When 20 , mL of 0.25 , N strong acid and 30 , | vedclass

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(A) The number of equivalents of acid $(N_1V_1)$ = $20 \, mL 	imes 0.25 \, N = 5 \, meq$.The number of equivalents of base $(N_2V_2)$ = $30 \, mL 	i

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$0.02 \, N$ basic

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(A) The number of equivalents of acid $(N_1V_1)$ = $20 \, mL 	imes 0.25 \, N = 5 \, meq$.The number of equivalents of base $(N_2V_2)$ = $30 \, mL 	imes 0.2 \, N = 6 \, meq$.Since the number of equivalents of base is greater than the acid,the resulting solution will be basic.Excess equivalents of base = $6 \, meq - 5 \, meq = 1 \, meq$.Total volume of the mixture = $20 \, mL + 30 \, mL = 50 \, mL$.Normality of the resulting solution = $\frac{	ext{Excess equivalents}}{	ext{Total volume}} = \frac{1 \, meq}{50 \, mL} = 0.02 \, N$.Therefore,the solution is $0.02 \, N$ basic.

When $20 \, mL$ of $0.25 \, N$ strong acid and $30 \, mL$ of $0.2 \, N$ strong base are mixed,the resulting solution will be.......

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