Question

3. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g

of a dibasic acid A. Calculate the relative molecular mass of acid A.
(2 marks)
​

Answer 1

The relative molecular mass of acid A : 50 g/mol

Further explanation

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g of a dibasic acid

Required

the relative molecular mass of acid A

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2

mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

0.2 x 40 x 1 = M₂ x V₂ x 2

M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A

The relative molecular mass of acid A (M) :

`\tt M_A=(mass )/(mol)=(0.2~g)/(4.10^(-3))=50~g/mol`