The normality of H_{2}SO_{4} in the solution | vedclass

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(A) The reaction is: $H_{2}SO_{4} + 2NaOH ightarrow Na_{2}SO_{4} + 2H_{2}O$. Number of milliequivalents of $H_{2}SO_{4} = 	ext{Molarity} 	imes 	e

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The reaction is: $H_{2}SO_{4} + 2NaOH ightarrow Na_{2}SO_{4} + 2H_{2}O$. Number of milliequivalents of $H_{2}SO_{4} = 	ext{Molarity} 	imes 	ext{n-factor} 	imes 	ext{Volume (mL)} = 0.1 	imes 2 	imes 100 = 20 \ mEq$. Number of milliequivalents of $NaOH = 0.1 	imes 1 	imes 50 = 5 \ mEq$. Since $H_{2}SO_{4}$ is in excess,the remaining milliequivalents of $H_{2}SO_{4} = 20 - 5 = 15 \ mEq$. Total volume of the solution $= 100 \ mL + 50 \ mL = 150 \ mL$. Normality of the resulting solution $= \frac{	ext{Remaining mEq}}{	ext{Total Volume (mL)}} = \frac{15}{150} = 0.1 \ N = 1 	imes 10^{-1} \ N$. Thus,the nearest integer is $1$.

The normality of $H_{2}SO_{4}$ in the solution obtained on mixing $100 \ mL$ of $0.1 \ M \ H_{2}SO_{4}$ with $50 \ mL$ of $0.1 \ M \ NaOH$ is $\times 10^{-1} \ N$. (Nearest Integer)

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