Question

Calculate the amount of stock reagent(s) and water required to make the following solutions.

A)2.5 L of 25 mM NaCl (from 250 mM stock), 1 mM EDTA (from 0.5 M stock)

B)200 mL of 0.25 M Sodium Citrate (from dry powder), 20 mM SDS (from 0.5 M stock), 10% glycerol

C)3.0 L of 25% Saroksyl

Answer 1

A) (i) We have a 250 mM stock NaCl solution and we wish to prepare a 2.5 L of a 25 mM NaCl solution.

Use the dilution equation:

M₁*V₁ = M₂*V₂ where M₁ = 250 mM; M₂ = 25 mM, V₂ = 2.5 L and V₁ = volume of stock solution required.

Therefore,

(250 mM)*V₁ = (25 mM)*(2.5 L)

===> V₁ = (25*2.5)/(250) L = 0.25 L = (0.25 L)*(1000 mL/1 L) = 250 mL.

Therefore, the volume of 250 mM stock NaCl solution required = 250 mL and the volume of water that needs to be added = (2.5 – 0.25) L = 2.25 L (ans).

(ii) We have a 0.5 M EDTA stock solution and we wish to prepare 1 mM = (1 mM)*(1 M/1000 mM) = 0.001 M EDTA solution. The final volume of the solution is 2.5 L (hopefully).

Use the dilution equation:

M₁*V₁ = M₂*V₂ where M₁ = 0.5 M; M₂ = 0.001 M, V₂ = 2.5 L and V₁ = volume of stock solution required.

Therefore,

(0.5 M)*V₁ = (0.001 M)*(2.5 L)

===> V₁ = (0.001*2.5)/(0.5) L = 0.005 L = (0.005 L)*(1000 mL/1 L) = 5.0 mL.

Therefore, the volume of 0.5 M stock EDTA solution required = 5 mL and the volume of water that needs to be added = (2.5 – 0.005) L = 2.495 L (ans).

B) (i) Molar mass of sodium citrate = 258.06 g/mol.

Mole(s) of sodium citrate = (200 mL)*(1 L/1000 mL)*(0.25 mol/L) [1 M = 1 mol/L] = 0.05 mole.

Mass of sodium citrate required = (0.05 mole)*(258.06 g/mol) = 12.903 g ≈ 12.9 g (ans).

(ii) We wish to prepare 200 mL of 20 mM SDS solution from 0.5 M stock SDS solution. We know that 20 mM = (20 mM)*(1 M/1000 mM) = 0.02 M.

Use the dilution equation:

M₁*V₁ = M₂*V₂ where M₁ = 0.5 M; M₂ = 0.02 M, V₂ = 200 mL and V₁ = volume of stock solution required.

Therefore,

(0.5 M)*V₁ = (0.02 M)*(200 mL)

===> V₁ = (0.02*200)/(0.5) mL= 8.0 mL

The volume of stock SDS solution required is 8.0 mL and the volume of water = (200 – 8) mL = 192 mL (ans).

(iii) We define percent strength of a solution as the grams of solute in 100 mL water. We have a 10% glycerol solution in 200 mL water.

Therefore, mass of glycerol required = (10 g/100 mL)*(200 mL) = 20 g (ans).

C) As per the definition of percent strength, 100 mL solution will contain 25 g Saroskyl.

Therefore,

3.0 L solution will contain (3.0 L)*(1000 mL/1 L)*(25 g/100 mL) = 750 g Saroskyl (ans).