[4-631] On dissolving 12.214 g of impure \( \mathrm{Ba}(\mathrm{OH})_{2} \) in water and complete the volume to 300 mL . If 25 mL of this solution neutralizes with 40 mL of 0.125 M hydrochloric acid. Calculate the percentage of \( \mathrm{Ba}(\mathrm{OH})_{2} \) in the sample. [ \( \mathrm{Ba}=137,0=16, \mathrm{H}= \) 1, \( \mathrm{Cl}=35.5 \) ] a) \( 44 \% \) b) \( 45 \% \) c) \( 42 \% \)

To tackle this problem, let’s first establish what’s really going on! When barium hydroxide \((\mathrm{Ba(OH)}_2)\) dissolves in water, it releases hydroxide ions \((\mathrm{OH}^-)\). The neutralization reaction with hydrochloric acid \((\mathrm{HCl})\) is essential to calculate how much barium hydroxide is present in your sample. Now, let's run through the steps: when 25 mL of the barium hydroxide solution neutralizes 40 mL of 0.125 M hydrochloric acid, we can calculate moles of HCl first, which equals \(0.125 \, \text{mol/L} \times 0.040 \, \text{L} = 0.005 \, \text{mol}\). Since the reaction between barium hydroxide and hydrochloric acid is: \[ \mathrm{Ba(OH)}_2 + 2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2 + 2 \mathrm{H_2O} \] Two moles of \(\mathrm{HCl}\) react with one mole of \(\mathrm{Ba(OH)}_2\). Thus, the moles of \(\mathrm{Ba(OH)}_2\) that reacted is: \[ 0.005 \, \text{mol} \div 2 = 0.0025 \, \text{mol} \] Given that this corresponds to 25 mL of solution, the concentration is: \[ \frac{0.0025 \, \text{mol}}{0.025 \, \text{L}} = 0.1 \, \text{M} \] Now, let’s consider the total moles of barium hydroxide in the 300 mL solution: \[ \text{Total moles} = 0.1 \, \text{M} \times 0.300 \, \text{L} = 0.03 \, \text{mol} \] Now, let’s bring in the molar mass of barium hydroxide: \[ \mathrm{Ba} = 137.0 \text{ g/mol}, \, \mathrm{O} = 16.0 \text{ g/mol}, \, \mathrm{H} = 1.0 \text{ g/mol} \] \[ \text{Molar mass of } \mathrm{Ba(OH)}_2 = 137.0 + 2(16.0 + 1.0) = 137.0 + 34.0 = 171.0 \, \text{g/mol} \] The mass of barium hydroxide in the solution is: \[ 0.03 \, \text{mol} \times 171.0 \, \text{g/mol} = 5.13 \, \text{g} \] Finally, calculate the percentage of barium hydroxide in the original sample: \[ \text{Percentage} = \left( \frac{5.13 \, \text{g}}{12.214 \, \text{g}} \right) \times 100 \approx 42.0\% \] So, the percentage of \( \mathrm{Ba(OH)}_2 \) in the sample is approximately \( 42\% \). The correct answer is **c) 42%**!