At 27^{circ} C, 100 mL of 0.5 M HCl is mixe | vedclass

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(B) Initial moles of $HCl$ $(n_1)$ = $0.5 \ M 	imes 0.1 \ L = 0.05 \ mol$. Initial moles of $NaOH$ $(n_2)$ = $0.4 \ M 	imes 0.1 \ L = 0.04 \ mol$. S

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$2$

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(B) Initial moles of $HCl$ $(n_1)$ = $0.5 \ M 	imes 0.1 \ L = 0.05 \ mol$. Initial moles of $NaOH$ $(n_2)$ = $0.4 \ M 	imes 0.1 \ L = 0.04 \ mol$. Since $HCl$ is a strong acid and $NaOH$ is a strong base, they neutralize each other: $n_{H^+} = n_1 - n_2 = 0.05 - 0.04 = 0.01 \ mol$. Total volume of the final solution = $100 \ mL + 100 \ mL + 800 \ mL = 1000 \ mL = 1 \ L$. Final concentration of $H^+$ ions = $\frac{0.01 \ mol}{1 \ L} = 0.01 \ M = 10^{-2} \ M$. $pH = -\log[H^+] = -\log(10^{-2}) = 2$.

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

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