The volume of 0.025M Ca(OH)2 solution which can neutralise 100 ml of 10–4 M H3PO4 is

3Ca(OH)2vml,0.025M+2H3PO4100ml,10-4M→ Ca3(PO4)2+6H2O V1M1n1=V2M2n2 V×0.0253=100×10-42 V=0.6 mL

The volume of 0.025M Ca(OH)2 solution which can neutralise 100 ml of 10–4 M H3PO4 is

3Ca(OH)2vml,0.025M+2H3PO4100ml,10-4M→ Ca3(PO4)2+6H2O V1M1n1=V2M2n2 V×0.0253=100×10-42 V=0.6 mL

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The volume of 0.025M Ca(OH)₂ solution which can neutralise 100 ml of 10^–4 M H₃PO₄ is

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2.8 ml

0.6 ml

10 ml

60 ml

answer is C.

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Detailed Solution

$\underset{vml, 0.025 M}{3 Ca {(OH)}_{2}} + 2 \underset{100 ml, 10^{- 4} M}{H_{3} {PO}_{4}} \to {Ca}_{3} {({PO}_{4})}_{2} + 6 H_{2} O$

$\frac{V_{1} M_{1}}{n_{1}} = \frac{V_{2} M_{2}}{n_{2}} \frac{V \times 0.025}{3} = \frac{100 \times 10^{- 4}}{2} V = 0.6 mL$

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The volume of 0.025M Ca(OH)2 solution which can neutralise 100 ml of 10–4 M H3PO4 is

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The volume of 0.025M Ca(OH)₂ solution which can neutralise 100 ml of 10^–4 M H₃PO₄ is