of a mixture containing Sodium hydroxide () and Sodium sulphate () is completely neutralized by the — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
$100\text{ mL}$ of a mixture containing Sodium hydroxide ($NaOH$) and Sodium sulphate ($Na_2SO_4$) is completely neutralized by the addition of exactly $10\text{ mL}$ of $0.5\text{ M }$ Sulphuric acid ($H_2SO_4$). Determine the exact mass of $NaOH$ originally present in the $100\text{ mL}$ solution.
💡 Solution & Explanation
Step 1: Identify the reactive components. $Na_2SO_4$ is a neutral salt and will not react with $H_2SO_4$. Only the $NaOH$ undergoes neutralization. Step 2: Apply the Law of Equivalence. Equivalents of Base = Equivalents of Acid. Equivalents of $NaOH = \frac{\text{Mass}}{\text{Equivalent Weight}} = \frac{W}{40}$. Equivalents of $H_2SO_4 = \text{Normality} \times \text{Volume in Litres}$. Normality of $H_2SO_4 = M \times \text{n-factor} = 0.5 \times 2 = 1.0\text{ N}$. Step 3: Solve for mass ($W$). $\frac{W}{40} = 1.0 \times \left(\frac{10}{1000}\right) = 0.01\text{ equivalents}$. $W = 0.01 \times 40 = 0.4\text{ g}$.