Question

Calculate the pH and % dissociation of a 0.25 M solution of propanoic acid at 25 degrees Celsius? (Ka= 1.34 x 10^-5)

Please include the steps to solve the problem! Thank you. :)

Answer 1

We interrogate the equilibrium reaction....

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

Explanation:

For which `K_a=(["H"_3"CCH"_2"CO"_2^(-)]["H"_3"O"^+])/(["H"_3"CCH"_2"CO"_2"H"])=1.34xx10^-5`

We let the amount of propanoic acid that dissociates be `x`.

And so `x^2/(0.25-x)=1.34xx10^-5`

We assume that `0.25-x~=0.25`...

And thus .............

`x_1=sqrt(1.34xx10^-5xx0.25)=1.83xx10^-3*mol*L^-1`

`x_2=sqrt(1.34xx10^-5xx(0.25-1.83xx10^-3))=1.82xx10^-3*mol*L^-1`

`x_3=sqrt(1.34xx10^-5xx(0.25-1.82xx10^-3))=1.82xx10^-3*mol*L^-1`...

And thus the successive approximations give a consistent answer, the same answer as if we used the quadratic equation....

But `x=[H_3O^+]=1.82xx10^-3*mol*L^-1`..

`pH=-log_10[H_3O^+]=-log_10(1.82xx10^-3)=+2.74`

`"% dissocation"="acid that dissociated"/"starting acid"xx100%`

`=(1.82xx10^-3*mol*L^-1)/(0.250*mol*L^-1)xx100%=0.73%`

And thus under `1%` of the acid dissociated...