Question

1) How many moles of acid are there in 23 ml of a 2.1 M HCl solution?

2) 35 ml of a 1.0 M HCl solution was added to a flask and titrated against a solution of 6.4 M NaOH. How many ml of the NaOH solution was required to neutralize the acid?

3) What volume (in liters) of a 1.9 M NaOH solution would neutralize 4.7 moles of H₂SO₄ (diprotic)?

4) Using the formula M₁V₁ = M₂V₂, calculate the new concentration of a solution, if 309 ml of a 1.7 M solution was diluted to 450 ml.

Answer 1

1. Concentration of HCl = No of moles present in one litre (1000 ml).

Volume of HCl = V (ml) = 23 ml

Molarity of HCl = 2.1 M

So that no of moles in one litre gives the concentration of HCl

No of moles = Molarity x Volume / 1000 = 2.1 x 23 / 1000

= 9.13 x 10^-5 moles

2. Volume of HCl V = 35 ml

Molarity M = 1.0 M

No of moles in one litre = concentration of HCl

= Molarity x Volume / 1000 = 35 x 1.0 / 1000 = 0.035 moles

The moles of HCl = The moles of NaOH = 0.035 because in stoichiometric equation one mole NaOH reacts with one of HCl

So that the no of moles = 0.035 moles

Molarity of NaOH M = 6.4 M

Volume of NaOH (V) =?

V = No of moles /Molarity x 1000

= 0.035 x 1000 / 6.4 = 5.469 ml.

3. The no of moles of H2SO4 = 4.7 moles

H2SO4 + 2 NaOH Na 2 SO4 + 2 H2O

The no of moles of NaOH = 2 x no of moles of H2SO4

= 4.7 x 2 = 9.4 moles

The Molarity of NaOH = 1.9 M

Volume of NaOH =?

= no of moles / Molarity x 1000 = 9.4/1.9 x 1000 = 4.947 litres

4. Volume V1 = 309 ml

Concentration M1 = 1.7

Volume of V2 = 450 ml

Concentration M2 =?

M2 = V1M1 / V2 = 1.7 x 309 / 450 = 1.165 M

The new concentration M2 = 1.165 M