#C_8H_9N_3O_2NH + H_2O rightleftharpoonsC_8H_9N_3O_2NH^(+) + HO^-#
We write the expression for base association in the usual way:
#K_b=([C_8H_9N_3O_2NH^+][HO^-])/([C_8H_9N_3O_2NH])=4.1xx10^-4#
If, #[HO^-]=x#, then clearly (because of the size of #K_b#), #[C_8H_9N_3O_2NH]" >> "[HO^-]#, i.e. #[C_8H_9N_3O_2NH]" >> "x#. Given that #[HO^-]=x#, #x=[C_8H_9N_3O_2NH^+]#, and #[C_8H_9N_3O_2N]=(0.910-x)#
i.e., so if #K_b=(x^2)/(0.910-x)#
#K_b~=(x^2)/(0.910)#, this is an approximation, which we must justify later.
#x_1=sqrt(K_bxx0.910)=sqrt(4.1xx10^-4xx0.910)=0.019#
Since we got an approximation for #x#, we can recycle this value thru the equilibrium expression, and see if our estimate of #x_2# varies markedly:
#x_2=sqrt(K_bxx
(0.910-0.019))=sqrt(4.1xx10^-4xx(0.910-0.019))=0.019*mol*L^-1#
Our approximation was indeed reasonable. And thus #x=[HO^-]=0.019*mol*L^-1#. What is #pH# for this solution?
