Given #0.10*mol*L^-1# concentrations of acetic acid, and benzoic acid, what are the #pH# of EACH solution?

Question

Naomi Aronson · Accepted Answer

You have not quoted #pK_a# for neither acetic acid nor benzoic acid....fortunately I know one....

We interrogate the equilibrium... #"H"_3"CCO"_2"H(aq)"+"H"_2"O(l)" rightleftharpoons"H"_3"CCO"_2^(-) + "H"_3"O"^+# For which #K_a=10^(-4.76)#...and so... #K_a=(["H"_3"CCO"_2^(-)]["H"_3"O"^+])/(["H"_3"CCO"_2"H(aq)"])=1.74xx10^-5# We put #["H"_3"O"^+]=["H"_3"CCO"_2^(-)]=x#... And so #1.74xx10^-5=x^2/(0.10-x)#... Now this is a quadratic in #x#, the which could be solved exactly...but because acetic acid is a WEAK acid, we ASSUME that #x# is small, and that #0.1">>"x#, and thus #(0.10-x)~=0.1# And so #x_1=sqrt(1.74xx10^-5xx0.10)=1.32xx10^-3*mol*L^-1#.. And now we gots a first approximation, we can plug this back into the first expression to get a second and third approximation... And so #1.74xx10^-5=x^2/(0.10-x_1)=1.74xx10^-5=x^2/(0.10-0.00132)# #=1.31xx10^-3*mol*L^-1#...since the approximations have converged I am prepared to accept this as the true value, and indeed it is the same value as if I had solved the quadratic equation... And so #["H"_3"O"^+]=1.31xx10^-3*mol*L^-1#... And ...... #pH=-log_10["H"_3"O"^+]=-log_10(1.31xx10^-3)=-(-2.88)=2.88#. Now I don't know the #pK_a# of benzoic acid...anyway you have had your three wishes, so you will have to find it, and solve the equation appropriately.... This site quotes #pK_a("benzoic acid")=4.202.#