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 K_{a} is

K_{a} = [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 K_{a} equation

K_{a} = x · x / (C -x)

K_{a} = x^{²}/(C - x)

(C - x)K_{a} = x^{²}

x^{²} = CK_{a} - xK_{a}

x² + K_{a}x - CK_{a} = 0

Solve for x using the quadratic equation

x = [-b ± (b^{²} - 4ac)^{½}]/2a

x = [-K_{a} + (K_{a}^{²} + 4CK_{a})^{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 K_{a} and C

K_{a} = 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

K

where

[H

[B

[HB] = concentration of undissociated acid molecules

for a reaction HB → H

Let x represent the concentration of H

Enter these values into the K

K

K

(C - x)K

x

x² + K

Solve for x using the quadratic equation

x = [-b ± (b

x = [-K

**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 K

K

C = 0.1 M

x = {-1x10^-7 + [(1x10^-7)

x = 0.0001

[H

If [H

c) [H

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