CH 101/Keys 10
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| CH 101 |
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| 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 |
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Excercises
An excess of the sparingly soluble solid barium hydroxide Ba(OH)2 is allowed to equilibrate with water at room temperature. Under these conditions the solubility product Ksp= 4.10-18
- Write a balanced chemical equation for the equilibrium
- This works much like dissolving a salt: inasmuch as the hydroxide dissolves it does so by dissociating into ions.
- As wiki does not allow me to draw proper equilibrium arrows I will just use < = > instead
- Ba(OH)2(s) (+ H2O) < = > Ba2+(aq) + 2.OH-(aq)
- Notice the coefficient 2, needed to balance the equation
- Notice that I faithfully do write the states: (s) and (aq), because of the activity issue: for (s) activity=1, whereas for dilute (aq) activity = concentration
- Write the expression for Ksp in terms of the concentrations.
- Ksp is a glorified equilibrium constant of the above dissolution equilibrium. You can always compose those by taking activities or the products over those of the reactants.
- Thus: Ksp = a(Ba2+).a(OH-)2 / a(Ba(OH)2
- Notice the square that results from the coefficient above.
- Notice that in the previous answer I faithfully wrote the states: (s) and (aq), because of the activity issue: for (s) activity=1, whereas for dilute (aq) activity = concentration
- Thus Ksp = [Ba2+].[OH-]2
- Calculate the [Ba2+] as well as the [OH-] of the resulting solution
- Hmmm, can we do that? Actually yes, if we assume that the solid is the only source of Ba2+ and OH-. In that case we can use the balanced equation and say that in a liter of solution, for every x moles of Ba2+ exactly 2x moles of OH- must be in solution. That means we can substitute [Ba2+]=x and [OH-]=2x in the formula:
- Thus Ksp = [Ba2+].[OH-]2 = (x)(2x)2
- Or: 4x3= 4.10-18
- Now do a bit of algebra:
-
4x3=4.10-18 - x3= (10-6)3
- This makes
- [Ba2+]= 10-6 [mol/lit]
- [OH-]= 2.10-6 [mol/lit]
- What is the pH of the solution.
- We did make a bit of an assumption is this calculation, because water itself also produces a bit of OH- through a different equilibrium
- H2O + H2O < = > H3O+ + OH-
Its equilibrium constant is:
- Kw= [H3O+][OH-] = 10-14
- Water on its own therefore has [H3O+]=[OH-] = 10-7
- Obviously the barium hydroxide swamps the naturally occurring hydroxyl production by a factor of 2.10-6/10-7 = 20, so our assumption was justified.
The Kw must still be obeyed though and that gives us the key to calculate the pH, by filling in the actual [OH-]=2.10-6 :
- Kw= [H3O+][OH-] = 10-14
- Kw= x * 2.10-6 = 10-14
- x= 10-14/2.10-6 = 5.10-9
To find the pH we need to take the negative log on base 10
- pH = -10log([H3O+]) = -10log(5.10-9) = 8.3
And yes that makes the solution a bit basic