100 , mL of 0.04 , N , HCl aqueous solution i | vedclass

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(C) The number of milliequivalents of $HCl$ is $N_1V_1 = 0.04 	imes 100 = 4 \, meq$.The number of milliequivalents of $NaOH$ is $N_2V_2 = 0.02 	imes

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$2$

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(C) The number of milliequivalents of $HCl$ is $N_1V_1 = 0.04 	imes 100 = 4 \, meq$.The number of milliequivalents of $NaOH$ is $N_2V_2 = 0.02 	imes 100 = 2 \, meq$.Since $HCl$ is a strong acid and $NaOH$ is a strong base,they neutralize each other. The remaining milliequivalents of $HCl$ are $4 - 2 = 2 \, meq$.The total volume of the resulting solution is $100 \, mL + 100 \, mL = 200 \, mL$.The normality of the resulting solution is $N_3 = \frac{2 \, meq}{200 \, mL} = 0.01 \, N = 10^{-2} \, N$.Since $HCl$ is a strong monoprotic acid,$[H^+] = 10^{-2} \, M$.The $pH$ is calculated as $pH = -\log [H^+] = -\log [10^{-2}] = 2$.

$100 \, mL$ of $0.04 \, N \, HCl$ aqueous solution is mixed with $100 \, mL$ of $0.02 \, N \, NaOH$ solution. The $pH$ of the resulting solution is

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