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2 years ago

What is the pH of a solution in which [H 3O +] = 3.8 × 10 -8 M?

Chemistry

1 answer:

7nadin3 [17]2 years ago

4 0

Answer:

<h3>The answer is 7.42 </h3>

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ { H_3O}^{+}]$

From the question we have

$pH = - log(3.8 \times {10}^{ - 8} ) \\ = 7.420216...$

We have the final answer as

Hope this helps you

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Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um

Andre45 [30]

Answer:

$\large \boxed{\text{0.0038 mol/L}}$

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1. Calculate the initial moles of acid and base

$\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}$

2. Calculate the moles remaining after the reaction

OH⁻ + H₃O⁺ ⟶ 2H₂O

I/mol: 0.0053 0.005 00

C/mol: -0.00500 -0.005 00

E/mol: 0.0003 0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

$\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}$