Calculate the [OH^-] of a solution having a [H_3 O^+] = 2.85 times 10^-6 M. A so

Question

Calculate the [OH^-] of a solution having a [H_3 O^+] = 2.85 times 10^-6 M. A solution is tested with a series of indicators. The solution turns blue when bromcresol green is added, it turns purple when bromcresol purple is added, and is colorless when phenolphthalein is added. What is the estimated pH of this solution? How does a pH indicator work? Provide an equilibrium equation to illustrate your answer. A solution is tested with a pH meter and a pH of 3.95 is recorded. Calculate the [H_3 O^+] and the [OH^-] for this solution.

Explanation / Answer

2)

[H3O+] = 2.85 x 10^-6 M

[H3O+][OH-] = Kw

2.85 x 10^-6 x [OH-] = 1.0 x 10^-14

[OH-] = 3.51 x 10^-9 M

5)

pH = 3.95

pH = -log [H3O+]

3.95 = -log [H3O+]

[H3O+] = 1.12 x 10^-4 M

[OH-] = 8.91 x 10^-11 M

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