5 mL of 0.4 N NaOH is mixed with 20 mL of | vedclass

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Question

(A) Calculate the milliequivalents $(meq)$ of $NaOH$ and $HCl$ present in the solution. $meq \ of \ NaOH = N_1 	imes V_1 = 0.4 	imes 5 = 2.0 \ meq$.

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Calculate the milliequivalents $(meq)$ of $NaOH$ and $HCl$ present in the solution. $meq \ of \ NaOH = N_1 	imes V_1 = 0.4 	imes 5 = 2.0 \ meq$. $meq \ of \ HCl = N_2 	imes V_2 = 0.1 	imes 20 = 2.0 \ meq$. Since the number of milliequivalents of the strong base $(NaOH)$ is equal to the number of milliequivalents of the strong acid $(HCl)$,the neutralization reaction is complete. The resulting solution contains only the salt $(NaCl)$ and water,making the solution neutral. Therefore,the $pH$ of the resulting solution is $7$.

$5 \ mL$ of $0.4 \ N$ $NaOH$ is mixed with $20 \ mL$ of $0.1 \ N$ $HCl$. The $pH$ of the resulting solution will be

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