Question

The volume of 0.025M Ca( OH)2 solution which can neutralize 100 ml of 10-4 H3PO4 is​

Answer 1

✨ Here is your answer
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3 Ca(OH)2 + 2 H3PO4 → Ca3(PO4)2 + 6 H2O

(100 ml) x (0.0001 M H3PO4) x (3 mol Ca(OH)2 / 2 mol H3PO4) / (0.025 M Ca(OH)2) = 0.6 mL
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Answer 2

Answer: 0.6 ml

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $H_3PO_4$ solution = $10^{-4}$

$V_1$ = volume of $H_3PO_4$ solution = 100 ml

$M_2$ = molarity of $Ca(OH)_2$ solution = 0.025 M

$V_2$ = volume of $Ca(OH)_2$ solution = ?

$n_1$ = valency of $H_3PO_4$ = 3

$n_2$ = valency of $Ca(OH__2$ = 2

$3\times 10^{-4}M\times 100=2\times 0.025\times V_2$

$V_2=0.6ml$

Therefore, the volume will be 0.6 ml.