Answer to Question #21499 in Physical Chemistry 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.
The formula for Ka is
Ka = [H+][B-]/[HB]

where
[H+] = concentration of H+ ions
[B-] = concentration of conjugate base ions
[HB] = concentration of undissociated acid molecules
for a reaction HB &rarr; 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 &middot; x / (C -x)
Ka = x&sup2;/(C - x)
(C - x)Ka = x&sup2;
x&sup2; = CKa - xKa
x&sup2; + Kax - CKa = 0

Solve for x using the quadratic equation

x = [-b &plusmn; (b&sup2; - 4ac)&frac12;]/2a

x = [-Ka + (Ka&sup2; + 4CKa)&frac12;]/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)&sup2; + 4(0.1)(1x10^-7)]&frac12;}/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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