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

STEP 1

1. Potassium hydrogen phthalate (KHP) is a monoprotic acid.
2. The molar mass of KHP is approximately 204.22 g/mol.
3. Calcium hydroxide, $\text{Ca(OH)}_2$ , is a strong base.
4. The reaction between KHP and $\text{Ca(OH)}_2$ is a 1:1 stoichiometry.
5. The reaction between $\text{Ca(OH)}_2$ and hydrobromic acid ( $\text{HBr}$ ) is a 1:2 stoichiometry.
6. We need to find the molarity of the calcium hydroxide solution and the molarity of the hydrobromic acid solution.

STEP 2

1. Calculate the moles of KHP.
2. Determine the molarity of the calcium hydroxide solution.
3. Use the calcium hydroxide solution to find the molarity of the hydrobromic acid solution.

STEP 3

Calculate the moles of KHP.
Given mass of KHP = 0.970 g
Molar mass of KHP = 204.22 g/mol
Moles of KHP = $\frac{0.970 \, \text{g}}{204.22 \, \text{g/mol}}$
$\text{Moles of KHP} = 0.00475 \, \text{mol}$

STEP 4

Determine the molarity of the calcium hydroxide solution.
Since the reaction is 1:1, moles of $\text{Ca(OH)}_2$ = moles of KHP = 0.00475 mol
Volume of $\text{Ca(OH)}_2$ solution = 35.4 mL = 0.0354 L
Molarity of $\text{Ca(OH)}_2$ = $\frac{\text{Moles of } \text{Ca(OH)}_2}{\text{Volume in L}}$
$\text{Molarity of } \text{Ca(OH)}_2 = \frac{0.00475 \, \text{mol}}{0.0354 \, \text{L}}$
$\text{Molarity of } \text{Ca(OH)}_2 = 0.134 \, \text{M}$

STEP 5

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 $\text{HBr} = \frac{\text{Moles of } \text{HBr}}{\text{Volume in L}}$
$\text{Molarity of } \text{HBr} = \frac{0.00742 \, \text{mol}}{0.0244 \, \text{L}}$
$\text{Molarity of } \text{HBr} = 0.304 \, \text{M}$

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