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Question: What will be the normality of the salt solution obtained by neutralizing x ml y (N) HCl with y ml x ...

What will be the normality of the salt solution obtained by neutralizing x ml y (N) HCl with y ml x (N) NaOH, and finally adding (x+y) ml distilled water?
A.2(x+y)xyN\dfrac{2(x+y)}{xy}N
B.xy2(x+y)N\dfrac{xy}{2(x+y)}N
C. (2xyx+y)N\left( \dfrac{2xy}{x+y} \right)N
D. (x+yxy)N\left( \dfrac{x+y}{xy} \right)N

Explanation

Solution

The reaction between an acid and a base is a neutralization reaction which forms salts. Normality is the number of grams equivalent of a solute which are dissolved in per liter of any solution. Equivalent mass of any element is its mass upon valency.
Formula used: Formula for normality, N = no.ofmilliequivalentVol.ofsolution(ml)\dfrac{no.\,of\,milliequivalent}{Vol.\,of\,solution\,(ml)}

Complete answer:
As given, the neutralization reaction happens between HCl and NaOH, that has xy of HCl and xy of NaOH. The reaction is as,
HCl+NaOHNaCl+H2OHCl+NaOH\to NaCl+{{H}_{2}}O
Among this, first the concentration or number of HCl and NaOH are xy for both. Then after the reaction the products have been formed with amount of both products as xy, as shown,
HCl+NaOHNaCl+H2O milliequi.addedxyxy00 milliequi.left00xyxy \begin{aligned} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,HCl+NaOH\to NaCl+{{H}_{2}}O \\\ & milliequi.\,added\,\,\,xy\,\,\,\,\,\,\,\,\,\,\,xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\\ & milliequi.\,left\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy\,\,\,\,\,\,\,\,\,\,\,\,xy \\\ \end{aligned}
So, for taking out normality, the number of milliequivalent will be xy and that of the volume of solution will be twice of (x+y). Keeping these in the formula of normality, we will get,
Normality, N = xy(x+y)+(x+y)N\dfrac{xy}{(x+y)+(x+y)}N
Therefore, Normality, N = xy2(x+y)N\dfrac{xy}{2(x+y)}N
Hence, the normality of the solution is xy2(x+y)N\dfrac{xy}{2(x+y)}N.

So, option B is correct.

Note:
Number of milliequivalent in normality is taken by the mass of the substance in units of milligram. 1 mEq is 1/1000 of an equivalent of any substance. The relation between normality and molarity is normality = molarity×\times n factor, where n factor is the valency factor of any substance.