Question

Calculate the [H3O+] and pH of the polyprotic H3C6H5O7 with an initial concentration 0.125 M. I got 0.011, but apparently it is wrong even though I added the three dissociated concentrations. Help would be very much appreciated! Here are the Ka values: Ka1=7.4*10^-4, Ka2=1.7*10^-5 Ka3=4.0*10^-7

Answer 1

For the first dissociation of the acid

-------------- H3C6H5O7 ------- > H₂C₆H₅O₇^-(aq) + H⁺(aq) ; Ka1 = 7.4*10^-4

Init.con(M): 0.125 ---------------- 0 ---------------------- 0

Change(M):- 0.125x, --------- +0.125x, ---------- +0.125x

eqm.Con(M):0.125(1 - x), ----- 0.125x, ----------- 0.125x

Ka1 = 7.4x10^-4 = (0.125x * 0.125x) / 0.125(1 - x)

(1 - x) is nearly equals to 1

=> x = 0.07694 M

=> [H₂C₆H₅O₇^-(aq)] = [H⁺(aq)] = 0.125x = 0.125 * 0.077 = 0.00962 M

For the second dissociation of H3C6H5O7

-------------- H₂C₆H₅O₇^-(aq) ------ > HC₆H₅O₇^2-(aq) + H⁺(aq) ; Ka2 = 1.7*10^-5

Init.con(M): 0.00962 ------------------ 0 ---------------------- 0.00962

Change(M):- 0.00962y, ----------- +0.00962y, ---------- +0.00962y

eqm.Con(M):0.00962(1 - y), ------ 0.00962y, ------------ 0.00962(y + 1)

Ka1 = 1.7x10^-5 = (0.00962y * 0.00962(y + 1) / 0.00962(1 - y)

(1 - y) and (1+y) are nearly equals to 1

=> y = 0.00177

We can neglect the third dissociation as the Ka3 value is very small

Hence [H⁺(aq)] = 0.00962(y + 1) M = 0.00962(0.00177 + 1) = 0.00964 M (answer)

pH = - log[H⁺(aq)] = - log(0.00964 M) = 2.016 (answer)