Making
Molar
Solutions
From
Liquids
(More accurately, from
stock solutions)
Making molar solutions from liquids
Not all compounds are in a solid form
Acids are purchased as liquids (“stock
solutions”). Yet, we still need a way to make
molar solutions of these compounds.
The Procedure is similar:
Use pipette to measure moles (via volume)
Use volumetric flask to measure volume
Now we use the equation M1V1 = M2V2
1 is starting (concentrated conditions)
2 is ending (dilute conditions)
Reading a pipette
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
Practice using a pipette
• Always keep pipette vertical
• To rinse: take up water, remove green filler,
rotate pipette, replace filler, expel water
• If filler can not take up or expel enough
liquid, remove, place finger over pipette,
turn knob, replace filler.
• Take up water to 0 mark. Measure 3.2 mL
into 10 mL cylinder. (one per person)
• If drop is hanging off, touch to cylinder
• Repeat with 1.7 mL and 5.1 mL
__________________________________
• If done correctly you should get 10 mL in
graduated cylinder
The Dilution formula
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 L
M1V1 = M2V2, M1V1/V2 = M2
M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M
Why does the formula work?
Because we are equating mol to mol:
V1 = 1 L
M1 = 3 M
M1V1 = 3 mol
V2 = 6 L
M2 = 0.5 M
M2V2 = 3 mol
Practice problems
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 L
M1V1 = M2V2, M1V1/M2 = V2
V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L
Q – 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 Do 1 – 8 on
handout. Try 6
# L = 1 L + 0.5 L = 1.5 L
two ways
# mol/L = 4 mol / 1.5 L = 2.67 mol/L
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?
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?
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 (1-6, 6 two ways)
1.
M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL
V1 = 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 L
M2 = 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
Dilution problems (4 - 6)
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/L
Or, using M1V1 = M2V2,
M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L
Dilution problems (7, 8)
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.
Practice making molar solutions
• Calculate # of mL of 1 M HCl required to make
100 mL of a 0.1 M solution of HCl
• Get volumetric flask, pipette, plastic bottle,
100 mL beaker, 50 mL beaker, eyedropper.
Rinse all with tap water. Dry 50 mL beaker
• Place about 20 mL of 1 M HCl in 50 mL beaker
• Rinse pipette, with small amount of acid
• Fill flask about 1/4 full with distilled water
• Add correct amount of acid with pipette. Mix.
• Add water to line (use eyedropper at the end)
• Place solution in plastic bottle
• Label bottle. Place at front of the room.
• Rinse and return all other equipment.
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