Question #56324

a)calculate the [OH-] of a solution having a [H3O+]=2.85x10^-6 M

b)how does a pH indicator work? Provide an equilibrium equation to illustrate your answer.

c)a solution is tested with a pH of 3.95 is recorded. Calculate the [H3O+] and the [OH-] for this solution

d)a solution is tested with a series of indicators the solution turns blue when bromcresol green is added, it turns pruple when bromcresol purple is added, and is colorless when phenolphthalein is added. What is the estimated pH of this solution?
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Expert's answer

2015-11-16T09:52:38-0500

Answer on Question #56324 - Chemistry - General Chemistry

Question:

a) calculate the [OH][\mathrm{OH} - ] of a solution having a [H3O+]=2.85×106M[\mathrm{H}_3\mathrm{O} + ] = 2.85\times 10^{\wedge} - 6\mathrm{M}

b) how does a pH indicator work? Provide an equilibrium equation to illustrate your answer.

c) a solution is tested with a pH of 3.95 is recorded. Calculate the [H3O+][\mathrm{H}_3\mathrm{O} + ] and the [OH-] for this solution

d) 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?

Answers:

a)

Kw=[H3O+][OH]\mathrm{K_w = [H_3O^+]*[OH^-]} from this, [OH]=Kw/[H3O+]=1.00×1014/2.85×106=3.5×109M[\mathrm{OH}^{-}] = \mathrm{K_{w} / [H_{3}O^{+}] = 1.00\times 10^{\wedge} - 14 / 2.85\times 10^{\wedge} - 6 = 3.5\times 10^{\wedge} - 9M}

[OH]=3.5×109M\mathrm{[OH^{-}] = 3.5\times 10^{\wedge} - 9M}

b)

pH indicator is a chemical compound which reacts with hydronium ions (H3O+)(\mathrm{H}_3\mathrm{O}+) or hydrogen ions (H+)(\mathrm{H}+) , and visually changes its colour, due to the different ionic forms present in the solution.



c)

pH=log[H3O+]\mathrm{pH} = -\log [\mathrm{H}_3\mathrm{O}^+ ] from this, [H3O+]=1.122×104M[\mathrm{H}_3\mathrm{O}^+ ] = 1.122\times 10^{\wedge} - 4\mathrm{M}

[OH]=Kw/[H3O+]=1.00×1014/1.122×104=9x11M\mathrm{[OH^{-}] = K_w / [H_3O^+] = 1.00\times 10^{\wedge} - 14 / 1.122\times 10^{\wedge} - 4 = 9x^{\wedge} - 11M}

d)

The solution is basic. Because all the stated indicators can give the corresponding colours only in the basic solutions (pH from 5.4 to 8.2).

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