Question

50 ml of n a o h solution was exactly neutralized by 38.6 ml at 0.213 molarity H2 S o4 what is the molarity of n a o h​

Answer 1

0.32 molarity of NaOH

Answer 2

Answer:

The molarity of $NaOH$ for the given solution is equal to $0.328M$.

Explanation:

Given the neutralization reaction:

   $2NaOH+H_{2}SO_{4}$  →   $Na_{2}SO_{4}+2H_{2}O$

According to law of equivalence:

               $N_{1}V_{1}=N_{2}V_{2}$                                  ..............(1)

N₁ and V₁ are normality and volume of$NaOH$ respectively.

N₂ and V₂ are normality and volume of $H_{2}SO_{4}$ respectively.

Given Molarity of H₂SO₄ $M_{2}=0.213M$

So, Normality of H₂SO₄ $(N_{2})$ $=2*0.213=0.426N$

where n-factor of H₂SO₄ =2

$V_{1}=50ml, V_{2}=38.6ml$

Put the values of N₂, V₂ and V₁ in eq.(1),  

$N_{1}(50)=(0.426)(38.6)\\N_{1}=\frac{16.4436}{50}\\ N_{1}=0.328N$

Consider $M_{1}$ is the molarity of NaOH,

$Normality=(Molarity)(n-factor)\\N_{1}=M_{1}(n-factor)\\0.328=M_{1}*1\\M_{1}=0.328M$(n-factor of NaOH =1)