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(D) To have a $pH$ of $1.0$,the concentration of $[H^+]$ ions must be $0.1 \, M$. For option $D$: $n_{HCl} = M 	imes V = (1/5) 	imes 0.075 = 0.015 \

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$75 \, mL$ of $(M/5) \, HCl + 25 \, mL$ of $(M/5) \, NaOH$

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(D) To have a $pH$ of $1.0$,the concentration of $[H^+]$ ions must be $0.1 \, M$. For option $D$: $n_{HCl} = M 	imes V = (1/5) 	imes 0.075 = 0.015 \, mol$. $n_{NaOH} = M 	imes V = (1/5) 	imes 0.025 = 0.005 \, mol$. Remaining $n_{HCl} = 0.015 - 0.005 = 0.010 \, mol$. Total volume $= 75 \, mL + 25 \, mL = 100 \, mL = 0.1 \, L$. $[H^+] = \frac{0.010 \, mol}{0.1 \, L} = 0.1 \, M$. $pH = -\log[H^+] = -\log(0.1) = 1.0$.

Which of the following solutions will have a $pH$ close to $1.0$?

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