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.
1
Expert's answer
2013-01-28T10:20:57-0500
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 → 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
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