25 ml of "4 N HNO"_(3), 15 ml of "1 MH"_(2)SO_(4) and 20 ml of HCl of unknown concentration are mixed and made upto 1 litre. 20 ml of this solution required 26 ml of Ba(OH)_(2) solution. The Ba(OH)_(2) solution is prepared by dissolving 4.725 g of Ba(OH)_(2) per 250 ml of pure water. Calculate the normality of HCl solution. [Hint : Take hydrated form of Barium Hydroxide, i.e. Ba(OH)_(2)·8 H_(2)O]

25 ml of "4 N HNO"_(3), 15 ml of "1 MH"_(2)SO_(4) and 20 ml of HCl of

1.	25 ml of "4 N HNO"_(3), 15 ml of "1 MH"_(2)SO_(4) and 20 ml of HCl of unknown concentration are mixed and made upto 1 litre. 20 ml of this solution required 26 ml of Ba(OH)_(2) solution. The Ba(OH)_(2) solution is prepared by dissolving 4.725 g of Ba(OH)_(2) per 250 ml of pure water. Calculate the normality of HCl solution. [Hint : Take hydrated form of Barium Hydroxide, i.e. Ba(OH)_(2)·8 H_(2)O]
Answer» Solution :Equivalent WEIGHT of `BA(OH)_(2).8H_(2)O=(315)/(2)=157.5` Normality of `Ba(OH)_(2).8H_(2)O` solution `=(4.725)/(157.5xx0.25)=0.1198` EQUIVALENTS of `Ba(OH)_(2).8H_(2)O` solution used `=(26xx0.1198)/(1000)` NUMBER of equivalents of acid `=20xx((25xx4)+(15xx2)+(20xxa))/(1000xx1000)=20xx(130+20a)/(1000)` Number of equivalents of `Ba(OH)_(2)·8H_(2)O` solution = Number of equivalents of acid `20xx(130+20a)/(1000)=(26xx0.1198)/(1000), a=1.287N`