[4-631] On dissolving 12.214 g of impure Ba(OH)_{2} in water and complete the volume to 300mL . If 25 mL of this solution neutralizes with 40mL of 0.125 M hydrochloric acid. Calculate the percentage of Ba(OH)_{2} in the sample. [ Ba=137, O=16, H=1, Cl=35.5 ] a) 44 %, b) 45 %, c) 42 %, d) 50%

To find the percentage of \( \mathrm{Ba}(\mathrm{OH})_{2} \) in the sample, we first need to determine how many moles of hydrochloric acid (HCl) are used in the reaction. Given: - Volume of HCl = 40 mL = 0.040 L - Molarity of HCl = 0.125 M Moles of HCl: \[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.040 \times 0.125 = 0.005 \text{ moles} \] The balanced equation for the neutralization reaction is: \[ \mathrm{Ba(OH)}_{2} + 2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_{2} + 2 \mathrm{H}_{2}\mathrm{O} \] From this equation, we see that 1 mole of \( \mathrm{Ba(OH)}_{2} \) reacts with 2 moles of HCl. Therefore, the moles of \( \mathrm{Ba(OH)}_{2} \) reacting with 0.005 moles of HCl is: \[ \text{Moles of } \mathrm{Ba(OH)}_{2} = \frac{0.005}{2} = 0.0025 \text{ moles} \] Now, we need to find the mass of \( \mathrm{Ba(OH)}_{2} \): Molar mass of \( \mathrm{Ba(OH)}_{2} \): - Ba = 137 g/mol - O = 16 g/mol × 2 = 32 g/mol - H = 1 g/mol × 2 = 2 g/mol Total molar mass = 137 + 32 + 2 = 171 g/mol Mass of \( \mathrm{Ba(OH)}_{2} \): \[ \text{Mass} = \text{Moles} \times \text{Molar mass} = 0.0025 \times 171 = 0.4275 \text{ g} \] Now we can calculate the percentage of \( \mathrm{Ba(OH)}_{2} \) in the sample: Given mass of the impure sample = 12.214 g. Percentage: \[ \text{Percentage} = \left( \frac{0.4275}{12.214} \right) \times 100 \approx 3.5 \% \] It appears there was a miscalculation in assumptions; we need to check the liquid volume used. Now, for the calculations of the original 300 mL solution: - Moles of \( \mathrm{Ba(OH)}_{2} \) in 25 mL: Finding its molarity in the original solution: \[ \text{Molarity of solution} = \frac{0.0025 \text{ moles}}{0.025 \text{ L}} = 0.1 \text{ M} \] Then, Moles in 300 mL = \( 0.1 \times 0.300 = 0.03 \text{ moles}\) Mass in original solution: \[ 0.03 \times 171 = 5.13 g \] Percentage in the sample: \[ \left( \frac{5.13 g}{12.214 g} \right) \times 100 \approx 42.05 \% \] Thus, the answer is: **c) 42%**