300 mL of deci-molar solution of Ba(OH)_{2(a | vedclass

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Question

(B) Step $1$: Calculate the initial millimoles of $Ba(OH)_2$.$Molarity = 0.1 \ M$,$Volume = 300 \ mL$.$Millimoles \ of \ Ba(OH)_2 = 0.1 	imes 300 = 3

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Step $1$: Calculate the initial millimoles of $Ba(OH)_2$.$Molarity = 0.1 \ M$,$Volume = 300 \ mL$.$Millimoles \ of \ Ba(OH)_2 = 0.1 	imes 300 = 30 \ mmol$.$Milliequivalents \ of \ Ba(OH)_2 = 30 	imes 2 = 60 \ meq$.Step $2$: Calculate the milliequivalents of $HCl$ added.$Molarity = 0.2 \ M$,$Volume = 100 \ mL$.$Milliequivalents \ of \ HCl = 0.2 	imes 100 	imes 1 = 20 \ meq$.Step $3$: Calculate remaining $Ba(OH)_2$ after reaction with $HCl$.$Remaining \ meq \ of \ Ba(OH)_2 = 60 - 20 = 40 \ meq$.Step $4$: Titrate remaining $Ba(OH)_2$ with $H_2SO_4$.$Normality \ of \ H_2SO_4 = 0.5 \ M 	imes 2 = 1 \ N$.$Equivalence \ point: meq \ of \ Ba(OH)_2 = meq \ of \ H_2SO_4$.$40 = 1 	imes V$.$V = 40 \ mL$.

$300 \ mL$ of deci-molar solution of $Ba(OH)_{2(aq)}$ was titrated with $100 \ mL$ of $0.2 \ M \ HCl_{(aq)}$ solution. The final solution had to be titrated with $V \ mL$ of semi-molar $H_2SO_{4(aq)}$ solution for end-point. Correct option is $V = ....... \ mL$

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