Question

What is the pH of a `"0.065 M"` aqueous solution of acetic acid? `K_a = 1.8 xx 10^(-5)`, and acetic acid, `"CH"_3"COOH"`, reacts with water as shown below: `"CH"_3"COOH"(aq) + "H"_2"O"(l) rightleftharpoons "CH"_3"COO"^(-)(aq) + "H"_3"O"^(+)(aq)`

Answer 1

`pH=2.98`

Explanation:

As with all these problems, we first write out the equilibrium equation to inform our reasoning...

`HOAc(aq) + H_2O(l) rightleftharpoons ""^(-)OAc + H_3O^+`

Now `K_"eq"=([H_3O^+][""^(-)OAc])/([HOAc(aq)])`

Now INITIALLY, `[HOAc]=0.065*mol*L^-1`..and we PROPOSE that `x*mol*L^-1` of the acid dissociates....

And so we rewrite the equilibrium expression...

`K_"eq"=1.74xx10^-5=x^2/(0.065-x)`

...and thus `x=sqrt(1.74xx10^-5xx(0.065-x))`

We ASSUME that `0.065">>"x`..and so...

`x~=sqrt(1.74xx10^-5xx0.065)`

`x_1=1.06xx10^-3`..and now we gots an approximation for `x`, we can resubstitute this value back into the equilibrium expression, and see how `x` evolves....

`x_2=sqrt(1.74xx10^-5xx(0.065-1.06xx10^-3))=1.05xx10^-3`

`x_3=1.05xx10^-3*mol*L^-1`...

Since the approximations have converged I am prepared to accept this value...this method of successive approximations is usually more efficient than pfaffing about with the quadratic equation...

And since `x=[H_3O^+]=1.05xx10^-3*mol*L^-1`...

`pH=-log_10[H_3O^+]=-log_10(1.05xx10^-3)=-(-2.98)=??`