The p^H of a mixture containing 10 mL of 0.1 | vedclass

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(D) Step $1$: Calculate the milliequivalents of $NaOH$ and $H_2SO_4$. Milliequivalents of $NaOH = N 	imes V = 0.1 \ N 	imes 10 \ mL = 1 \ meq$. Mill

Vedclass · Accepted Answer

$> 7$

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(D) Step $1$: Calculate the milliequivalents of $NaOH$ and $H_2SO_4$. Milliequivalents of $NaOH = N 	imes V = 0.1 \ N 	imes 10 \ mL = 1 \ meq$. Milliequivalents of $H_2SO_4 = N 	imes V = 0.05 \ N 	imes 10 \ mL = 0.5 \ meq$. Step $2$: Determine the nature of the resulting mixture. Since $NaOH$ is a strong base and $H_2SO_4$ is a strong acid,they react as follows: $H^+ + OH^- ightarrow H_2O$. Remaining $OH^-$ milliequivalents $= 1 \ meq - 0.5 \ meq = 0.5 \ meq$. Step $3$: Calculate the concentration of $OH^-$ in the final mixture. Total volume $= 10 \ mL + 10 \ mL = 20 \ mL$. $[OH^-] = \frac{0.5 \ meq}{20 \ mL} = 0.025 \ N = 2.5 	imes 10^{-2} \ M$. Step $4$: Calculate $pOH$ and $p^H$. $pOH = -\log[OH^-] = -\log(2.5 	imes 10^{-2}) = 2 - \log(2.5) = 2 - 0.3979 = 1.6021$. $p^H = 14 - pOH = 14 - 1.6021 = 12.3979$. Since $12.3979 > 7$,the correct option is $D$.

The $p^H$ of a mixture containing $10 \ mL$ of $0.1 \ N$ $NaOH$ and $10 \ mL$ of $0.05 \ N$ $H_2SO_4$ is ............

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