Question

1. How many mL of 6M HCl is required to neutralize a)15 mL 5% aqueous NaOH solution.

b) 20 mL saturated sodium bicarbonate.

Answer 1

(a) 5% aqueous NaOH has 5gm NaOH in 100ml of solution.

Molarity of NaOH = 5g/40g/mol ×(1000/100) = 1.25 M

M1V1 = M2V2

1.25M × 15ml = 6M × V₂

V₂ = 3.125ml

3.125ml of 6M HCl is required to neutralize 15ml of 5% aqueous NaOH solution . (Answer)

(b)

Solubility of NaHCO₃ is not given. I have taken the standard data of 96g/L solubility at 20℃ .

Molarity of NaHCO₃ =( 96g/84g/mol)/L = 1.143mol/L

M1V1 = M2V2

6M×V₁ = 20ml ×1.143M

V₁ = 3.81ml (answer)