If 0.1 , M, 100 , mL NaOH solution is mixed w | vedclass

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(B) Step $1$: Calculate the milliequivalents $(meq)$ of $NaOH$ and $H_2SO_4$.$meq 	ext{ of } NaOH = M 	imes V 	imes n	ext{-factor} = 0.1 	imes 10

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$7.0$

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(B) Step $1$: Calculate the milliequivalents $(meq)$ of $NaOH$ and $H_2SO_4$.$meq 	ext{ of } NaOH = M 	imes V 	imes n	ext{-factor} = 0.1 	imes 100 	imes 1 = 10 \, meq$.Step $2$: Calculate the milliequivalents $(meq)$ of $H_2SO_4$.$meq 	ext{ of } H_2SO_4 = M 	imes V 	imes n	ext{-factor} = 0.05 	imes 100 	imes 2 = 10 \, meq$.Step $3$: Since the $meq$ of the base $(NaOH)$ equals the $meq$ of the acid $(H_2SO_4)$,the reaction results in complete neutralization.$NaOH + \frac{1}{2} H_2SO_4 ightarrow \frac{1}{2} Na_2SO_4 + H_2O$.Step $4$: The resulting solution contains only salt $(Na_2SO_4)$,which is a salt of a strong acid and a strong base,making the solution neutral.Therefore,the $pH$ of the resulting solution is $7.0$.

If $0.1 \, M, 100 \, mL$ $NaOH$ solution is mixed with $0.05 \, M, 100 \, mL$ $H_2SO_4$ solution,the $pH$ of the resulting solution is:

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