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Use the solubility curve provided on the right to determine the answers to the following questions: Solubility Curves for Selected Solutes How many grams of solute are required to saturate 100 g of water in each of the following solutions? a) KCL at 80"C b) KCIOz at 90"C c) NaNOs at 10"C SOz at 20 "C e) NH-Cl at 70"C 150 140 130 120 110 NaNO; 100 KNO, 80 1 Hcl NHacl What is each of the solutions below: NH; Kcll saturated, unsaturated or 40 supersaturated? All of the solutes are Nacl KcIOg mixed with 100 g of water: 40 g of NaCl at 50"C b. 30 g of NHs at 30"C SO 70 8 of HCI at 20"C 30 40 50 60 70 90 100 d, 80 g of KNOs at 60"C Temperature (PC) 80 g of NH-Cl at 80"C
(ii) nature of the solvent ((intermolecular forces) (iii) temperature (solubility curves) (i) Using less than the tabulated mass of solute in 100 mL water results in an unsaturated solution. (ii) Using the same, or greater, mass of solute than that tabulated in 100 mL of water results in a saturated solution. (i) mass of solute in grams per 100 mL of water (w/v % or m/v concentration) (ii) mass of solute in grams per 100 g of water (weight ratio percentage concentration, or, mass ratio percentage concentration) Water is a commonly used solvent, so it is very useful to construct a table of solubilities based on the mass of a solute that will dissolve in a given volume of water.(i) nature of the solute ((intermolecular forces)
Solubility tables (tables of solubility) usually tabulate the maximum mass in grams of the solute that can be dissolved in 100 mL of water.
This data in the table is relevant only to the temperature given for the table. Commonly, solubility data is given for 25°C.
At 25°C and 101.3 kPa (1 atm) the density of water is 1.00 g mL-1
That is, 1 g of water has a volume of 1 mL at 25°C
Therefore, 100 g of water will have a volume of 100 × 1 mL = 100 mL at 25°C
So at 25°C and 101.3 kPa, the solubility of a solute in water given as mass in grams per 100 g water is the same as the solubility of the solute given as mass in grams per 100 mL of water.
That is, at 25°C: solubility in g/100 g water = solubility in g/100 mL water
Example: the solubility of sodium chloride (NaCl) in water at 25°C is 36 g/100 mL , or, 36 g/100 g
A solute is usually considered to be soluble in water if more than 1.0 g can be dissolved in 100 mL of water.
A solute is usually considered to be insoluble in water if less than 0.1 g can be dissolved in 100 mL of water.
It is important to note that "insoluble" does NOT necessarily mean that the solute cannot be dissolved, it only means that an extremely small amount of the solute can be dissolved in the solvent.
If the solubility of a substance is 0 g/100 mL, then, and only then, can we say that none of the solute dissolves in the solvent.
A solute is usually considered to be slightly soluble, or sparingly soluble, in water if between 0.1 and 1.0 g can be dissolved in 100 mL of water.
The solubility table below gives the maximum mass of solute in grams that can be dissolved in 100 mL of water at 25°C.
silver nitrate | AgNO3(s) | 245 | soluble |
sodium hydroxide | NaOH(s) | 80 | |
sucrose | C12H22O11(s) | 70 | |
lead(II) nitrate | Pb(NO3)2(s) | 60 | |
ammonia | NH3(g) | 48 | |
glucose | C6H12O6(l) | 45 | |
ammonium chloride | NH4Cl(s) | 39 | |
sodium chloride | NaCl(s) | 36 | |
potassium chloride | KCl(s) | 35 | |
magnesium sulfate | MgSO4(s) | 18 | |
copper(II) sulfate | CuSO4(s) | 14 | |
bromine | Br2(g) | 3.5 | |
chlorine | Cl2(g) | 0.6 | slightly soluble |
carbon dioxide | CO2(g) | 0.15 | |
calcium hydroxide | Ca(OH)2(s) | 0.12 | |
iodine | I2(s) | 0.03 | insoluble |
barium sulfate | BaSO4(s) | 0.001 | |
silver chloride | AgCl(s) | 0.0002 | |
sulfur | S(s) | 0 |
Sodium chloride, NaCl, is a soluble salt. From the solubility table above we see that the solubility of sodium chloride is 36 g/100 mL water at 25°C.
- 1 g of NaCl(s) will dissolve in 100 mL of water at 25°C.
