Question

What is the pOH of a 0.80M solution of acetic acid, HC2H3O2, with a dissociation constant for a weak acid of 1.8 x 10^-5 ?

Answer 1

Consider the equilibrium,

`CH_3COOH rightleftharpoons H^(+) + CH_3COO^(-)`

where,

`K_"a" = ([H^+][CH_3COO^-])/([CH_3COOH]) approx 1.8*10^-5`

Now, using an ICE table,

`CH_3COOH rightleftharpoons H^(+) + CH_3COO^(-)`

`K_"a" = x^2/(0.80-x) = 1.8*10^-5`

`=> x =[H^+] approx 3.79*10^-3`

Moreover, recall,

`K_"w" = [H^+][OH^-] = 1.0*10^-14`

Hence,

`"pOH" = -log(K_"w"/([H^+])) approx 11.58`

Answer 2

`pOH=11.58.....`

Explanation:

We investigate the equilibrium....

`"H"_3"CCO"_2"H(aq)" +"H"_2"O(l)" rightleftharpoons"H"_3"CCO"_2^(-) + "H"_3"O"^+`

....for `K_a=(["H"_3"CCO"_2^(-)]["H"_3"O"^+])/(["H"_3"CCO"_2"H(aq)"])=1.80xx10^-5`

Now if `x*mol*L^-1` acid dissociates...then...

`x^2/(0.80-x)=1.80xx10^-5`

And so....`x=sqrt(1.8xx10^-5xx(0.80-x))`

`~=sqrt(1.8xx10^-5xx(0.80))`...given the assumption that `(0.80-x)~=0.80`

`x_1=3.79xx10^-3*mol*L^-1`...and we plug this approx..back in..

`x_2=3.79xx10^-3*mol*L^-1`...and I prepared to accept this as the true value because the approximations have swiftly converged.

But `x=[H_3O^+]`...`pH=-log_10(3.79xx10^-3)=2.42`

And `pOH=14-2.42=??`