Question

Determine the pH and the concentrations of all aqueous species (Na+ , H+ , OH– , H3PO4, H2PO4 – , HPO4 2– , and PO4 3– ) in a 0.500 M Na3PO4 solution. (Note that this is a sodium phosphate solution, not a phosphoric acid solution!)

Answer 1

Na3PO4 is completely dissociated to 3 Na+ and PO₄^3- when dissolved in water.

-------------Na3PO4 ------ > 3Na⁺ + PO₄^3-  

---Initial: 0.500 M ----------- 0 ------- 0

Aft.dissn: 0 M -------------- 3*0.500M, 0.500 M

Hence concentration of Na⁺, [Na⁺] = 3*0.500M = 1.50 M (answer)

Now 0.500 M  PO₄^3- will undergo hydrolysis to form  HPO₄^2-(first hydrolysis)

----------------- PO₄^3- + H2O ------ > HPO₄^2- + OH^- , Kb = Kw/Ka3 = 10^-14 / 4.80*10^-13 = 2.08*10^-2

Init.conc(M):0.500

Eqm.con(M):0.500(1 - x), ------------- 0.500x, 0.500x

Kb =  2.08*10^-2 = (0.500x * 0.500x) / 0.500(1 - x)

=> x = 0.184

Hence concentration of PO₄^3-, [PO₄^3-] = 0.500(1 - x) = 0.408 M (answer)

After first hydrolysis, [HPO₄^2-] = [OH^-] = 0.500x = 0.500 * 0.184 M = 0.092 M

Chemical reaction for the hydrolysis of HPO₄^2- is ( second hydrolysis)

----------------- HPO₄^2- + H2O ------ > H₂PO₄^- + OH^- , Kb = Kw/Ka2 = 10^-14 / 6.34*10^-8 = 1.58*10^-7

Init.conc(M):0.092, ------------------------ 0 ----------- 0.092

Eqm.con(M):0.092(1 - y), ------------- 0.092y, 0.092(y+1)

Kb = 1.58*10^-7 = [0.092y * 0.092(y+1)] / 0.092(1 - y)

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

=> y = 1.58*10^-7 / 0.092 = 1.72*10^-6

Hence equilibrium concentration of HPO₄^2-, [HPO₄^2-] = 0.092(1 - y) 0.092 M (answer)

After second hydrolysis, [H₂PO₄^-] = 0.092y = 0.092*1.72*10^-6 = 1.58*10^-7 M

[OH^-] = 0.092(y+1) 0.092 M

Now the chemical reaction for the hydrolysis of H₂PO₄^- is (third hydrolysis)

----------------- H₂PO₄^- + H2O ------ > H₃PO₄ + OH^- , Kb = Kw/Ka1 = 10^-14 / 7.11*10^-3 = 1.41*10^-12

Init.conc(M):1.58*10^-7 , ----------------- 0 ---------- 0.092

Eqm.con(M):1.58*10^-7(1 - z), ------- 1.58*10^-7z, 0.092

Kb = 1.41*10^-12 = [1.58*10^-7z, 0.092] / 1.58*10^-7(1 - z)

(1 - z) are nearly equals to 1. Hence

=> z = 1.41*10^-12 / 0.092 = 1.53*10^-11

Hence equilibrium concentration of H₂PO₄^-, [H₂PO₄^-]= 1.58*10^-7(1 - z) 1.58*10^-7M (answer)

Hence equilibrium concentration of H₃PO₄, [H₃PO₄] = 1.58*10^-7z = 1.58*10^-7*1.53*10^-11 = 2.42*10^-18 M (answer)

[OH^-] 0.092 M (answer)

[H+] = 10^-14 / 0.092 = 1.09*10^-13 M (answer)

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