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Question: 20 ml of \(\dfrac{N}{10}HCl\) is mixed with 40 ml \(\dfrac{N}{10}NaOH \) solution. the normality of ...

20 ml of N10HCl\dfrac{N}{10}HCl is mixed with 40 ml N10NaOH\dfrac{N}{10}NaOH solution. the normality of the resultant acidic/basic solution will be:
(a) N20\dfrac{N}{20}
(b) N30\dfrac{N}{30}
(c) N40\dfrac{N}{40}
(d) N15\dfrac{N}{15}

Explanation

Solution

By the term normality we means the no of moles of gram equivalent to the total volume of solution and we know the normalities and volume of the HCl and NaOH solutions and now we can easily calculate the normality of the solution by applying the formula as; normality=N2V2−N1V1total volume of the solution normality=\dfrac{{{N}_{2}}{{V}_{2}}-{{N}_{1}}{{V}_{1}}}{\begin{aligned} & total\text{ }volume \\\ & of\text{ }the\ solution \\\ \end{aligned}}. Now solve it.

Complete step by step answer:
First of all, let’s discuss what is normality. By the term normality we mean the number of moles of gram equivalent of the solute to the total volume of the solution in litres and it is denoted by the symbol as N. Now, considering the numerical ;
Normality of HCl solution =N10=0.1N\dfrac{N}{10}=0.1N (given)
And it is taken as N1{{N}_{1}}.
Similarly, Normality of NaOH solution =N10=0.1N\dfrac{N}{10}=0.1N (given)
And it is taken as N2{{N}_{2}}.
Now the volume of HCl solution=20 ml (given)
And it is taken as V1{{V}_{1}}.
Similarly, Normality of NaOH solution =40 ml (given)
And it is taken as V2{{V}_{2}}.
Now, the total volume of the solution i.e. of HCl and NaOH=V1+V2=20+40=60 ml{{V}_{1}}+{{V}_{2}}=20+40=60\text{ }ml
Now, we can easily calculate the normality of the solution by applying the formula as;
normality=N2V2−N1V1total volume of the solution normality=\dfrac{{{N}_{2}}{{V}_{2}}-{{N}_{1}}{{V}_{1}}}{\begin{aligned} & total\text{ }volume \\\ & of\text{ }the\ solution \\\ \end{aligned}} (1)
We know the values of N1{{N}_{1}},N2{{N}_{2}},V1{{V}_{1}}and V2{{V}_{2}}and the total volume of the solution and by putting these values in equation (1), we can calculate the normality of the solution as;
normality=[0.1×40−0.1×20]N60  =[4−2]N60  =2N60  =N30 \begin{aligned} & normality=\dfrac{[0.1\times 40-0.1\times 20]N }{60} \\\ & \text{ =}\dfrac{[4-2]N}{60} \\\ & \text{ =}\dfrac{2N}{60} \\\ & \text{ =}\dfrac{N}{30} \\\ \end{aligned}
Hence, when 20 ml of N10HCl\dfrac{N}{10}HCl is mixed with 40 ml N10NaOH\dfrac{N}{10}NaOH solution , then the normality of the resultant acidic/basic solution is N30\dfrac{N}{30}.

So, option (b) is correct.

Note: Don’t get confused in the word’s molarity and normality. By the term molarity we mean the number moles of the solute to the total volume of the solution in litres whereas normality means the number moles of gram equivalent of the solute to the total volume of the solution in litres. So, molarity=no of moles of the solutetotal volume of the solution in litersmolarity=\dfrac{no\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{total\text{ }volume\text{ }of\text{ }the\text{ }solution\text{ }in\text{ }liters} And, normality=no of moles of gram equivalent of the solutetotal volume of the solution in litersnormality=\dfrac{no\text{ }of\text{ }moles\text{ }of\ gram\text{ }equivalent\text{ }of\text{ }the\text{ }solute}{total\text{ }volume\text{ }of\text{ }the\text{ }solution\text{ }in\text{ }liters}