CH 101/Keys 8
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| CH 101 |
|---|
| Exercises 1 |
| Exercises 2 |
| Exercises 3 |
| Exercises 4 |
| Exercises 5 |
| Exercises 6 |
| Exercises 7 |
| Exercises 8 |
| Exercises 9 |
| Exercises 10 |
| Exercises 11 |
| Exercises 12 |
| Exercises 13 |
| Exercises 14 |
| Exercises 15 |
|
Back to CH 101 |
Contents |
Exercises
Some Ksp values (same standard state)
- AgBr 5.2 x 10−13
- AgI 8.3 x 10−17
Exercise 1
- How much NaCl(s) (by weight) can be added to a 100 ml solution of 3mM AgNO3 before AgCl starts precipitating?
- AgNO3 ==> Ag+ +NO3- is a strong electrolyte
- thus [Ag+]= 3mM
- Ksp for AgCl= 1.8 x 10−10
- The maximum chloride concentration is therefore 1.8 x 10−10/ 3 10−3 = .6 x 10−7
- In 100 ml this amounts to 6 x 10−9 mol.
- 6 x 10−9 [mol]. of NaCl = 6 x 10−9 [mol] *(23+35.5) [g/mol] = .35 [μg]
- How much NaBr(s) (by weight) can be added to a 10 l solution of .2mM AgNO3 before AgBr starts precipitating?
- 5.2 x 10−13 / .2 x 10−3 = 26 x 10−10 is the max. concentration
- In moles: 26 x 10−10 x 10 l = 2.6 10−8 moles of NaBr allowed
- 2.6 10−8 moles * 103 [g/mol] = 2.6 [μg]
- How much NaI(s) (by weight) can be added to a 300 ml solution of 1μM AgNO3 before AgI starts precipitating?
- 8.3 x 10−17 / 10−6 = 8.3 x 10−11 max conc.
- in moles of I-: 8.3 x 10−11 *.3 = 2.1 x 10−11 mole
- in weight: 3.7 [ng]
Exercise 2
- What is the mass of silver dissolved in 3000 l of a solution saturated in AgCl in pure water?
- The [Ag+] is the square root of Ksp = 1.35 x 10−5
- Number of moles dissolved in 3000 l: 3000*1.35 x 10−5 = 4 x10−2 [mol]
- By mass: 4 x10−2 [mol] * 107.8 [g/mol] = 4.3 [g]
- What is the mass of silver dissolved in 3000 l of a solution saturated in AgI in pure water?
- Could you advise a mining company what to do with solutions containing low concentrations of silver?
- Reduce by evaporation and add salt! NaBr and NaI would work better but are probably too expensive. Sulfide would work great too.
Exercise 3
- How many moles of nitric acid do we need to add to a 10 liter drum of 18 megohm water to raise the [H+] to 10-5 [mol/l]? What is the resulting [OH-]? (Hint: First ignore the contribution [H+]native from the water itself. Then do the calculation. Then argue that the assumption was valid because [H+]native is equal to [OH-]). How many times larger is [H+] compared to [H+]native?
- We need 10x10-5 mol= 0.1 mmol nitric acid to acidify the drum (not a lot..)
- The resulting [OH-] = 10-14/10-5 = 10-9
- This is also the native proton concentration from the water dissociation reaction
- There are 10,000 times more protons from the nitric acid than from the water dissociation.