Question

Q.1 What volume (in mL) of a solution 0.190 M in Ca(OH)2 must we use to neutralize 27.4 mL of a solution 0.363M in H3PO4.

Answer 1

2H3PO4 + 3Ca(OH)2 ----------> Ca3(PO4)2 + 3HO

moles of H3PO4 reacted = 0.363 x 27.4 /1000 = 0.0099

according to balanced reaction

2 moles H3PO4 reacts with 3 moles Ca(OH)2

0.0099 moles H3PO4 reacts with 0.0099 x 3 / 2 = 0.0148

0.0148 = 0.190 X V

V = 0.078 L

V = 78.0 mL

volume of Ca(OH)2 required = 78.0 mL