In which case is the change in pH maximum? | vedclass

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(D) Let us analyze the change in $pH$ for each case:$(A)$ Initial $pH = 2$,$[H^+] = 10^{-2} \ M$. After dilution to $100 \ mL$,$[H^+] = (1 \ mL 	imes

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$100 \ mL$ of $pH = 2$ solution is mixed with $100 \ mL$ of $pH = 12$ solution

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(D) Let us analyze the change in $pH$ for each case:$(A)$ Initial $pH = 2$,$[H^+] = 10^{-2} \ M$. After dilution to $100 \ mL$,$[H^+] = (1 \ mL 	imes 10^{-2} \ M) / 100 \ mL = 10^{-4} \ M$. New $pH = 4$. Change $\Delta pH = 4 - 2 = 2$.$(B)$ Initial $[OH^-] = 0.01 \ M$. Adding $0.1 \ mol$ $NaOH$ to $0.1 \ L$ solution significantly increases $[OH^-]$. The solution becomes highly basic,$pH$ changes from $12$ to nearly $14$. Change $\Delta pH \approx 2$.$(C)$ Initial $[OH^-] = 0.01 \ M$. Diluting $900 \ mL$ to $1000 \ mL$,$[OH^-] = (900 	imes 0.01) / 1000 = 0.009 \ M$. $pOH = -\log(0.009) \approx 2.04$. $pH = 13.96 - 2.04 = 11.96$. Change $\Delta pH \approx 0.04$.$(D)$ Mixing $100 \ mL$ of $pH = 2$ $([H^+] = 10^{-2} \ M)$ and $100 \ mL$ of $pH = 12$ $([OH^-] = 10^{-2} \ M)$. Since moles of $H^+$ $(10^{-3})$ equal moles of $OH^-$ $(10^{-3})$,they neutralize to form water. The resulting solution is neutral,$pH = 7$. Initial $pH$ was $2$ (or $12$). Change $\Delta pH = 7 - 2 = 5$ or $12 - 7 = 5$.Comparing all,the change in case $(D)$ is the maximum.

In which case is the change in $pH$ maximum?

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