Question

Calculate the final concentration of each of the following diluted solutions.

0.50 L of a 4.0 M HNO3 solution is added to water so that the final volume is 8.0 L .

Water is added to 0.35 L of a 6.0 M KOH solution to make 2.0 L of a diluted KOH solution.

A 65.0 mL sample of 8.0 % (m/v) NaOH is diluted with water so that the final volume is 200.0 mL .

A 8.0 mL sample of 50.0 % (m/v) acetic acid (HC2H3O2) solution is added to water to give a final volume of 25 mL .

Answer 1

Answer – 1) We are given, M1 = 4.0 M HNO₃ , V1 = 0.50 L , M2 = ? , V2 = 8.0 L

We know dilution law

M₁V₁ = M₂V₂

So, M₂ = M₁V₁ /V₂

            = 4.0 M * 0.50 L / 8.0 L

            = 0.25 M

So, concentration HNO3 solution is 0.25 M

2) We are given, M1 = 6.0 M KOH , V1 = 0.35 L , M2 = ? , V2 = 2.0 L

We know dilution law

M₁V₁ = M₂V₂

So, M₂ = M₁V₁ /V₂

            = 6.0 M * 0.35 L / 2.0 L

            = 1.05 M

So, diluted concentration KOH solution is 1.05 M.

3) We are given 65.0 mL sample of 8.0 % (m/v) NaOH, means 8.0 g of NaOH in 100 mL , so

100 mL = 8.0 g of NaOH

So, 65 mL = ?

= 5.2 g of NaOH

So, 5.2 g of NaOH in 200 mL solution has concentration

= 5.2 g / 200 mL * 100 %

= 2.6 %

So, diluted concentration of NaOH is 2.6 %

4) We are given 8.0 mL sample of 50.0 % (m/v) acetic acid, means 50.0 g of acetic acid in 100 mL

100 mL = 50.0 g of HC₂H₃O₂

So, 8.0 mL = ?

= 4.0 g of HC₂H₃O₂

Now we need to calculate the initial concentration of HC₂H₃O₂ and for the we need to convert the mass to moles as follow –

Moles of HC₂H₃O₂= 4.0 g of HC₂H₃O₂/ 60.05 g.mol^-1

                                 = 0.0666 moles

So, molarity of HC₂H₃O₂= 0.0666 moles / 0.008 L

                                     = 8.33 M

So, M1 = 8.33 M HC₂H₃O₂, V1 = 8.0 mL , M2 = ? , V2 = 25 mL

We know dilution law

M₁V₁ = M₂V₂

So, M₂ = M₁V₁ /V₂

            = 8.33 M * 8.0 mL / 25 mL

            = 2.7 M

Or we can calculate the in percent like 4 g in 25 mL

= 4 g / 25 mL * 100 %

= 16 %

So, final concentration of HC₂H₃O₂ is 16 %