75 mL of 0.2 M HCl is mixed with 25 mL of | vedclass

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(A) Step $1$: Calculate the total moles of $HCl$ in the mixture.$n_1 = M_1 	imes V_1 = 0.2 \ M 	imes 0.075 \ L = 0.015 \ mol$$n_2 = M_2 	imes V_2 =

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(A) Step $1$: Calculate the total moles of $HCl$ in the mixture.$n_1 = M_1 	imes V_1 = 0.2 \ M 	imes 0.075 \ L = 0.015 \ mol$$n_2 = M_2 	imes V_2 = 1 \ M 	imes 0.025 \ L = 0.025 \ mol$Total moles $n = n_1 + n_2 = 0.015 + 0.025 = 0.040 \ mol$Step $2$: Calculate the total volume of the final solution.$V_{total} = 75 \ mL + 25 \ mL + 300 \ mL = 400 \ mL = 0.4 \ L$Step $3$: Calculate the final molarity of $HCl$.$M_{final} = \frac{n}{V_{total}} = \frac{0.040 \ mol}{0.4 \ L} = 0.1 \ M$Step $4$: Calculate the $pH$.$[H^+] = 0.1 \ M = 10^{-1} \ M$$pH = -\log_{10}[H^+] = -\log_{10}(10^{-1}) = 1$

$75 \ mL$ of $0.2 \ M \ HCl$ is mixed with $25 \ mL$ of $1 \ M \ HCl$. To this solution,$300 \ mL$ of distilled water is added. What is the $pH$ of the resultant solution?

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