QuestionMay 8, 2025

3. If the pH of a calcium hydroxide solution is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are in the solution?

3. If the pH of a calcium hydroxide solution is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are in the solution?
3. If the pH of a calcium hydroxide solution is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are in the solution?

Solution
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Answer

0.413 \, \text{g} Explanation 1. Calculate the hydroxide ion concentration [OH^-] Use pOH = 14 - pH. Then, [OH^-] = 10^{-pOH}. pOH = 14 - 11.64 = 2.36 [OH^-] = 10^{-2.36} \approx 4.37 \times 10^{-3} \, \text{M} 2. Relate [OH^-] to calcium hydroxide concentration Each Ca(OH)_2 dissociates into 2 OH^- ions. Thus, [Ca(OH)_2] = \frac OH^- {2}. [Ca(OH)_2] = \frac{4.37 \times 10^{-3}}{2} \approx 2.185 \times 10^{-3} \, \text{M} 3. Calculate moles of Ca(OH)_2 Moles = Molarity \times Volume (in liters). \text{Moles of } Ca(OH)_2 = 2.185 \times 10^{-3} \times 2.55 \approx 5.57 \times 10^{-3} \, \text{moles} 4. Convert moles to grams Mass = Moles \times Molar Mass. The molar mass of Ca(OH)_2 is 40.08 + 2(16.00 + 1.01) = 74.10 \, \text{g/mol}. Mass = 5.57 \times 10^{-3} \times 74.10 \approx 0.413 \, \text{g}

Explanation

1. Calculate the hydroxide ion concentration [$OH^-$]<br /> Use $pOH = 14 - pH$. Then, $[OH^-] = 10^{-pOH}$.<br />$pOH = 14 - 11.64 = 2.36$ <br />$[OH^-] = 10^{-2.36} \approx 4.37 \times 10^{-3} \, \text{M}$ <br /><br />2. Relate $[OH^-]$ to calcium hydroxide concentration<br /> Each $Ca(OH)_2$ dissociates into 2 $OH^-$ ions. Thus, $[Ca(OH)_2] = \frac{[OH^-]}{2}$.<br />$[Ca(OH)_2] = \frac{4.37 \times 10^{-3}}{2} \approx 2.185 \times 10^{-3} \, \text{M}$ <br /><br />3. Calculate moles of $Ca(OH)_2$<br /> Moles = Molarity $\times$ Volume (in liters).<br />$\text{Moles of } Ca(OH)_2 = 2.185 \times 10^{-3} \times 2.55 \approx 5.57 \times 10^{-3} \, \text{moles}$ <br /><br />4. Convert moles to grams<br /> Mass = Moles $\times$ Molar Mass. The molar mass of $Ca(OH)_2$ is $40.08 + 2(16.00 + 1.01) = 74.10 \, \text{g/mol}$.<br />Mass = $5.57 \times 10^{-3} \times 74.10 \approx 0.413 \, \text{g}$
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