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P_(4)O_(6) reacts with water according to equation, P_(4)O_(6) + 6 H_(2)O rarr 4 H_(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. |
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Answer» <P> Solution :The chemical equations for the reactions involved are :`{:(""P_(4)O_(6) + 6H_(2)O rarr 4 H_(3)PO_(4)""("Hydrolysis reaction")),(ul(H_(3)PO_(3) + 2 NaOH rarr [4 Na_(2)HPO_(3)+ 2H_(2)O]xx 4) " "("Neutralisation reaction")),(" "underset("= 220 g")underset(4 xx 31 + 6 xx 16)(P_(4)O_(6))+underset("= 320 g")underset(8 xx 40)(8 NaOH) rarr 4 Na_(2)HPO_(3)+2H_(2)O""("Overall reaction")):}` Now 220 g of `P_(4)O_(6)` require NaOH for neutralization = 320 g `therefore` 1.1 g of `P_(4)O_(6)` will require NaOH `= (320)/(220) xx 1.1 = 1.6 g` Now 1000 ML of 0.1 MHCl contain NaOH `= 40 xx 0.1 = 4 g` In other words, 4 g of NaOH are present in 0.1 M NaOH = 1000 mL `therefore` 1.6 g of NaOH will be present in 0.1 M NaOH `=(1000)/(4) xx 1.6 =400 mL = 0.4 L` |
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