Calculate the pH of a mixture obtained by mix | vedclass

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(B) Let the volume of each solution be $V \ L$. Total volume of the mixture = $2V \ L$. Moles of $H^+$ from $HCl$ = $0.01 	imes V = 0.01V \ mol$. Mol

Vedclass · Accepted Answer

$12.65$

Vedclass · Answer

(B) Let the volume of each solution be $V \ L$. Total volume of the mixture = $2V \ L$. Moles of $H^+$ from $HCl$ = $0.01 	imes V = 0.01V \ mol$. Moles of $OH^-$ from $NaOH$ = $0.1 	imes V = 0.1V \ mol$. Since $OH^-$ is in excess,moles of $OH^-$ remaining = $0.1V - 0.01V = 0.09V \ mol$. Concentration of $[OH^-]$ in the mixture = $\frac{0.09V}{2V} = 0.045 \ M$. $pOH = -\log[OH^-] = -\log(0.045) = -(\log(4.5 	imes 10^{-2})) = -(0.653 - 2) = 1.347$. $pH = 14 - pOH = 14 - 1.347 = 12.653 \approx 12.65$.

Calculate the $pH$ of a mixture obtained by mixing equal volumes of $0.01 \ M \ HCl$ and $0.1 \ M \ NaOH$.

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