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

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(D) The number of equivalents of acid is $N_1 V_1 = 0.25 	imes 20 = 5 \, 	ext{meq}$.The number of equivalents of base is $N_2 V_2 = 0.2 	imes 30 =

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$0.02 \, N$ basic

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(D) The number of equivalents of acid is $N_1 V_1 = 0.25 	imes 20 = 5 \, 	ext{meq}$.The number of equivalents of base is $N_2 V_2 = 0.2 	imes 30 = 6 \, 	ext{meq}$.Since the number of equivalents of base is greater than the number of equivalents of acid,the resulting solution will be basic.The excess equivalents of base = $6 - 5 = 1 \, 	ext{meq}$.The total volume of the mixture = $20 + 30 = 50 \, mL$.The normality of the resulting solution = $\frac{	ext{Excess equivalents}}{	ext{Total volume}} = \frac{1 \, 	ext{meq}}{50 \, mL} = 0.02 \, N$.Therefore,the resulting solution is $0.02 \, N$ basic.

If $20 \, mL$ of $0.25 \, N$ strong acid and $30 \, mL$ of $0.2 \, N$ strong base are mixed,then the resulting solution is

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