Question

The volume of 0.025M $H_3{PO}_4$ require to neutralise 25ml of 0.03M $Ca{\left(OH\right)}_2$ is :

• A. $40ml$
• B. $20ml$
• C. $10ml$
• D. $80ml$

Answer 1

The correct option is B $20ml$

Assuming total dissociation and using stoichiometry,

$Equivalent{s}_{H_{3}P{O}_4}$ = $Equivalent{s}_{Ca(OH{)}_2}$

$N_1{V}_1=N_2{V}_2$

$3\times 0.025\times v=2\times 0.03\times 25$

$\Rightarrow v=20ml$

Option B is correct.