PLEASE only answer with COMPLETE, QUALITY answers. Thanks! Given a diprotic acid

Question

PLEASE only answer with COMPLETE, QUALITY answers. Thanks!

Given a diprotic acid, H2A, with two ionization constants of Ka2=5.56 times 10-11,calculate the PH and molar concentrations of H2A, HA-, and A2-, and A-2 for each of the solution below. the solution below. A0.176M solution of H2A 0.176 M solution of Na2A For part (a), treat H2A as a monoprotic weak acid with ka=ka1. For part (b), note that HA- is the intermediate from of the diprotic acid. For part (c), Treat A2- as a monobasic base with Kb=kb1=kwlka2.

Explanation / Answer

a) H2A <-----> HA- + H+

Ka1 = [HA-][H+]/[H2A] ,

2.35 x10^-4 = x^2/(0.176-x)   , x^2 +2.35x10^-4x -4.136 x10^-5 =0

solving we get x=[HA-]=[H+]= 0.0063

now HA- <-----> H+ + A2- ,

Ka2 = 5.56 x10^-11 = x^2/(0.0063-x)

x =[H+]=[A2-] = 5.9 x10^-7 ,

hence now [A2-] =5.9 x10^-7 M ,

total [H+] = 5.9x10^-7 + 0.0063 = 0.0063 M,

[H2A]= 0.176-0.0063 = 0.1697 M

pH = -log(0.0063) = 2.2

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