Expressing the Concentration of a Solute in a Solution

Several Ways of Expressing Concentration1. mol/L the most important

way of expressing concentration in chemistry. Why?

2. % (m/v)

3. % (m/m) to be covered in 4. % (v/v) future lessons

5. ppm

6. ppb

1. Molarity: mol solute per L solution

units are:

mol/L; mol·L-1; M or

C = n/V,

where n is # mol solute

V is volume of solution (L)

C is molarity of solute (mol/L)

NOTE: Solution =

solute + solvent

How to Prepare a Solution of Known Molarity

That is, how would you prepare 250 mL of a solution of potassium nitrate, KNO3, whose concentration (molarity) is 0.246 mol/L?

C = n/V,

C = 0.246 mol/L

V = 0.250 L

n = C*V = (0.246 mol·L-1)(0.250 L)

n = 0.0615 mol of KNO3 is required.

What now? Can we measure moles directly?

Need to convert mol to mass (g).

mass = (0.0615 mol of KNO3)*(101.1 g/mol)

mass KNO3 = 6.22 g

What must we do to get 6.22 g KNO3 in

250 mL of solution?

Do we simply add 6.22 g KNO3 to 250 mL of water?

No—solution is solute + solvent.

How do we deal with this?

We need special glassware . . .

Volumetric FlaskA volumetric flask is

specially

designed to contain (TC) a certain volume of solution.

How to use a Volumetric Flask

Put exact mass of solute in vol. flask; add some water to dissolve. Then fill to the line with water, inverting flask frequently to ensure good mixing.

(Use a to addfinal few drops of water.)

Volumetric flasks come in many standard sizes:

Sample Problems

1. How would you prepare 100 mL of a solution that is 1.38 mol/L Na2SO4. Include calculations and a description of how you would prepare the solution; mention any special glassware.

C = 1.38 mol/LV = 0.100 Ln = C*V = (1.38 mol/L)(0.100 L)

= 0.138 mol . . . now convert this tograms

mass = (0.138 mol)*(142.1 g/mol)

= 19.6 g Na2SO4

Transfer 19.6 g Na2SO4 to a 100 mL volumetric flask; dissolve; fill to the line with water. Use a medicine dropper to put the bottom of the meniscus on the line.

the line

2. What is the molarity of a solution obtained by dissolving 6.58 g of glucose, C6H12O6 in 50.0 mL of solution?

C = n/V

n = 6.58 g/(180.2 g/mol)

= 0.0365 mol

C = 0.0365 mol/0.0500 L

= 0.730 mol/L (or M)

Now do the practice with a volumetric flask. Assignment due at end of class.

HW. p 381 #41 to 50; do half a dozen

(Quiz next class on this lesson. Includes volumetric flask practice and the HW.)