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?

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Answer 1 · 2018-01-30T18:57:32

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

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#