Problem #1:
The osmotic pressure of a 1.0 x 10¯2 M solution of cyanic acid (HOCN) is 217.2 torr at 25 °C. Calculate the Ka for HOCN.
The solution:
Examine the osmosis equation:
π = i M R T
for what is given and what is not given in the problem.
We have:
1) osmotic pressure (π) = 217.2 torr
2) molarity (M) = 1.0 x 10¯2-molar
3) a constant (R) = 0.08206 L atm mol¯1 K¯1
4) temperature (T) = 25 °C = 298 K
We lack only the van 't Hoff factor. So let's calculate it. Notice I've included a conversion of torr to atm.
(217.2 torr / 760.0 torr/atm) = x (1.0 x 10¯2 mol L¯1) (0.08206 L atm mol¯1 K¯1) (298 K)
This all works out to:
x = 1.17
Notice that ALL the units cancel.
I want to pause for a moment and ponder the physical meaning of i and exactly what 1.17 means. Keep in mind that i is associated with the degree of dissociation.
Imagine, if you will the making of 0.010 M HOCN. Someone weighs out the necessary grams of HOCN to equal 0.010 mole of it. Then they dissolved it in some distilled water and added sufficient water to then make 1.0 L of the solution. The person was very careful at each step, used appropriate glassware that was carefully cleaned, etc, etc.
What the person created is called the analytical concentration. If we evaporated all the water, we would recover 0.010 mol of HOCN. However, in the solution it is a different story and i = 1.17 is what tells us that.
The 1.17 tells us that the HOCN molecules we put in (call it 100%) is actually creating an osmotic pressure that would require 117% of what we put in, IF the molecules did not dissociate. In fact, they did dissociate, enough so that the solution is acting like it is 0.0117 M and not 0.010 M.
Let's say that again. 0.010 mole of HOCN was put into the solution, Some, not all, of the molecules fell apart into H3O+ and OCN¯. When all the individual non-water particles are counted up, it all comes out to 0.0117 M.
But how will that help us? Aha! Onward!
The next step is to ponder the Ka expression:
Ka = ( [H3O+] [OCN¯] ) / [HOCN]
For me, the key is to see that, since the Ka value is the unknown, ALL the other values must be known. So then the question becomes, how to figure out [H3O+], [OCN¯] and [HOCN]? For this, we turn to the equation of dissociation:
HOCN <===> H3O+ + OCN¯
We see that [H3O+] = [OCN¯] upon dissociation. Let us call that value 'y.' Therefore the [HOCN] is 0.010 - y and here comes a key statement:
All three values sum up to 0.0117
Therefore:
y + y + 0.010 - y = 0.0117y = 0.0017
Now we have all the data to solve the last step, plugging into the Ka expression and solving it. I get 3.5 x 10¯4. I'll leave you to check my answer.
Problem #2:
A 0.035 M aqueous nitrous acid (HNO2) solution has an osmotic pressure of 0.93 atm at 22.0 °C. Calculate the percent ionization of the acid.
The solution:
Examine the osmosis equation:
π = i M R T
Let us calculate the van 't Hoff factor:
0.93 atm = i (0.035 mol L¯1) (0.08206 L atm mol¯1 K¯1) (295 K)i = 1.1
This tells use that the true concentration (when all three types of non-water particles are added up) is 0.0384 M.
Next, consider the dissociation equation:
HNO2 <==> H+ + NO2¯
We need to get [H+].
Let [H+] = x; therefore [NO2¯] = x and [HNO2] = 0.035 - x. (If you're not sure where the x, x and 0.035 - x came from, write down the dissociation equation for HNO2.) Now, we need to add up the molarities of the three solute particles to equal 0.0384.
Therefore:
x + x + 0.035 - x = 0.0384x = 0.0034 M
We are now ready to calculate the percent ionization:
(0.0034 / 0.035) times 100An alternate solution to problem #2.percent ionization = 9.8%
Problem #3:
A solid mixture contains MgCl2 and NaCl. When 0.5000 g of this mixture is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25.0 °C is observed to be 0.3990 atm. What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution.)
The solution:
1) Calculate the apparent molarity of the solution:
0.3990 atm = (x) (0.08206) (298.15)x = 0.01630822 M
Comment: in reality, the above calculated the van 't Hoff factor times the analytical concentration. Using the provided data, we cannot individually get to either the van't Hoff factor or the analytical concentration. However, neither individual value is required.
