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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» <html><body><p></p>Solution :`15 mL` of `1 M H_(2) SO_(4) + 25 mL of 4 M HNO_(3) + 20 mL of <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> M <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a>` <br/> `:. N_(1) V_(1) + N_(2) + V_(2) + N_(3) V_(3) = N_(4) V_(4)``(V_(4) = 1000 mL)` <br/> `15 xx 2 + 25 xx 4 + 20 X = N_(4) xx 1000` <br/> `:. N_(4) = ((130 + 20 X)/(1000))` <br/> mEq of mixture of acid = mEq of `Ba(OH)_(2)` <br/> `Mw of Ba (OH)_(2). 8H_(2) O = 137.4 + 34 + 18 xx 8 = 315.4` <br/> `Ew = (315)/(2) = 157.7 g` <br/> `N of Ba(OH)_(2) .8H_(2) O = (W_(2) xx 1000)/(Ew_(2) xx V_(<a href="https://interviewquestions.tuteehub.com/tag/sol-1216281" style="font-weight:bold;" target="_blank" title="Click to know more about SOL">SOL</a>) ("in" mL))` <br/> `= (4.725 xx 1000)/(157.7 xx 250)` <br/> `0.1198 N ~~ 0.12 N` <br/> mEq of acid mix = mEq of `Ba(OH)_(2)` <br/> `20 xx N_(4) = 26 xx 0.12` <br/> `N_(4) = (26 xx 0.12)/(20) = 0.156 N` <br/> `implies (130 + 20 X)/(1000) = 0.156` <br/> `:. X = (0.156 xx 1000 - 130)/(20) = 1.3` <br/> `N` or `M HCl = 1.3`</body></html>


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