When you need to produce a variety of diluted solutions of a solute, you can dil

Question

When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solution has a significantly higher concentration of the given solute (typically 101 to 104 times higher than those of the diluted solutions). The high concentration allows many diluted solutions to be prepared using minimal amounts of the stock solution.

What volume of a 6.38  M stock solution do you need to prepare 250.  mL of a 0.0794  M solution of HNO3?

The absorbance of a cationic iron(II) sample solution was measured in a spectrophotometer, but the instrument returned an error because the absorbance was too high. The sample was then diluted by using a pipette to take 100.0 L of the sample and injecting it into a cuvette already containing 2.00 mL of water. The absorbance value of the diluted solution corresponded to a concentration of 8.48×106  M . What was the concentration of the original solution?

Explanation / Answer

We can use dilution formula here which is

M1V1 = M2V2

where M1 and V1 are the concentration and volume of stock solution and

M2 , V2 are the concentration and volume of diluted solution.

Let's write down the given

M1 = 6.38 M

V1 = ?

M2 = 0.0794 M

V2 = 250 mL

Substituting in the formula, we get

6.38 M * V1 = 0.0794 M * 250 mL

V1 = 0.0794 M * 250 mL / 6.38 M

V1 = 3.11 mL

We need 3.11 mL of stock solution

________________________________________________________________________________________________

here, the volume of original solution we used , V1 = 100 uL

It was diluted by adding to 2 mL water. Therefore final volume becomes

100 uL + 2 mL

Before adding above values, we need the same units for both of them.

Let's convert 2 mL to uL

2mL * 1 L/ 1000 mL * 1000000 uL/ 1 L = 2000 uL

Therefore final volume, V2 = 2000 uL + 100 uL = 2100 uL

concentration of this diluted solution was recorded as , M2 = 8.48 x 10^-6 M

We need to find M1

Let's use dilution formula again

M1 * 100 uL = 8.48 x 10^-6 M * 2100 uL

M1 =  8.48 x 10^-6 M * 2100 uL / 100 mL

M1 = 1.781 x 0^-4 M

Concentration of original solution was 1.78 x 10^-4 M

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