By adding 20 mL of 0.1 N HCl to 20 mL of | vedclass

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(B) Step $1$: Calculate the number of gram equivalents of $HCl$ and $KOH$.$g \ eq. \ of \ HCl = N 	imes V(L) = 0.1 	imes 0.020 = 2 	imes 10^{-3} \

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$1.3$

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(B) Step $1$: Calculate the number of gram equivalents of $HCl$ and $KOH$.$g \ eq. \ of \ HCl = N 	imes V(L) = 0.1 	imes 0.020 = 2 	imes 10^{-3} \ g \ eq.$$g \ eq. \ of \ KOH = N 	imes V(L) = 0.001 	imes 0.020 = 2 	imes 10^{-5} \ g \ eq.$Step $2$: Calculate the remaining $HCl$ after neutralization.$g \ eq. \ of \ HCl \ 	ext{left} = 2 	imes 10^{-3} - 2 	imes 10^{-5} = 2 	imes 10^{-3} - 0.02 	imes 10^{-3} = 1.98 	imes 10^{-3} \ g \ eq.$Step $3$: Calculate the concentration of $H^+$ ions in the final volume.Total volume $= 20 \ mL + 20 \ mL = 40 \ mL = 0.040 \ L$.$[H^+] = \frac{1.98 	imes 10^{-3} \ g \ eq.}{0.040 \ L} = 4.95 	imes 10^{-2} \ M$.Step $4$: Calculate the $pH$.$pH = -\log[H^+] = -\log(4.95 	imes 10^{-2}) = 2 - \log(4.95) \approx 2 - 0.6946 \approx 1.305$.Thus,the $pH$ is approximately $1.3$.

By adding $20 \ mL$ of $0.1 \ N \ HCl$ to $20 \ mL$ of $0.001 \ N \ KOH$,the $pH$ of the resulting solution will be:

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