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(B) For $pH = 1$,the concentration of $[H^{+}]$ must be $10^{-1} \ M$ or $0.1 \ M$ $(M/10)$.For option $B$: $75 \ mL \ \frac{M}{5} \ HCl + 25 \ mL \ \

Vedclass · Accepted Answer

$75 \ mL \ \frac{M}{5} \ HCl + 25 \ mL \ \frac{M}{5} \ NaOH$

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(B) For $pH = 1$,the concentration of $[H^{+}]$ must be $10^{-1} \ M$ or $0.1 \ M$ $(M/10)$.For option $B$: $75 \ mL \ \frac{M}{5} \ HCl + 25 \ mL \ \frac{M}{5} \ NaOH$.$25 \ mL \ \frac{M}{5} \ NaOH$ will neutralize $25 \ mL \ \frac{M}{5} \ HCl$.Remaining $HCl$ volume = $75 \ mL - 25 \ mL = 50 \ mL$ of $\frac{M}{5} \ HCl$.Total volume of the solution = $75 \ mL + 25 \ mL = 100 \ mL$.New concentration of $HCl = \frac{M}{5} 	imes \frac{50 \ mL}{100 \ mL} = \frac{M}{10} = 0.1 \ M$.Since $HCl$ is a strong acid,$[H^{+}] = 0.1 \ M$.$pH = -\log_{10}[H^{+}] = -\log_{10}(10^{-1}) = 1$.

Following four solutions are prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations. The $pH$ of which one of them will be equal to $1$?

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