2a) We know the following: MgCl2 yields 3 particles per formula unit and its molar mass is 95.211 g mol¯1. Let us assume the sample is 100% MgCl2. Therefore:
(0.5000 g / 95.211 g mol¯1) / 1.000 L = 0.052515 M0.052515 M times 3 = 0.0157545 M (in terms of total particles)
2b) We know the following: NaCl yields 2 particles per formula unit and its molar mass is 58.443 g mol¯1. Let us assume the sample is 100% NaCl. Therefore:
(0.5000 g / 58.443 g mol¯1) / 1.000 L = 0.008555345 M3) Let x = percent MgCl2 (expressed as a decimal). Therefore:0.008555345 M times 3 = 0.0171107 M (in terms of total particles)
(% MgCl2) (0.0157545 M) + (% NaCl) (0.0171107 M) = 0.01630822 M(x) (0.0157545 M) + (1 - x) (0.0171107 M) = 0.01630822 M
0.0013562x = 0.00080248
x = 0.5917
Sample is 59.17% MgCl2.
Comment: note the analytical concentration times the van 't Hoff factor in 2a and 2b. This puts all three concentrations used in the calculation in #3 on the same footing.
Comment: note the use of 'x' and '1-x.' There are only two compounds in the solid mixture. If the percentage of one is 'x,' then the other must be '1-x.' These two ('x' and '1-x) add up to 1, which is 100% expressed as a decimal.
Problem #4:
A solution is prepared by dissolving 35.0 g of hemoglobin in enough water to make up 1.00 L in volume. The osmotic pressure of the solution is found to be 10.0 mmHg at 25.0 °C. Calculate the molar mass of hemoglobin.
Solution:
1) calculate the molarity (NOT the moles) of the solution (notice the conversion of mmHg to atm):
π = i M R T(10.0 mmHg / 760.0 mmHg atm¯1) = (1) (x) (0.08206 L atm mol¯1 K¯1) (298 K)
x = 0.00053807 M
(Note that the van 't Hoff factor is assumed to be one, a VERY safe assumption with hemoglobin.)
2) Determine moles of solute:
M = moles / liters of solution0.00053807 mol L¯1 = x / 1.00 L
x = 0.00053807 mol
3) Calculate the molar mass:
molar mass = g / molx = 35.0 g / 0.00053807 mol
x = 65,047 g/mol (Yes, this is a correct answer.)
Problem #5: Osmosis is the process responsible for carrying nutrients and water from groundwater supplies to the upper parts of trees. The osmotic pressures required for this process can be as high as 18.6 atm. What would the molar concentration of the tree sap have to be to achieve this pressure on a day when the temperature is 27.0 degrees Celsius?
Solution:
Use the van 't Hoff Equation:
π = iMRT18.6 atm = (1) (x) (0.08206 L atm / mol K) (300 K)
x = 0.756 M (to three sf)
Problem #6: Isotonic saline solution, which has the same osomotic pressure as blood, can be prepared by dissolving 0.923 grams of NaCl in enough water to produce 100.0 mL of solution. What is the osmotic pressure, in atmospheres of this solution at 25.0 °Celsius?
Solution:
1) Determine the molarity of the NaCl:
MV = g/molar mass(x) (0.1000 L) = 0.923 g / 58.443 g mol¯1
x = 0.15793166 M
2) Use the van 't Hoff Equation:
π = iMRTx = (2) (0.15793166 mol / L) (0.08206 L atm / mol K) (298 K)
x = 7.72 atm (to three sf)
Note: due to ion pairing, the experimentally determined van 't Hoff factor for NaCl equals 1.8. Using this value, rather than the theoretical value of 2, gives 6.95 atm.
Problem #7: For rehydration theraphy for cholera patients, the World Health Organization uses an aqueous solution with the following concentration: 3.50 g NaCl/L, 2.50 g NaHCO3/L, 1.50 g KCl/L and 20.0 g glucose/L. Calculate the osmolarity of the resulting solution.
Solution: The osmolarity of a solution is equal to the molarity times the the van 't Hoff factor (the number of particles produced in solution per formula unit).
1) Molarity of each of the four substances (I kept some guard digits):
NaCl ⇒ 3.50 g / 58.443 g/mol = 0.0598874 mol
NaHCO3 ⇒ 2.50 g / 84.0059 g/mol = 0.0297598 mol
KCl ⇒ 1.50 g / 74.551 g/mol = 0.020120
glucose ⇒ 20.0 g / 180.1548 g/mol = 0.1110156 mol
2) Multiply each molarity by the van 't Hoff factor of that substance:
NaCl ⇒ 0.0598874 mol/L x 2 = 0.1197748 mol/L
NaHCO3 ⇒ 0.0297598 mol/L x 2 = 0.0595196 mol/L
KCl ⇒ 0.020120 mol/L x 2 = 0.040240 mol/L
glucose ⇒ 0.1110156 mol/L x 1 = 0.1110156 mol/L
3) Add for the final answer:
0.1197748 + 0.0595196 +0.040240 + 0.1110156 = 0.33055 moles of particles per liter (to three sig figs, this is 0.330)
Problem #8: You add 10.0 grams of NaCl to one liter of water and 20.0 grams of glucose to another liter of water. Which one has a higher osmolarity?
