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1 year ago

Chemistry

College

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

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

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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The

volumeof the 1.75 MH₃PO₄ solutionrequired to neutralize 50 ml of 3.5 M Ca(OH)₂ solution is66.67 mL## Balanced equation

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

mole ratioof the acid,H₃PO₄ (nA) = 2mole ratioof the base,Ca(OH)₂ (nB) = 3## How to determine the volume of H₃PO₄

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 mLThus, the

volume of the H₃PO₄solution needed is66.67 mLLearn more about

titration:https://brainacademy.pro/question/14356286