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

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(D) $1$. 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

$> 7$

Vedclass · Answer

(D) $1$. 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 milliequivalents of $OH^-$ ions: $1 - 0.5 = 0.5 \ meq$. $5$. Total volume of the solution: $10 \ mL + 10 \ mL = 20 \ mL$. $6$. Concentration of $OH^-$ ions: $[OH^-] = \frac{0.5 \ meq}{20 \ mL} = 0.025 \ N = 2.5 	imes 10^{-2} \ M$. $7$. Calculate $pOH$: $pOH = -\log(2.5 	imes 10^{-2}) = 2 - \log(2.5) = 2 - 0.3979 = 1.6021$. $8$. Calculate $pH$: $pH = 14 - pOH = 14 - 1.6021 = 12.3979$. $9$. Since $pH > 7$,the correct option is $D$.

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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