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25ml of a solution of $Na_2CO_3$ having a specific gravity of $1.25ml^{-1}$ required 32.9ml of a solution of HCl containing 109.5g of the acid per litre for complete neutralization .Calculate the volume of 0.84N $H_2SO_4$ that will be completely neutralized by 125g of $Na_2CO_3$ solution.


1 Answer

$N_{HCl}=\large\frac{109.5}{36.5\times 1}$$=3$
$Na_2CO_3$ is completely neutralized by HCl
Meq of $Na_2CO_3$=Meq of $HCl$
$N\times 25=32.9\times 3$
Weight of $Na_2CO_3$ solution=125g
Volume of $Na_2CO_3$ solution =$\large\frac{125}{1.25}$$=100ml$
$\therefore$ Meq of $H_2SO_4$=Meq of $Na_2CO_3$
$0.84\times V=100\times 3.948$
Volume of $H_2SO_4$ required =470ml
Hence (b) is the correct answer.
answered Jan 7, 2014 by sreemathi.v

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