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Answer on Physical Chemistry Question for Celebilover

Question #21499
What are the [H+] and [OH-] in a 0.1moldm^-3 solution of a weak acid (Ka = 1x10^-7)?

a) [H+] = 0.1, [OH-] = 1x10^-13
b) [H+] = 0.001, [OH-] = 1x10^-11
c) [H+] = 0.0001, [OH-] = 1x10^-10
d) [H+] = 1x10^-6, [OH-] = 1x10^-8

The answer is C but I want to know why.
Expert's answer
The formula for Ka is
Ka = [H+][B-]/[HB]

[H+] = concentration of H+ ions
[B-] = concentration of conjugate base ions
[HB] = concentration of undissociated acid molecules
for a reaction HB → H+ + B-

Let x represent the concentration of H+ that dissociates from HB, then [HB] = C - x where C is the initial concentration.

Enter these values into the Ka equation

Ka = x · x / (C -x)
Ka = x²/(C - x)
(C - x)Ka = x²
x² = CKa - xKa
x² + Kax - CKa = 0

Solve for x using the quadratic equation

x = [-b ± (b² - 4ac)½]/2a

x = [-Ka + (Ka² + 4CKa)½]/2

**Note** Technically, there are two solutions for x. Since x represents a concentration of ions in solution, the value for x cannot be negative.

Enter values for Ka and C

Ka = 1x10^-7
C = 0.1 M

x = {-1x10^-7 + [(1x10^-7)² + 4(0.1)(1x10^-7)]½}/2
x = 0.0001

[H+] = 0.0001

If [H+] = 0.0001, [OH-] = 1x10^-10

c) [H+] = 0.0001, [OH-] = 1x10^-10

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