Solution:
1) Calculate the molarity of each substance.
2) Multiply each solution's molarity by the van 't Hoff factor for the solute.
3) Compare.
Problem #9: A biochemical engineer isolates a bacterial gene fragment and dissolves a 17.6 mg sample of the material in enough water to make 31.5 mL of solution. The osmotic pressure of the solution is 0.340 torr at 25.0 °C.
(a) What is the molar mass of the gene fragment?
(b) If the solution density is 0.997 g/mL, how large is the freezing point depression for this solution?
Solution, part a:
1) Calculate molarity of sample:
(0.340 torr / 760.0 torr atm¯1) = (1) (x) (0.08206) (298 K)x = 1.829437 x 10¯5 M
2) Calculate moles of solute present in 31.5 mL:
(1.829437 x 10¯5 mol/L) (0.0315 L) = 5.7627 x 10¯7 mol
3) Calculate molecular weight:
0.0176 g / 5.7627 x 10¯7 mol = 3.05 x 104 g/mol
Solution, part b:
1) Calculate molality of solution:
mass of solution: 31.5 mL x 0.997 g/mL = 31.4055 gmass of water: 31.4055 g - 0.0176 g = 31.3879 g
molality calculation: 5.7627 x 10¯7 mol/ 0.0313879 kg = 1.83596 x 10¯5 molal
2) Calculate freezing point depression:
Δt = (1) (1.86 °C/m) (1.83596 x 10¯5 m)Δt = 3.42 x 10¯5 °C
In case you missed it, the point to part b is to show how difficult it would be to use freezing point depession to determine the molecular weight in this example. The amount of freezing point depression would take very specialized equipment to determine. Whereas, the change of 0.340 torr (or, 0.340 mm Hg) is very easy to measure using equipment commonly available in most universities and even some well-equiped high schools.
Problem #10: Consider a 4% starch solution and a 10% starch solution seperated by a semipermeable membrane, which starch solution will decrease in volume as osmosis occurs?
a) 4%
b) 10%
c) Neither exerts osmotic pressure
d) They exerts equal osmotic pressure
e) They exert opposite osmotic pressure
Answer: Diffusion is a entropically favorable spontaneous movement of molecules to create equal distributions. Without a membrane present, starch would diffuse from the 10% solution toward the 4% solution, and water would diffuse from the 4% solution (where water is more abundent) to the 10% solution (where water is more scarce). However, since starch molecules are too large to pass through a semipermiable membrane, only the water will move across the membrane, causing the 4% solution to decrease in volume.
11) A 0.102 g sample of an unknown compound dissolved in 100.0 mL of water has an osmotic pressure of 28.1 mmHg at 20.0 °C. Calculate the molar mass of the compound.
12) To determine the molar mass of an unknown protein, 1.00 mg was dissolved in 1.00 mL of water. At 25.0 °C the osmotic pressure of this solution was found to be 1.47 x 10-3 atm. What is the molar mass of this protein?
13) Arrange the following solutions in order of decreasing osmotic pressure:
(a) 0.10 M urea
(b) 0.06 M NaCl
(c) 0.05 M Ba(NO3)2
(d) 0.06 M sucrose
14) The colligative properties of a 1.26-molar solution of calcium chloride has the same colligative proerties as a _____ M solution of sucrose.
15) What is the value of the van't Hoff factor for a 0.64 molar aqueous solution of acetone?
16) 2 solutions, a 0.1% (m/v) albumin solution (compartment A) and a 2% (m/v) albumin solution (compartment B), are separated by a semi-permeable membrane. (Albumins are colloidal proteins).
a) Which compartment will have the higher osmotic pressure?Answer:
b) Which compartment will lose water?
c) If some NaCl is added to compartment A, will the sodium and chloride ions stay in compartment A, or will some cross the membrane to compartment B?
a) compartment B due to the higher albumin concentration in B.
b) compartment A. Since it has the lower albumin concentration, it has the higher water "concentration."
c) As long as the membrane isn't selective against Na+ and Cl¯ ions, then they will flow to B so they attain equilibrium. Why can the Na+ and Cl¯ flow through the membrane but not albumin? Albumin is much lager than the NaCl ions. That's why in part B, compartment A loses water to compartment B instead of compartment B losing albumin to compartment A.