A diprotic acid, H2A, has acid dissodation constants of Kr = 3.48x 10-4 and Ka2

Question

A diprotic acid, H2A, has acid dissodation constants of Kr = 3.48x 10-4 and Ka2 64x 10-12 Calculate! Map pH and molar concentrations of H2A, HAT, and A2 at equilibrium for each of the solutions below. (a) a 0.171 M solution of H2A HA-] pH Number 2.113 Number Number 0.163 7.719 x 103 M 2.64 x 10-12M Incorrec (b) a 0.171 M solution of NaHA See the lower panel for more information on how to solve part (a). pH A- Number Number er Number (c) a 0.171 M solution of Na2A PH H,A HA Nur Number Number Number

Explanation / Answer

H2A <------> HA- + H+;..................Ka1 = 3.48*10^-4
HA- <------> A(^2-) + H+;...............Ka2 = 2.64*10^-12

For the concentrations of H+ and HA- you consider only the first reaction since the HA- product will only slightly dissociate (Ka2 is very small).

If x moles of H2A dissociate, then x moles of HA- and H+ are produced, and the concentration of H2A is reduced by x:

Answer (a) 0.171 M Solution of H2A

[H2A] = 0.171 - x
[H+] = x
[HA-] = x

Ka1 = [H+]*[HA-]/[H2A] = x²/(0.0.171 - x)

3.48*10^-4*(0.171 - x) = x²
x² + 3.48*10^-4*x - 0.171*3.48*10^-4 = 0

x² + 3.48*10^-4*x - 0.0000595

[H+] = x
pH = -log(x) = - log(0.0242)=1.61
[H2A] = 0.171 – 0.0242 = 0.1468 M

For the second reaction,

Ka2 = [A(^2-)]*[H+]/[HA-]

= [A(^2-)]*x/x

[A(^2-)]* = 2.64*10^-4

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