1. Salts and Solubility Activity 2
Learning Goals: Students will be able to:
Write the dissolving reaction for salts
Describe a saturated solution microscopically and macroscopically with supporting illustrations
Calculate solubility in grams/100ml
Distinguish between soluble salts and slightly soluble salts macroscopically.
Trish Loeblein July 2008

2. 1. Which is correct for dissolving barium iodide in water ?
BaI2(s)  Ba(aq) + 2I(aq)
BaI(s)  Ba(aq) + I(aq)
BaI2(s)  Ba+2(aq) + 2I-(aq)
BaI2  Ba I- C

3. 2. Sue used Salts to learn about “saturated solution”
2. Sue used Salts to learn about “saturated solution”. Which image best shows a saturated solution? A B
C is best answer because there are just few stuck to wall
A does not show complete dissolving so it hard to tell if they will all stay in solution
B could be, but C shows more particles do dissolve
D is past the saturation point. C D

4. 3. Waldo added salt to a test tube of water to learn about “saturated solution”. Which image best shows a saturated solution?
B because it looks like all the others are past the saturation point.
C (same reasons as previous question)
A B C D

5. 4. If you used the sim to test silver chloride, you would see 80 Ag+ ions dissolved in 1E-17 liters. What is the solubility in 100 ml of water?
.0019 grams/100 ml water
grams/100 ml water
.0014 grams/100 ml water
grams/100 ml water
I wanted this to be a “real problem”. Here’s how I determined how many particles would be correct:
K sp at 25 degrees =1.8×10–10 so 1.34E-5 moles/liter for each ion and the default volume is 1E-16L for slightly soluble. This would have given 807 particles; I decided to change the volume since the students might remember that the sim would show that many particles.
1.34E-5 moles/liter *6.02E23 particles/mole * 1E-17 L= 80.8 particles in 1E-E-17 L
Ksp, for AgCl is 1.8 x 10-10, which indicates that one liter of water will dissolve grams of AgCl. From
There are many variations for the value of Ksp and solubility published; I chose this reference since it both for the same Temperature, but you’ll see by my calculations on the next slide that I have a very different solubility. I decided that they probably did not have the same temperature after all and decided not to bother since I had already checked several sources for values.

6. The calculation for AgCl example: 80 AgCl /6. 02E23 AgCl/mole). (143
The calculation for AgCl example: 80 AgCl /6.02E23 AgCl/mole) *(143.5grams/mole) = 1.9E –20 grams 1.9E –20 grams/(1E-17L) = grams/L grams/L* .1L/100ml= g/100ml
If you forget to use the molecular mass and use just silver at 108 you get g/ml B

7. Sodium chloride Silver Chloride This is not an identifying test
5. You knew a salt was either sodium chloride or silver chloride. If you put 1 gram in 10 ml of water in a test tube, and it looked like this
Which is it?
Sodium chloride
Silver Chloride
This is not an identifying test
B Since the solubility of AgCl is .0024g/L then only about 2.4E-5 would dissolve in 10 ml. Sodium chloride is 35 g in 100 ml (3.5g in 10 ml) so it would dissolve.