Question

The volume of $0.025M{H}_3{PO}_4$ required to complete neutralise $25ml$ of $0.03M{Ca\left(OH\right)}_2$ is:

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

Answer 1

Answer: (B)

$20ml$

Solution:- (B) 20 mL

$2H_{3}P{O}_4+3Ca{\left(OH\right)}_2⟶{Ca}_3{\left(P{O}_4\right)}_2+6H_{2}O$

2 moles of $H_{3}P{O}_4$ reacts with 3 moles of $Ca{\left(OH\right)}_2$

As we know that,

No. of moles $=V_{\left(\text{in L}\right)}\times M$

$\Rightarrow V=\frac{\text{no. of moles}}{M}$

whereas,

M = Molarity

V = Volume

Given that:-

Molarity of $Ca{\left(OH\right)}_2=0.03M$

Volume of $Ca{\left(OH\right)}_2=25mL=2.5\times {10}^{-2}L$

$\therefore$ No. of moles of $Ca{\left(OH\right)}_2=2.5\times {10}^{-2}\times 0.03=7.5\times {10}^{-4}\text{moles}$

No. of moles of $H_{3}P{O}_4$ required to react with 3 moles of $CA{\left(OH\right)}_2=2$

No. of moles of $H_{3}P{O}_4$ required to react with $7.5\times {10}^{-4}$ moles of $CA{\left(OH\right)}_2=\frac{2}{3}\times 7.5\times {10}^{-4}=5.0\times {10}^{-4}$

As no. of moles of $H_{3}P{O}_4$ required is $5.0\times {10}^{-4}$.

Given that molarity of $H_{3}P{O}_4=0.025M$

$\therefore V=\frac{5\times {10}^{-4}}{0.025}=0.02L=20mL$

Hence, 20 mL volume of 0.025M $H_{3}P{O}_4$ required to completely neutralise 25ml of 0.03M $Ca{\left(OH\right)}_2$.