Question

what volume of 0.100 m sulfuric acid could be neutralized by 40.00 ml of 0.100 m potassium hydroxide solution?

Explanation:

Step1: Write neutralization reaction

$$\text{H}_2\text{SO}_4 + 2\text{KOH} ightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$$

Mole ratio: $\frac{n(\text{H}_2\text{SO}_4)}{n(\text{KOH})} = \frac{1}{2}$

Step2: Calculate moles of KOH

Moles = Molarity × Volume (in L)
$n(\text{KOH}) = 0.100\ \text{mol/L} \times 0.04000\ \text{L} = 0.004000\ \text{mol}$

Step3: Find moles of $\text{H}_2\text{SO}_4$

From mole ratio: $n(\text{H}_2\text{SO}_4) = \frac{1}{2}n(\text{KOH})$
$n(\text{H}_2\text{SO}_4) = \frac{1}{2} \times 0.004000\ \text{mol} = 0.002000\ \text{mol}$

Step4: Calculate volume of $\text{H}_2\text{SO}_4$

Volume = $\frac{\text{Moles}}{\text{Molarity}}$
$V(\text{H}_2\text{SO}_4) = \frac{0.002000\ \text{mol}}{0.100\ \text{mol/L}} = 0.02000\ \text{L}$
Convert to mL: $0.02000\ \text{L} \times 1000 = 20.00\ \text{mL}$

Answer:

20.00 mL