QuestionA student is asked to standardize a solution of calcium hydroxide. He weighs out 0.970 g potassium hydrogen phthalate $\left(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}\right.$, treat this as a monoprotic acid). It requires 35.4 mL of calcium hydroxide to reach the endpoint, A. What is the molarity of the calcium hydroxide solution? $\square$ M
This calcium hydroxide solution is then used to titrate an unknown solution of hydrobromic acid. B. If 27.7 mL of the calcium hydroxide solution is required to neutralize 24.4 mL of hydrobromic acid, what is the molarity of the hydrobromic acid solution? $\square$

Studdy Solution
Use the calcium hydroxide solution to find the molarity of the hydrobromic acid solution.
Volume of $\text{Ca(OH)}_2$ used = 27.7 mL = 0.0277 L
Moles of $\text{Ca(OH)}_2$ used = Molarity $\times$ Volume = $0.134 \, \text{M} \times 0.0277 \, \text{L}$
$$\text{Moles of } \text{Ca(OH)}_2 = 0.00371 \, \text{mol}$$
Since the reaction between $\text{Ca(OH)}_2$ and $\text{HBr}$ is 1:2, moles of $\text{HBr} = 2 \times \text{Moles of } \text{Ca(OH)}_2$
$$\text{Moles of } \text{HBr} = 2 \times 0.00371 \, \text{mol} = 0.00742 \, \text{mol}$$
Volume of $\text{HBr}$ solution = 24.4 mL = 0.0244 L
Molarity of $\text{HBr} = \frac{\text{Moles of } \text{HBr}}{\text{Volume in L}}$
$$\text{Molarity of } \text{HBr} = \frac{0.00742 \, \text{mol}}{0.0244 \, \text{L}}$$
$$\text{Molarity of } \text{HBr} = 0.304 \, \text{M}$$