What will be the normality of the salt soluti | vedclass

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Question

(B) The neutralization reaction is $HCl + NaOH \longrightarrow NaCl + H_2O$.Number of milliequivalents of $HCl = x 	imes y$.Number of milliequivalent

Vedclass · Accepted Answer

$\frac{xy}{2(x+y)} \ N$

Vedclass · Answer

(B) The neutralization reaction is $HCl + NaOH \longrightarrow NaCl + H_2O$.Number of milliequivalents of $HCl = x 	imes y$.Number of milliequivalents of $NaOH = y 	imes x$.Since the milliequivalents are equal, the solution is neutral and the number of milliequivalents of $NaCl$ formed is $xy$.Total volume of solution = $x \ mL$ $(HCl)$ + $y \ mL$ $(NaOH)$ + $(x+y) \ mL$ (distilled water) = $2(x+y) \ mL$.$	ext{Normality} = \frac{	ext{Total milliequivalents of salt}}{	ext{Total volume in } mL} = \frac{xy}{2(x+y)} \ N$.

What will be the normality of the salt solution obtained by neutralising $x \ mL$ $y \ (N)$ $HCl$ with $y \ mL$ $x \ (N)$ $NaOH$, and finally adding $(x+y) \ mL$ distilled water?

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