Titration of 37.22 mL of Ca(OH)₂ solution requires 31.19 mL of 0.219 M H₃PO₄ solution according to the following equation: 3 Ca(OH)₂ (aq) + 2 H₃PO₄ (aq) $\rightarrow$Ca₃(PO₄)₂ (s) + 6 H₂O (l) What is the molarity of the base in this reaction?

Question

Quizplus · Accepted Answer

The balanced equation shows that 3 moles of Ca(OH)2 react with 2 moles of H3PO4. Therefore, the number of moles of Ca(OH)2 in 37.22 mL of solution is:$$37.22 	ext{ mL} 	imes \frac{1 	ext{ L}}{1000 	ext{ mL}} 	imes \frac{0.275 	ext{ mol}}{1 	ext{ L}} = 0.0102 	ext{ mol}$$The number of moles of H3PO4 in 31.19 mL of solution is:$$31.19 	ext{ mL} 	imes \frac{1 	ext{ L}}{1000 	ext{ mL}} 	imes \frac{0.219 	ext{ mol}}{1 	ext{ L}} = 0.00682 	ext{ mol}$$Therefore, the molarity of the base is:$$\frac{0.0102 	ext{ mol}}{0.03722 	ext{ L}} = 0.275 	ext{ M}$$

Titration of 37 →\rightarrow→ Ca3(PO4)2 (S) + 6 H2O (L) What Is the Molarity

Titration of 37 $\rightarrow$ Ca₃(PO₄)₂ (S) + 6 H₂O (L)
What Is the Molarity