Question

$100\,ml$ of $0.5N\,NaOH$ solution is added to $10\,ml$ of $3\,N \,{H_2}S{O_4}$ solution and $20\,ml$ of $1\,N\,HCl$ solution. The mixture is:
A. Acidic
B. Alkaline
C. Neutral
D. None of these

Answer 1

Hint:Normality is a measure of concentration equal to the gram equivalent weight per liter of solution. The gram liter equivalent weight is the measure of the reactive capacity of a molecule. The normality of a solution in simple terms can be defined as the molar concentration divided by an equivalence factor.

Complete answer:
When acid and base react with each other, neutralization reaction occurs. The acid in the solution reacts with the base and forms salt and water. If a strong acid and strong base reacts with each other, then a neutral salt will be formed and therefore a neutral solution will form.
Now we know that $m_{eq} = volume \times normality$
Given here in the question
Normality of $NaOH$ = $0.5\,N$ , Volume of $NaOH$ = $100\,ml$
$m_{eq}$ of $NaOH$= $0.5 \times 100 = 50$
Normality of $HCl$= $1\,N$ , Volume of $HCl$ = $20\,ml$
$m_{eq}$ of $HCl$= $20 \times 1 = 20$
Normality of ${H_2}S{O_4}$= $3\,N$, Volume of ${H_2}S{O_4}$= $10\,ml$
$m_{eq}$ of ${H_2}S{O_4}$= $3 \times 10 = 30$
Now, $HCl$ and ${H_2}S{O_4}$ are acids in the given solution. So, on calculating total $m_{eq}$ of acids
Total $m_{eq}$ of acid = 20 + 30 = 50
And total $m_{eq}$ of base = 50
Since the $m_{eq}$ of acid and base are the same, hence all base has been neutralized by the acidic solution therefore the solution will be neutral.

Therefore, the correct option is C. Neutral

Note:
If the total $m_{eq}$ of acid will be more than base, then that solution would have been acidic in nature, on the other hand if the total $m_{eq}$ of base would be more than acid, then the solution would have been basic in nature.