The pH of the solution containing 10 mL of 0 | vedclass

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(C) $1$. Calculate the milliequivalents of $NaOH$: $n_{NaOH} = 10 \ mL 	imes 0.1 \ N = 1 \ meq$. $2$. Calculate the milliequivalents of $H_2SO_4$: $n

Vedclass · Accepted Answer

$> 7$

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(C) $1$. Calculate the milliequivalents of $NaOH$: $n_{NaOH} = 10 \ mL 	imes 0.1 \ N = 1 \ meq$. $2$. Calculate the milliequivalents of $H_2SO_4$: $n_{H_2SO_4} = 10 \ mL 	imes 0.05 \ N = 0.5 \ meq$. $3$. Since $NaOH$ is a strong base and $H_2SO_4$ is a strong acid,they neutralize each other. $4$. Remaining $NaOH = 1 \ meq - 0.5 \ meq = 0.5 \ meq$. $5$. Total volume of the solution = $10 \ mL + 10 \ mL = 20 \ mL$. $6$. Concentration of $[OH^-] = \frac{0.5 \ meq}{20 \ mL} = 0.025 \ N = 2.5 	imes 10^{-2} \ M$. $7$. $pOH = -\log(2.5 	imes 10^{-2}) = 2 - \log(2.5) = 2 - 0.3979 = 1.6021$. $8$. $pH = 14 - pOH = 14 - 1.6021 = 12.3979$. $9$. Since $pH > 7$,the correct option is $C$.

The $pH$ of the solution containing $10 \ mL$ of $0.1 \ N$ $NaOH$ and $10 \ mL$ of $0.05 \ N$ $H_2SO_4$ would be

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