Question

During your reaction, you added 0.3 mL concentrated H2SO4 (18.4 M) as the catalyst. At the end of the reaction, you need to add base to neutralize it. How much volume (in mL) of 10% Na2CO3 to neutralize all the acid present

Answer 1

Answer:

58.72 mL

Explanation:

The chemical equation for the neutralization reaction is :

H₂SO₄(aq) + Na₂CO₃(s)  --------------> Na₂SO₄(aq) + H₂O(l) + CO₂(g)

where;

M₁ = Molarity of H₂SO₄

M₂= Molarity of Na₂CO₃

V₁= Volume of H₂SO₄

V₂ = Volume of Na₂CO₃

Given that :

M₁ = 18.4 M

V₁= 0.3 mL

10% Na₂CO₃ means 100 g of solution contain 10 g of Na₂CO₃

i.e. 10 g Na₂CO₃ dissolved and diluted to 100 mL water.

Molar mass of Na₂CO₃ = 106 g/mol

106 g Na₂CO₃ dissolved in 100 mL will give 0.1 M Na₂CO₃ solution.

However;

If, 106 g Na₂CO₃ ≡ 0.1 M Na₂CO₃

Then, 10 g Na₂CO₃ ≡  'A' M  of Na₂CO₃

By cross multiplying; we have:

106 × A = 10 × 0.1

106 × A = 1

A = (1/106) M/100 mL

A = 10 x (1/106)) M/L

A = (10/106) M

A = 0.094  M

Therefore,the molarity of 10% Na₂CO₃ solution is 0.094 M.

For the Neutralization equation, we have:

M₁V₁ = M₂V₂

18.4×0.3 = 0.094×V₂

Making V₂  the subject of the formula;we have:

$V_2 = \frac{18.4*0.3}{0.094}$

V₂ = 58.72 mL