Question

77.60 ml of .210 M NaOH are required to neutralize 25.05 ml

Answer 1

Answer:

Molarity =  0.650 M

Normality = 1.30 N

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 25.05mL

M₂ = 0.210 M

V₂ = 77.60 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂

M₁ * 25.05mL =  0.210 M * 77.60 mL

M₁ = 0.650 M

Hence, the molarity of the acid = 0.650 M

Normality of a diprotic acid can be calculated by multiplying 2 with the value of molarity ,

Hence,

Normality = M * 2 = 0.650 * 2 = 1.30 N