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suter [353]
3 years ago
8

Calculate the concentration of OH- in a solution that contains 3.9 x 10-4 M H3O+ at 25°C. Identify the solution as acidic, basic

or neutral. Calculate the concentration of OH- in a solution that contains 3.9 x 10-4 M H3O+ at 25°C. Identify the solution as acidic, basic or neutral. 3.9 × 10-4 M, neutral 2.7 × 10-2 M, basic 2.6 × 10-11 M, basic 2.7 × 10-2 M, acidic 2.6 × 10-11 M, acidic
Chemistry
1 answer:
aalyn [17]3 years ago
8 0

Answer:

[OH⁻] = 2,6x10⁻¹¹

Acidic

Explanation:

The kw in water is:

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

kw = [OH⁻] [H₃O⁺] = 1,00x10⁻¹⁴

If concentracion of H₃O⁺ is 3,9x10⁻⁴M:

[OH⁻] [3,9x10⁻⁴M] = 1,00x10⁻¹⁴

<em>[OH⁻] = 2,6x10⁻¹¹</em>

pH is defined as - log[H₃O⁺]. If pH>7,0 the solution is basic, if pH<7,0 solution is acidic, if pH=7,0 solution is neutral.

In this problem,

pH = - log [3,9x10⁻⁴M] = <em>3,4</em>

As pH is < 7.0, the solution is <em>acidic</em>

<em></em>

I hope it helps!

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<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

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\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

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\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

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[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

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