30 cc of frac{M}{3} HCl,20 cc of frac{M}{2 | vedclass

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(B) Total milliequivalents of $H^{+}$ ions from $HCl$ and $HNO_{3}$:$= (30 	imes \frac{1}{3}) + (20 	imes \frac{1}{2}) = 10 + 10 = 20 \ mEq$.Total m

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$2$

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(B) Total milliequivalents of $H^{+}$ ions from $HCl$ and $HNO_{3}$:$= (30 	imes \frac{1}{3}) + (20 	imes \frac{1}{2}) = 10 + 10 = 20 \ mEq$.Total milliequivalents of $OH^{-}$ ions from $NaOH$:$= 40 	imes \frac{1}{4} = 10 \ mEq$.Since $H^{+}$ reacts with $OH^{-}$ to form water,the remaining milliequivalents of $H^{+}$:$= 20 - 10 = 10 \ mEq$.The final volume of the solution is $1 \ dm^{3} = 1000 \ mL$.Concentration of $H^{+}$ ions $[H^{+}] = \frac{10 \ mEq}{1000 \ mL} = 10^{-2} \ M$.$pH = -\log[H^{+}] = -\log(10^{-2}) = 2$.

$30 \ cc$ of $\frac{M}{3} \ HCl$,$20 \ cc$ of $\frac{M}{2} \ HNO_{3}$ and $40 \ cc$ of $\frac{M}{4} \ NaOH$ solutions are mixed and the volume was made up to $1 \ dm^{3}$. The $pH$ of the resulting solution is

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