Question

x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M Ba(OH)₂ solution is used when the HCl solution was titrated against it. The numerical value of x is_______ × 10⁻¹. (Nearest integer) Given: Molar mass of HCl and Ba(OH)₂ are 36.5 and 171.0 g mol⁻¹ respectively.}

Hint:

$Ba(OH)_2$ is a diacidic base. Always multiply its molarity by 2 to get the normality (equivalents) because it releases 2 $OH^-$ ions per molecule.

Answer 1

Correct Answer: 1825

Step 1: Understanding the Concept:
In a titration, the equivalents of acid must equal the equivalents of base at the end point.
Step 2: Detailed Explanation:
Equivalents of $Ba(OH)_2 = \text{Molarity} \times \text{Volume (L)} \times n\text{-factor}$ $n\text{-factor for } Ba(OH)_2 = 2$. Equivalents of $Ba(OH)_2 = 0.1 \times 0.025 \times 2 = 0.005 \text{ eq}$. Equivalents of $HCl = \text{Equivalents of } Ba(OH)_2 = 0.005$. Since $n\text{-factor for } HCl = 1$, Moles of $HCl = 0.005 \text{ mol}$.
Step 3: Calculating Mass:
Mass of $HCl = \text{Moles} \times \text{Molar mass}$ Mass $= 0.005 \times 36.5 = 0.1825 \text{ g}$. In mg: $0.1825 \times 1000 = 182.5 \text{ mg}$. To express as $x \times 10^{-1}$: $1825 \times 10^{-1} \text{ mg}$. $x = 1825$.
Step 4: Final Answer:
The value of x is 1825.