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

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(B) The reaction between $HCl$ (a strong acid) and $KOH$ (a strong base) is a neutralization reaction: $HCl + KOH \rightarrow KCl + H_2O$. Since the n

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The reaction between $HCl$ (a strong acid) and $KOH$ (a strong base) is a neutralization reaction: $HCl + KOH ightarrow KCl + H_2O$. Since the number of equivalents of $HCl$ $(20 \ mL 	imes 0.1 \ N = 2 \ mEq)$ is equal to the number of equivalents of $KOH$ $(20 \ mL 	imes 0.1 \ N = 2 \ mEq)$,the reaction results in complete neutralization. The resulting salt,$KCl$,is formed from a strong acid and a strong base,which does not undergo hydrolysis. Therefore,the solution remains neutral,and the $pH = 7$.

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

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