For the diprotic weak acid h2a, ka1 = 4.0 * 10-6 and ka2 = 7.2 * 10-9. what is the ph of a 0.0450 m solution of h2a? what are the equilibrium concentrations of h2a and a2 - in this solution?

Question

Chemistry

Alejandra Hester

29 October, 12:03

For the diprotic weak acid h2a, ka1 = 4.0 * 10-6 and ka2 = 7.2 * 10-9. what is the ph of a 0.0450 m solution of h2a? what are the equilibrium concentrations of h2a and a2 - in this solution?

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Answers (2)

Know the Answer?

Marina Oconnell · Answer 1

Marina Oconnell 29 October, 14:06

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That's right how you did it

Woofie · Answer 2

For the first reaction:

the ICE table:

H2A ↔ HA - + H+

initial 0.045 m 0 0

change - X + X + X

Equ (0.045-X) X X

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

4x10^-6 = X^2 / (0.045-X) by solving for X

∴X = 4.2x10^-4

∴[ H2a] = 0.045 - (4.2x10^-4) = 0.045 m

[HA-] = [H+] = 4.2x10^-4

when PH = - ㏒[H+]

by substitution:

PH = - ㏒ (4.2x10^-4)

= 3.377

For the second reaction:

by using ICE table:

HA - ↔ A^2 - + H+

when

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

by substitution

7.2 x 10^-9 = [A^2-] * (4.2x10^-4) / (4.2x10^-4)

∴[A^2-] = (7.2x10^-9) * (4.2x10^-4) / (4.2x10^-4)

∴ [A^2-] = 7.2x10^-9 m