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)#

1 Answer
Mar 19, 2018

Answer:

#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)=??#