Question

a) What volume of a 6.40  M stock solution do you need to prepare 500.  mL of a 0.0347  M solution of HNO3?

b) 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 μL of the sample and injecting it into a cuvette already containing 2.00 mL of water (total volume is 2.00 mL + 100.0 μL). The absorbance value of the diluted solution corresponded to a concentration of 8.60×10⁻⁶  M . What was the concentration of the original solution?

Answer 1

a) moles of HNO3 in 500ml of 0.0347 M HNO3 = 0.5 L x 0.0347 M = 0.01735 moles

volume of 6.40 M needed

6.40 M = 0.01735 mol / x

x = 0.0027 L

This calculates the volume in liters. Multiplying the answer by 1000 provides the required mL value:

0.0027 L x (1000 mL / L) = 2.7 mL

b) A=xbC

A=Absorbance x=absorbance b=length of solution (constant)
C=Concentration units
Absorbance=-log(Light total/Light through sample)

A1 / x1C1 = A2 / x2C2

concentration of the original solution = 9.02x10^-6 M