# 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.

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

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

_{a}isK

_{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}equationK

_{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}= 0Solve for x using the quadratic equation

x = [-b ± (b

^{²}- 4ac)^{½}]/2ax = [-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 CK

_{a}= 1x10^-7C = 0.1 M

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

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

[H

^{+}] = 0.0001If [H

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

^{+}] = 0.0001, [OH^{-}] = 1x10^-10Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment