At 27^{circ}C,100 mL of 0.4 M HCl is mixed | vedclass

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(A) Moles of $HCl = 0.4 \ M 	imes 0.1 \ L = 0.04 \ 	ext{moles}$.Moles of $NaOH = 0.5 \ M 	imes 0.1 \ L = 0.05 \ 	ext{moles}$.$HCl$ and $NaOH$ reac

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$12$

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(A) Moles of $HCl = 0.4 \ M 	imes 0.1 \ L = 0.04 \ 	ext{moles}$.Moles of $NaOH = 0.5 \ M 	imes 0.1 \ L = 0.05 \ 	ext{moles}$.$HCl$ and $NaOH$ react in a $1:1$ ratio according to the equation: $HCl + NaOH ightarrow NaCl + H_2O$.Since $0.05 \ 	ext{moles}$ of $NaOH$ react with $0.04 \ 	ext{moles}$ of $HCl$,$NaOH$ is in excess.Remaining moles of $NaOH = 0.05 - 0.04 = 0.01 \ 	ext{moles}$.Total volume of the solution = $100 \ mL + 100 \ mL + 800 \ mL = 1000 \ mL = 1 \ L$.Concentration of $[OH^-] = \frac{0.01 \ 	ext{moles}}{1 \ L} = 0.01 \ M = 10^{-2} \ M$.$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.Since $pH + pOH = 14$ at $27^{\circ}C$,$pH = 14 - 2 = 12$.

At $27^{\circ}C$,$100 \ mL$ of $0.4 \ M$ $HCl$ is mixed with $100 \ mL$ of $0.5 \ M$ $NaOH$ solution. To the resultant solution,$800 \ mL$ of distilled water is added. What is the $pH$ of the final solution?

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