kg = 103 g eg. 1.6 kg = 1.6 x 103 gmg = 10-3 g 6 mg = 6 x 10-3 gμg = 10-6 g 0.6 mg = 6 x 10-4 g

Review of FormulasReview of Formulas

Making Molar

Solutions From Liquids(More accurately, from

stock solutions)

What are molar solutions?What are molar solutions?

A molar solution is one that expresses “concentration” in moles per volume

Molar solutions are prepared using:• a balance to weigh solids (in grams)• a pipette to measure small liquid volumes (μL/mL)• a volumetric flask to measure large volumes (L)

and for mixing

Molar Volume is measured in mol/L, ∴ we can use the equation c = n/V

• mol/L can be abbreviated as M or [ ]

Calculations with molar solutionsCalculations with molar solutionsQ: How many moles of NaCl are required to make

7.5 L of a 0.10 M solution?

But in the lab we weigh grams not moles, so …Q: How many grams of NaCl are required to make

7.5 L of a 0.10 M solution?

M=n/L, n = 0.10 M x 7.5 L = 0.75 mol# mol NaCl = 7.5 L x 0.10 mol NaCl

1 L= 0.75 mol

# g NaCl =7.5 L x 0.10 mol NaCl

1 L=43.83 gx 58.44 g NaCl

1 mol NaCl

Practice QuestionsPractice Questions1. How many grams of nitric acid are present in 1.0 L of a

1.0 M HNO3 solution?2. Calculate the number of grams needed to produce 1.00

L of these solutions: a) 1.00 M KNO3 b) 1.85 M H2SO4 c) 0.67 M KClO3

3. Calculate the # of grams needed to produce each:a) 0.20 L of 1.5 M KCl b) 0.160 L of 0.300 M HClc) 0.20 L of 0.09 mol/L AgNO3

d) 250 mL of 3.1 mol/L BaCl2 4. Give the molarity of a solution containing 10 g of each

solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)25. Describe how 100 mL of a 0.10 mol/L NaOH solution would be made.

63 g

101 g181 g 82 g

a) 22 g b) 1.75 gc) 3 g d) 0.16 kg

a) 0.041 mol/Lb) 0.054 mol/L

C, V, n, m if these variables are in the question (or required for your answer) use the following steps

e.g. Given c and V. Find m.Find # of moles (c=n/v)Convert to grams (n= m/M)

Preparing Standard SolutionsPreparing Standard Solutions This is a volumetric flask

It is used for preparing solutions (but not for storing them)

A standard solution is one with an accurate, known concentration. This is also known as a stock solution. • These are used as reactant solutions•They usually have a higher concentration than is needed for creating solutions and therefore must be diluted

Preparing Standard SolutionsPreparing Standard SolutionsAfter diluting a solution, the concentration of the solution changes. To calculate the new molar concentration, we use the equation:

c1V1 = c2V2

C = concentration; V = volume1 = initial (concentrated); 2 = final (diluted)

Don’t forget the equation for molar concentration! (c = n/V)

Identify each volume to two decimal places(values tell you how much you have expelled)

4.48 - 4.50 4.86 - 4.87 5.00

Reading a pipetteReading a pipette

E.g. if we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L?

M1 = 3 mol/L, V1 = 1 L, V2 = 6 LM1V1 = M2V2, M1V1/V2 = M2

M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M

M1V1 = 3 mol

V1 = 1 LM1 = 3 M

V2 = 6 LM2 = 0.5 M

M2V2 = 3 mol

The Dilution FormulaThe Dilution Formula

Q – What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl?

M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 LM1V1 = M2V2, M1V1/M2 = V2

V2 = (12 mol/L x 1 L) / (0.5 L) = 24 LQ – 1 L of a 3 M HCl solution is added to 0.5 L of a 2 M

HCl solution. What is the final concentration of HCl? (hint: first calculate total number of moles and total number of L)# mol = (3 mol/L)(1 L) + (2 mol/L)(0.5 L)

= 3 mol + 1 mol = 4 mol# L = 1 L + 0.5 L = 1.5 L# mol/L = 4 mol / 1.5 L = 2.67 mol/L

ExamplesExamples

1. How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?

2. You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L?

3. 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M?

Dilution problems Dilution problems

1. M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mLV1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)V1 = 0.03125 L = 31.25 mL

2. M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 LM2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L)M2 = 1.2 M

3. M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ?V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M)V2 = 0.4 L or 400 mL

4. What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl?

5. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl?

6. What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water?

Dilution problems (cont’d)Dilution problems (cont’d)

Mixing two solutions together, need to find new concentration

1.Calculate the amount of moles in the final solution (n=m/M) n1+n2=n(final)

2.Calculate the final volume after mixing the two solutions together

3.From #1 & 2 calculate final concentration

4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L) = 0.8 mol + 1.8 mol = 2.6 mol # L = 0.4 L + 0.6 L # mol/L = 2.6 mol / 1 L = 2.6 mol/L

5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L) = 1.5 mol + 0.4 mol = 1.9 mol # mol/L = 1.9 mol / 5 L = 0.38 mol/L

6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L) = 1.5 mol + 0 mol = 1.5 mol # mol/L = 1.5 mol / 5 L = 0.3 mol/LOr, using M1V1 = M2V2,M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L

7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution?

8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out, what is the concentration of the remaining HF?

Dilution problems (cont’d)Dilution problems (cont’d)

7. M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ?V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M)V2 = 16 L

8. The concentration remains 0.2 M, both volume and moles are removed when the solution is poured out. Remember M is mol/L. Just like the density of a copper penny does not change if it is cut in half, the concentration of a solution does not change if it is cut in half.

Homework

• Read Table 8.7 and Table 8.8 thoroughly to prepare for Friday’s lab – MANDATORY!!