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  • P_{4}O_{6} reacts with water according to equation P_{4}O_{6}+6H_{2}O\rightarrow 4H_{3}PO_{3}. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P_{4}O_{6} in H_{2}O.

P_{4}O_{6}  reacts with water according to equation P_{4}O_{6}+6H_{2}O\rightarrow 4H_{3}PO_{3}.

Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of  P_{4}O_{6} in H_{2}O.

Answers (1)

                                                         P_{4}O_{6}+6H_{2}O\rightarrow 4H_{3}PO_{3}

4H_{3}PO_{3}+8NaOH\rightarrow 4Na_{2}HPO_{3}+8H_{2}O (H_{3}PO_{3} can be neutralised with NaOH )

                             P_{4}O_{6}(1 \; mol)+8NaOH (8\; mol)\rightarrow 4Na_{2}HPO_{3}+2H_{2}O

                                               Moles of P_{4}O_{6} = \frac{1.1}{220}=0.005\; mol

Acid formed by 1 mol of P_{4}O_{6} Require 8 mol NaOH

Acid formed by 0.005 mol of P_{4}O_{6} required =8\times 0.005=0.04\; mol

0.1 M NaOH means 0.1 mol of NaOH is present in 1000 mL solution

0.4 M of NaOH is present in solution =\frac{1000}{0.1}\times 0.04=400mL

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