15 mL 1 N H_(2) SO_(4), 25 mL of 4 N HNO_(3), and 20 mL of X M HCl were mixed and made up to 1000 mL. Prepared by dissolving 4.725 g of pure Ba(OH)_(2). 8H_(2) O in water made up to 0.25 litre. What is the molarity of HCl solution (i.e. find X)

15 mL 1 N H_(2) SO_(4), 25 mL of 4 N HNO_(3), and 20 mL of X M HCl wer

1.	15 mL 1 N H_(2) SO_(4), 25 mL of 4 N HNO_(3), and 20 mL of X M HCl were mixed and made up to 1000 mL. Prepared by dissolving 4.725 g of pure Ba(OH)_(2). 8H_(2) O in water made up to 0.25 litre. What is the molarity of HCl solution (i.e. find X)
Answer» Solution :`15 mL` of `1 M H_(2) SO_(4) + 25 mL of 4 M HNO_(3) + 20 mL of X M HCL` `:. N_(1) V_(1) + N_(2) + V_(2) + N_(3) V_(3) = N_(4) V_(4)``(V_(4) = 1000 mL)` `15 xx 2 + 25 xx 4 + 20 X = N_(4) xx 1000` `:. N_(4) = ((130 + 20 X)/(1000))` mEq of mixture of acid = mEq of `Ba(OH)_(2)` `Mw of Ba (OH)_(2). 8H_(2) O = 137.4 + 34 + 18 xx 8 = 315.4` `Ew = (315)/(2) = 157.7 g` `N of Ba(OH)_(2) .8H_(2) O = (W_(2) xx 1000)/(Ew_(2) xx V_(SOL) ("in" mL))` `= (4.725 xx 1000)/(157.7 xx 250)` `0.1198 N ~~ 0.12 N` mEq of acid mix = mEq of `Ba(OH)_(2)` `20 xx N_(4) = 26 xx 0.12` `N_(4) = (26 xx 0.12)/(20) = 0.156 N` `implies (130 + 20 X)/(1000) = 0.156` `:. X = (0.156 xx 1000 - 130)/(20) = 1.3` `N` or `M HCl = 1.3`