Question

if 18.82 ml of a standard 0.1436 m naoh solution is required to neutralize 36.31 ml of \\(\text{h}_2\text{so}_4\\), what is the molarity of the acid solution? be sure your answer has the correct number of significant figures.

Explanation:

Write balanced neutralization equation

$$ 2\text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) ightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l) $$

Calculate moles of sodium hydroxide

$$ n_{\text{NaOH}} = 0.01882\text{ L} \times 0.1436\text{ M} = 2.702552 \times 10^{-3}\text{ mol} $$

Calculate molarity of sulfuric acid

$$ n_{\text{H}_2\text{SO}_4} = \frac{1}{2} \times n_{\text{NaOH}} = 1.351276 \times 10^{-3}\text{ mol} $$

$$ M_{\text{H}_2\text{SO}_4} = \frac{1.351276 \times 10^{-3}\text{ mol}}{0.03631\text{ L}} \approx 0.037215 \text{ M} $$

Answer:

If \(18.82\text{ mL}\) of a standard \(0.1436\text{ M}\) \(\text{NaOH}\) solution is required to neutralize \(36.31\text{ mL}\) of \(\text{H}_2\text{SO}_4\), what is the molarity of the acid solution? Be sure your answer has the correct number of significant figures.

<blank>0.03722</blank> M