Question

TOPICAL ASSINMENT 2

1. If 25cm' of O.IM H,SO, solution neutralised a solution containing 1.06g anhydrous sodium carbonate in
250cm. of solution, calculate the molarity and the volume of the sodium carbonate solution used. (3mks)​

Answer 1

The molarity = 0.04 M

The volume = 6.25 ml

Further explanation

Given

25cm³ of 0.1 M H₂SO₄

1.06g Na₂CO₃ in 250cm³ of solution

Required

the molarity and the volume of Na₂CO₃

Solution

Molarity of Na₂CO₃ :

`\tt (1.06/106~mol)/(0.25~L)=0.04~M`

Reaction

H₂SO₄ + Na₂CO₃ ⇒Na₂SO₄ + H₂O + CO₂

mol H₂SO₄ = 25 x 0.1 = 0.25 mlmol= 2.5 x 10⁻⁴ moles of Na₂CO₃

The volume of solution used :

`\tt V=(n)/(M)=(2.5.10^(-4))/(0.04)=6.25.10^(-3)~L=6.25~ml`