Question

8.00ml of 1.25M lithiukm hydroxide is reacted with sulfuric acid. It is found that 52.87mL of the sulfuric acid is required to completely neutralize the lithium hydroxide. What is the approximate molarity of sulfuric acid?

Answer 1

Answer: The molarity of sulfuric acid is 0.0946 M

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is LiOH

We are given:

$n_1=2\\M_1=?M\\V_1=52.87mL\\n_2=1\\M_2=1.25M\\V_2=8.00mL$

Putting values in above equation, we get:

$2\times M_1\times 52.87=1\times 1.25\times 8.00\\\\M_1=\frac{1\times 1.25\times 8.00}{2\times 52.87}=0.0946M$

Hence, the molarity of sulfuric acid is 0.0946 M