Question

Phosphorous acid, H3PO4(aq) is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the pH, and the concentration of all species in a 0.200M phosphoric acid solution. Please show how to derive the answer. I will rate you. Thank you!

Find: [H3PO4]

[H2PO4^-]

[HPO^2-]

[HPO4^2-]

[PO4^3-]

[H^+]

[OH^-]

pH=

Answer 1

H3PO4 (aq) <---> H2PO4- (aq) + H+ (aq)   , Ka1 = 0.00711

Ka = [H+][H2PO4-]/[H3PO4]         at equilibrium H3PO4 =0.2-X , [H+] = [H2PO4-] = X

0.00711 = ( X) ( X) / ( 0.2-X) ,   hence X^2 +0.00711X - 0.00144 = 0

X = [H2PO4-] = 0.03456 M ,

H2PO4- (aq) <---> HPO42-(aq) + H+(aq) , Ka2 =6.34 x 10^ -8

6.34 x 10^ -8 = [H+] [HPO42-]/[H2PO4-]

6.34 x 10^ -8 = X^2 / ( 0.0.3456-X)

X = 4.68 x 10^-5 = [HPO42-]

now HPO42- <---> PO43- (aq) + H+ (aq) , Ka3 = 4.8 x 10^ -13

4.8 x 10^ -13 = ( X^2) / ( 4.68 x 10^ -5 -X)

X = 4.74 x 10^ -8 M = [PO43-]

Hence [H3PO4] = 0.2-0.03456 = 0.1655 M

[H2PO4-] = 0.03456 - 4.68x10^-5 = 0.0345 M

[PO43-] = 4.74 x 10^ -8 M ,

[H+] = 0.03456+ 4.68 x 10^-5 + 4.74 x 10^ -8 = 0.0346 M

pH = -log ( 0.0346) = 1.46

[OH-] = 10^ -14 /[H+] = 10^ -14/0.0346 = 2.89 x 10^ -13 M