The volume of the 1.75 M H₃PO₄ solution required to neutralize 50 ml of 3.5 M Ca(OH)₂ solution is 66.67 mL

Balanced equation

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

• The mole ratio of the acid, H₃PO₄ (nA) = 2
• The mole ratio of the base, Ca(OH)₂ (nB) = 3

How to determine the volume of H₃PO₄

• Molarity of acid, H₃PO₄ (Ma) = 1.75 M
• Volume of base, Ca(OH)₂ (Vb) = 50 mL
• Molarity of base, Ca(OH)₂ (Mb) = 3.5 M

• Volume of acid, H₃PO₄ (Va) =?

MaVa / MbVb = nA / nB

(1.75 × Va) / (3.5 × 50) = 2/3

(1.75 × Va) / 175= 2/3

Cross multiply

1.75 × 3 × Va  = 175 × 2

5.25 × Va = 350

Divide both side by 5.25

Va = 350 / 5.25

Va = 66.67 mL

Thus, the volume of the H₃PO₄ solution needed is 66.67 mL

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