation:

We interrogate the equilibrium...

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

Now `K_b=10^(-4.76)=1.74xx10^-5`

And so `1.74xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])`

And if `x*mol*L^-1` ammonium dissociates...

...we gots `1.74xx10^-5=x^2/(0.75-x)`

We make the usual approximation....that `0.75">>>"x`..

`1.74xx10^-5~=x^2/(0.75)`

And so `x=sqrt(1.74xx10^-5xx0.75)`

`x_1=3.61xx10^-3`, and we recycle that approximation into the original expression, i.e. we got an estimate of `x`....

`x_2=3.60xx10^-3`,

`x_3=3.60xx10^-3`,..and since the approximations have converged...this is the true value....

But `x=[HO^-]=3.60xx10^-3*mol`...`pOH=2.44`...`pH=11.56`