Question

`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 1

`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`