#K_b# for #"acetoaminophen"# is #4.1xx10^-4#. What is #[HO^-]# for a solution of #"acetoaminophen"# at #0.910*mol*L^-1# concentration?

1 Answer
Mar 9, 2017

Answer:

#[HO^-]=0.019*mol*L^-1#

Explanation:

#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?