The pH of a solution obtained by mixing 50 m | vedclass

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(C) Number of equivalents $(N.eq.)$ for $HCl = 0.4 \ N 	imes 0.050 \ L = 0.02 \ eq$.Number of equivalents $(N.eq.)$ for $NaOH = 0.2 \ N 	imes 0.050

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(C) Number of equivalents $(N.eq.)$ for $HCl = 0.4 \ N 	imes 0.050 \ L = 0.02 \ eq$.Number of equivalents $(N.eq.)$ for $NaOH = 0.2 \ N 	imes 0.050 \ L = 0.01 \ eq$.Since $HCl$ is in excess,the remaining $H^{+}$ equivalents $= 0.02 - 0.01 = 0.01 \ eq$.Total volume of the solution $= 50 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$.Concentration of $[H^{+}] = \frac{0.01 \ eq}{0.1 \ L} = 0.1 \ M$.$pH = -\log[H^{+}] = -\log(0.1) = 1$.

The $pH$ of a solution obtained by mixing $50 \ mL$ of $0.4 \ N \ HCl$ and $50 \ mL$ of $0.2 \ N \ NaOH$ is

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