Question

How many milliliters of 1M \[{H_2}S{O_4}\]will be neutralized by 10ml of 1M NaOH solution?
A.10
B.20
C.2.5
D.5

Answer 1

Hint: In the question, as molarity and volume are given, we can find the answer using the normality equation. If you know the molarity of an acid or base solution, you can convert it into normality by multiplying the molarity with the number of hydrogen ions in the case of acids and hydroxide ions in the case of bases.

Complete answer:
Here, we shall first know about normality. It is the ratio of the number of gram equivalents present in a liter of solution.
\[ \to Normality = \dfrac{{No\;of\;gram\;equivalents}}{{Volume\;of\;solution}}\]
Normality equation is given by, \[{N_1}{V_1} = {N_2}{V_2}\]. According to this equation, for the reacting acid and base, the product of their normality and volume are equal. In the question, as molarity is given, we shall convert it into normality.
\[ \to Normality = Molarity \times n\]
Here, ‘n’ refers to the basicity which means the number of hydrogen ions that can be replaced by an acid. As
\[{H_2}SO{}_4\] is given, the basicity will be 2 and that of NaOH will be 1 as only one hydroxide ion can be replaced. So the formula will be as follows
\[
   \Rightarrow n \times {M_{{H_2}S{O_4}}} \times {V_{{H_2}S{O_4}}} = n \times {M_{NaOH}} \times {V_{NaOH}} \\
   \Rightarrow 2 \times 1 \times {V_{{H_2}S{O_4}}} = 1 \times 1 \times 10 \\
   \Rightarrow {V_{{H_2}S{O_4}}} = \dfrac{{10}}{2} \\
   \Rightarrow {V_{{H_2}S{O_4}}} = 5 \\
\]
By the above explanation, it is clear that we require 5ml of 1M\[{H_2}S{O_4}\]to neutralize 10ml of 1N NaOH.

So the correct option is D.

Note:
You may get a doubt that, as both molarity and volume are given, we can directly use \[{M_1}{V_1} = {M_2}{V_2}\] instead of normality. If the given acid donates more than one proton, then you should always use the normality equation and if you go for molarity, the answer would be wrong.