When 100 mL of 0.3 N HCl solution is mixed | vedclass

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(D) The normality of a component in a mixture is calculated by the formula $N_{final} = \frac{N_1V_1 + N_2V_2}{V_{total}}$.Here,we are asked for the n

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$0.4$

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(D) The normality of a component in a mixture is calculated by the formula $N_{final} = \frac{N_1V_1 + N_2V_2}{V_{total}}$.Here,we are asked for the normality of $H_2SO_4$ specifically in the final mixture.Initial volume of $H_2SO_4$ $(V_2)$ = $200 \ mL$.Initial normality of $H_2SO_4$ $(N_2)$ = $0.6 \ N$.Total volume of the mixture $(V_{total})$ = $100 \ mL + 200 \ mL = 300 \ mL$.The amount of $H_2SO_4$ (in terms of milliequivalents) remains constant as it is not reacting with $HCl$.$N_{final} 	imes V_{total} = N_2 	imes V_2$.$N_{final} = \frac{0.6 \ N 	imes 200 \ mL}{300 \ mL} = \frac{120}{300} \ N = 0.4 \ N$.

When $100 \ mL$ of $0.3 \ N$ $HCl$ solution is mixed with $200 \ mL$ of $0.6 \ N$ $H_2SO_4$ solution,what is the normality of $H_2SO_4$ in the final mixture?

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