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 w

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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)…

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Given: pure HCl solution is titrated with $25.0 \, \text{mL}$ of $0.1 \, \text{M}$ Ba(OH) $_2$ . Molar masses are $36.5 \, \text{g mol}^{-1}$ for HCl and $171.0 \, \text{g mol}^{-1}$ for Ba(OH) $_2$ .

Find: the numerical value of $x$ in $x \times 10^{-1}$ mg.

At the end point of titration, equivalents of acid equal equivalents of base.

For Ba(OH) $_2$ , the $n$ -factor is $2$ .

\text{Equivalents of } Ba(OH)_2 = \text{Molarity} \times \text{Volume (L)} \times n\text{-factor}

= 0.1 \times 0.025 \times 2 = 0.005 \text{ eq}

Therefore, equivalents of HCl are also $0.005$ .

Since HCl has $n$ -factor $1$ , moles of HCl are $0.005 \, \text{mol}$ .

\text{Mass of HCl} = \text{Moles} \times \text{Molar mass}

= 0.005 \times 36.5 = 0.1825 \text{ g}

Converting to mg:

0.1825 \times 1000 = 182.5 \text{ mg}

Now,

182.5 \text{ mg} = 1825 \times 10^{-1} \text{ mg}

Hence, $x = 1825$ .

Therefore, the value of $x$ is 1825.

Given: HCl is neutralized by Ba(OH)2. Find: the integer value of x. The key idea is that Ba(OH)2 is a diacidic base, so each mole provides 2 mol of OH^-. Ba(OH)2 Ba^2+ + 2OH^- So the equivalent amount of base used is 0.1 0.025 2 = 0.005 This equals the equivalent amount of HCl neutralized. HCl H^+ + Cl^- For HCl, one mole gives one equivalent, so moles of HCl = 0.005 Mass of HCl: 0.005 36.5 = 0.1825 g In milligrams: 0.1825 g = 182.5 mg Matching with the form x 10^-1 mg, 182.5 = x 10^-1 x = 1825 Therefore, the value of x is 1825.

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