This solution will be an unsaturated aqueous solution of sodium chloride. - 25 g of NaCl(s) will dissolve in 100 mL of water at 25°C.
This solution will be an unsaturated aqueous solution of sodium chloride. - 36g of NaCl(s) will dissolve in 100 mL of water at 25°C.
This solution will be a saturated aqueous solution of sodium chloride, but there will be no visible solid NaCl in the vessel. - If 40 g of NaCl(s) is added to 100 mL of water at 25°C, only 36g of NaCl(s) will dissolve, 4 g of the NaCl(s) will not dissolve.
This solution will be a saturated aqueous solution of sodium chloride and there will be visible solid NaCl in the vessel.
We can say that this excess NaCl(s) precipitates out of the solution.
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Question 1: A student has been given 250 mL of water at 25°C and needs to add enough calcium hydroxide to make a saturated solution. The solubility of calcium hydroxide at 25°C is 0.12 g/100 mL water.
What is the minimum mass, in grams, of calcium hydroxide that the student must add to the water?
Response:
What is the question asking you to do?
Calculate the mass of calcium hydroxide in grams.
m(calcium hydroxide) = ? g
What information have you been given in the question?
volume of water = 250 mL (25°C)
solubility of calcium hydroxide = 0.12 g/100 mL water at 25°C
What is the relationship between what you have been given and what you need to find out?
m(calcium hydroxide,g)/250 mL : 0.12 g/100 mL
Perform the calculation
- Write the equation:
- Rearrange the equation by multiplying both sides of the equation by 250 mL:
- Solve the equation:
m(calcium hydroxide) 250 mL | = | 0.12 g 100 mL |
m(calcium hydroxide) × 250 mL 250 mL | = | 0.12 g × 250 mL 100 mL |
m(calcium hydroxide) | = | 0.12 g × 250 100 |
m(calcium hydroxide) | = | 0.12 g × 250 100 |
m(calcium hydroxide) | = | 0.30 g |
Is your answer plausible?
250 mL is 2.5 times more than 100 mL, so if 0.12 g calcium hydroxide dissolved in 100 mL water produces a saturated solution, then 0.12 × 2.5 = 3.0 g calcium hydroxide dissolved in 250 mL water will produce a saturated solutution.
State your solution to the problem.
The minimum mass of calcium hydroxide required is 0.30 g
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Question 2: A student adds 0.01 g of solid iodine to 20 mL of water at 25°C. The solubility of iodine is 0.03 g/100 mL water at 25°C.
How much iodine, in grams, should precipitate out of the solution?
Response:
What is the question asking you to do?
Calculate the mass of iodine precipitated in grams.
m(undissolved iodine) = ? g
What information have you been given in the question?
m(total iodine) = 0.01 g
volume of water = 20 mL (25°C)
solubility of iodine = 0.03 g/100 mL water at 25°C
What is the relationship between what you have been given and what you need to find out?
m(dissolved iodine)/20 mL : 0.03 g/100 mL
m(undissolved iodine) = m(total iodine) - m(dissolved iodine)
Step 1: Calculate mass of iodine that dissolves
- Write the equation:
- Rearrange the equation by multiplying both sides of the equation by 20 mL:
- Solve the equation:
m(dissolved iodine) 20 mL | = | 0.03 g 100 mL |
m(dissolved iodine) × 20 mL 20 mL | = | 0.03 g × 20 mL 100 mL |
m(dissolved iodine) | = | 0.03 g × 20 100 |
m(dissolved iodine) | = | 0.006 g |
Step 2: Calculate mass of iodine that does not dissolve
m(undissolved iodine) = m(total iodine) - m(dissolved iodine)
m(undissolved iodine) = 0.01 g - 0.006 g = 0.004 g
Is your answer plausible?
Since we have been asked to calculate the mass of iodine that precipitates out of solution, we know that the mass of iodine added must have exceeded 0.03 g/100 mL water. 0.03 g per 100 mL = (0.03 g / 100 = 0.0003 g) per 1 mL 0.0003 × 20 = 0.006 g per 20 mL 0.01 g is greater than 0.006 g so our answer is reasonable.
Since only 0.006 g can dissolve in 20 mL, the rest, 0.01 - 0.006 = 0.004 g must not have dissolved.
State your solution to the problem.
The mass of iodine that precipitates out of solution is 0.004 g
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