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Question

(A) Step $1$: Calculate the moles of $H^+$ and $OH^-$ ions.Moles of $H^+$ from $HCl = Molarity 	imes Volume (L) = 0.1 \ M 	imes 0.1 \ L = 0.01 \ mol

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Step $1$: Calculate the moles of $H^+$ and $OH^-$ ions.Moles of $H^+$ from $HCl = Molarity 	imes Volume (L) = 0.1 \ M 	imes 0.1 \ L = 0.01 \ mol$.Moles of $OH^-$ from $NaOH = Molarity 	imes Volume (L) = 0.05 \ M 	imes 0.2 \ L = 0.01 \ mol$.Step $2$: Determine the nature of the resulting solution.Since moles of $H^+ = 0.01$ and moles of $OH^- = 0.01$,the reaction $H^+ + OH^- ightarrow H_2O$ results in complete neutralization.Step $3$: Calculate the $pH$.For a neutral solution at $25^{\circ}C$,the concentration of $H^+$ ions is $10^{-7} \ M$,which results in a $pH$ of $7$.

$pH$ of a resulting solution prepared by mixing $100 \ mL$ $0.1 \ M$ $HCl$ and $200 \ mL$ $0.05 \ M$ $NaOH$ is

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