Question

You need to make a `1` `mu"M"` solution from a 1.0 M stock solution. If you start with 1.0 mL of the stock solution: (a) How much solvent is needed to make a solution with the needed concentration in one dilution. (b) How much solvent will be needed?

You need to make a `1` `mu "M"` (micromolar) solution from a 1.0 M stock solution. If you start with 1.0 mL of the stock solution: (a) How much solvent is needed to make a solution with the needed concentration in one dilution. (b) How much solvent will be needed to make a solution with the needed concentration if a series of three 1:100 (old volume: new volume) dilutions are done? From each dilution 1.0 mL of the new solution will be diluted until the final concentration is reached. What are the concentrations of the intermediate solutions?

Answer 1

(a) 1000 L
(b) 297 mL; 10 mmol/L; 100 µmol/L; 1 µmol/L.

Explanation:

(a) Single diution

You can use the dilution formula to calculate the volume:

`color(blue)(bar(ul(|color(white)(a/a)c_1V_1 = c_2V_2color(white)(a/a)|)))" "`

where `c_1` and `c_2` are the concentrations and `V_1` and `V_2` are the volumes.

You can rearrange this formula to get

`V_2 = V_1 × c_1/c_2`

Your data are:

`c_1 = "1.0 mol/L;" color(white)(mmml)V_1 = "1.0 mL"`
`c_2 =color(white)(l) 1 × 10^"-6" color(white)(l)"mol/L"; color(white)(l)V_2 = ?`

`V_2 = "1.0 mL" × (1.0 color(red)(cancel(color(black)("mol/L"))))/(1 × 10^"-6" color(red)(cancel(color(black)"mol/L"))) = 1 × 10^6 color(white)(l)"mL" = "1000 L"`,

(b) Serial dilutions

The general formula is for calculating a single dilution factor is

`"DF" = V_f/V_i`

(i) The first dilution.

For a 1:100 dilution, add 1 mL of the stock solution to 99 mL solvent.

`c = "1.0 mol/L" × (1color(red)(cancel(color(black)("mL"))))/(100 color(red)(cancel(color(black)("mL")))) = "0.01 mol/L = 10 mmol/L"`

(ii) The second dilution.

Add 1 mL of the 10 mmol/L solution to 99 mL solvent.

`c = "10 mmol/L" × (1color(red)(cancel(color(black)("mL"))))/(100 color(red)(cancel(color(black)("mL")))) = "0.10 mmol/L = 100 µmol/L"`

(iii) The third dilution.

Add 1 mL of the 100 µmol/L solution to 99 mL solvent.

`c = "100 µmol/L" × (1color(red)(cancel(color(black)("mL"))))/(100 color(red)(cancel(color(black)("mL")))) = "1 µmol/L"`

`"Total solvent used = 3 × 97 mL = 297 mL"`