Triprotic acid versus monoprotic strong base

What is the formula of the titration curve of a triprotic acid versus a strong base?

Consider a solution containing water, a triprotic acid of concentration C_aand a strong base of concentration C_bwith the following equilibria:

$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$

$H_3A(aq)\rightleftharpoons H^+(aq)+H_2A^-(aq)\; \; \; \; \; \; \; \; (20)$

$H_2A^-(aq)\rightleftharpoons H^+(aq)+HA^{2-}(aq)\; \; \; \; \; \; \; \; (21)$

$HA^{2-}(aq)\rightleftharpoons H^+(aq)+A^{3-}(aq)\; \; \; \; \; \; \; \; (22)$

$BOH(aq)\rightleftharpoons B^+(aq)+OH^-(aq)$

With reference to charge balance of the above equilibria, the sum of the number of moles of cations H⁺ and B⁺ must equal to that of anions H₂A^–, HA^2-, A^3- and OH^–. As the volume is of the solution is common to all ions,

$[H^+]+[B^+]=[OH^-]+[H_2A^-]+2[HA^{2-}]+3[A^{3-}]\; \; \; \; \; \; \; \; (23)$

Read this article to understand how to arrive at eq23.

With reference to eq20, eq21 and eq22, at any point of the titration, the sum of the number of moles of H₃A, H₂A^–, HA^2-and A^3– must equal to the number of moles of H₃A if it were undissociated. As the change in volume of the solution is common to all ions,

$[H_3A]+[H_2A^-]+[HA^{2-}]+[A^{3-}]=\frac{C_aV_a}{V_a+V_b}\; \; \; \; \; \; \; \; (24)$

where V_a and V_b are the volume of triprotic acid in the solution and the volume of strong monoprotic base in the solution respectively. The equilibrium constant equations of eq20, eq21 and eq22 are:

$[H_3A]=\frac{[H^+][H_2A^-]}{K_{a1}}\; \; \; \; \; \; \; \; (24a)$

$[H_2A^-]=\frac{[H^+][HA^{2-}]}{K_{a2}}\; \; \; \; \; \; \; \; (24b)$

$[HA^{2-}]=\frac{[H^+][A^{3-}]}{K_{a3}}\; \; \; \; \; \; \; \; (24c)$

Substitute eq24c in eq24b

$[H_2A^-]=\frac{[H^+]^2[A^{3-}]}{K_{a2}K_{a3}}\; \; \; \; \; \; \; \; (25)$

Substitute eq25 in eq24a

$[H_3A]=\frac{[H^+]^3[A^{3-}]}{K_{a1}K_{a2}K_{a3}}\; \; \; \; \; \; \; \; (26)$

Substitute eq26, eq25 and eq24c in eq24 and rearranging,

$[A^{3-}]=\frac{C_aV_aK_{a1}K_{a2}K_{a3}}{(V_a+V_b)([H^+]^3+[H^+]^2K_{a1}+[H^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}\; \; \; \; \; \; \; \; (27)$

Substitute eq23 in eq24a and eq24b

$[H_2A^-]=\frac{[H^+]^2C_aV_aK_{a1}}{(V_a+V_b)([H^+]^3+[H^+]^2K_{a1}+[H^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}\; \; \; \; \; \; \; \; (28)$

$[HA^{2-}]=\frac{[H^+]C_aV_aK_{a1}K_{a2}}{(V_a+V_b)([H^+]^3+[H^+]^2K_{a1}+[H^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}\; \; \; \; \; \; \; \; (29)$

Substitute eq7, eq27, eq28, eq29, K_w= [H⁺][OH^–] and [H⁺] = 10^-pH in eq23,

$10^{-pH}+\frac{C_bV_b}{V_a+V_b}=\frac{K_w}{10^{-pH}}+A+B+C\; \; \; \; \; \; \; \; (30)$

where

$A=\frac{10^{-2pH}C_aV_aK_{a1}}{(V_a+V_b)(10^{-3pH}+10^{-2pH}K_{a1}+10^{-pH}K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}$

$B=\frac{2\times 10^{-pH}C_aV_aK_{a1}K_{a2}}{(V_a+V_b)(10^{-3pH}+10^{-2pH}K_{a1}+10^{-pH}K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}$

$C=\frac{3C_aV_aK_{a1}K_{a2}K_{a3}}{(V_a+V_b)(10^{-3pH}+10^{-2pH}K_{a1}+10^{-pH}K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}$

Eq30 is the complete pH titration curve for a triprotic acid (either strong or weak) versus monoprotic strong base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of H₃PO₄(K_a1= 7.11 x 10^-3, K_a2= 6.00 x 10^-8, K_a3= 4.80 x 10^-13) with 0.100 M of NaOH, we have the following:

The 1^st and 2^nd stoichiometric points are titrated using methyl orange and thymolphthalein as indicators respectively, while the 3^rd stoichiometric point is non-discernible. Another way to titrate phosphoric acid is to first react it with excess cation to completely precipitate the phosphate salt, thereby releasing all H⁺, which is then titrated with a strong base:

$2H_3PO_4(aq)+3Ca^{2+}(aq)\rightarrow Ca_3(PO_4)_2(s)+6H^+(aq)$

Triprotic acid versus monoprotic strong base

next article: Polyprotic weak acid versus monoprotic strong base

Previous article: Diprotic weak acid versus monoprotic strong base

Content page of complete ph titration curve

Content page of advanced chemistry

Main content page

Leave a Reply Cancel reply