Question
Question: Volume of \(0.1M\,{H_2}S{O_4}\) solution required to neutralize \(40\,mL\) of \(0.1M\,NaOH\) solutio...
Volume of 0.1MH2SO4 solution required to neutralize 40mL of 0.1MNaOH solution will be:
A. 40mL
B. 39mL
C. 20mL
D. 10mL
Solution
Acid (having pH less than 7) when reacts quantitatively with base (pH having more than 7), they form salt and water. And this reaction is well known as neutralization reaction. The pH of the neutralized reaction depends upon the acid strengths of reactants.
Complete answer:
The neutralization reaction can be represented as
Acid + base → salt + water
Simplest example of this reaction is when hydrochloric acid reacts with sodium hydroxide
HCl+NaOH→NaCl+H2O
The products formed are salt that is sodium chloride and water. Since both the acid and base are strong so this reaction is a strong acid- strong base neutralization reaction. In strong acid – strong base neutralization reaction, the pH of the resulting solution lies around pH 7. And the solution will be neutral.
For the above reaction-
H2SO4+2NaOH→Na2SO4+2H2O
Here we can see that the basicity of sulphuric acid is 2. That means for 1mol of sulphuric acid we need 2mol of base to completely neutralize the acid.
We know that, NacidVacid=NbaseVbase
Given ,
Nacid=0.1×2=0.2N
Nbase=0.1N
Vbase=40mL
We have to calculate the volume of acid required.
Putting all the values in the above equation
⇒0.2×Vacid=1.0×40 ⇒Vacid=20mL
So the volume of sulphuric acid required to neutralize the base is 20mL
**Hence the correct option will be C. 20mL
Note:**
But if we have strong acid - weak base or strong base - weak acid then the pH of the resulting solution will depend on the salt that has formed. Either the salt formed will be acidic salt or basic salt. The formation of acidic salt when acid will be stronger, and basic salt when base will be stronger.