Question

10B 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 6.31  M stock solution do you need to prepare 100.  mL of a 0.2227  M solution of HNO3?

Express the volume to three significant figures with the appropriate units.

10C 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 5.92×10⁻⁶  M . What was the concentration of the original solution?

Express the concentration to three significant figures with the appropriate units.

11A 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 5.92×10⁻⁶  M . What was the concentration of the original solution?

Express the concentration to three significant figures with the appropriate units.

11B A 429-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 18.1 g CO2. What was the concentration of the HCl solution?

Answer 1

10 B) What volume of a 6.31  M stock solution do you need to prepare 100.  mL of a 0.2227  M solution of HNO3?

M1 = 6.31 M

V1 = ?

M2 = 0.2227

V2 = 100 mL

we know that M1V1 =M2V2

V1 = M2V2/M1

= 0.2227 M x 100 mL / 6.31 M

= 3.53 mL

V1 = 3.53 mL

Therefore,

3.53 mL of 6.31  M stock solution is needed to prepare 100 mL of a 0.2227  M solution of HNO3.

11 B) A 429-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 18.1 g CO2. What was the concentration of the HCl solution?

Na2CO3 + 2HCl ------------> 2 NaCl + H2O + CO2

2mol 1mol

2 x 36.5 g = 73 g 44 g

? 18.1 g

? = 73 g x (18.1 g/ 44 g)

= 30.03 g HCl

Hence, mass of HCl required = 30.03 g

Given that volume of HCl = 429 mL = 0.429 L

Concentration of HCl = (mass/ molar mass) x (1/volume in Litres)

=  (30.03 g/ 36.5 g/mol) x (1/0.429 L)

= 1.92 M

Therefore,

concentration of the HCl solution = 1.92 M