Most hydroxide salts (except for the strong bases) are sparingly soluble. This allows us to use acid/base chemistry and/or buffers to control solubility.
This also allows us to use precipitation reactions implement separation of ions from solution. This requires that the solubility of the salts chosen to precipitate is quite different.
Suppose a solution is 0.25 M Fe3+ and 0.25 M Ni2+ and solid sodium hydroxide is added to the mixture. Answer the following questions:
1. What pH is required to just start precipitation of the iron(III)?
2. What pH is required to just start precipitation of the more nickel(II)?
3. What is the concentration of the less soluble metal ion when the more soluble ion just starts to precipitate?
4. What percentage of the less soluble ion has precipitated when the more soluble ion is just starting to precipitate?
Reactions:
Fe3+(aq) + 3 OH–(aq) → Fe(OH)3(s)
Ni2+(aq) + 2 OH–(aq) → Ni(OH)2(s)
Find the Ksp values for each sparingly soluble salt from the Table of Solubility Product Constants.
Fe(OH)3, Ksp = 4×10–38
Ni(OH)2, Ksp = 2.0×10–15
Question 1: What is the pH when the iron just starts to precipitate?
Fe(OH)3(s)
→ ← Fe3+(aq) + 3 OH–(aq)
Ksp = [Fe3+]e[OH– ]e3 = 4×10–38
Initial0.250
Change+x+3x
Equilibrium0.25 + x3x
x is expected to be small so 0.25 + x ~ 0.25
4×10–38 = [0.25][3x]3
x = 2×10–13
So [OH– ] = 3(2×10–13) = 6×10–13 M just as precipitation of iron(III) hydroxide starts.
pOH = –log(6×10–13) = 12.2 so pH = 1.8
Question 2: What is the pH when the nickel just starts to precipitate?
Ni(OH)2(s)
→ ← Ni2+(aq) + 2 OH–(aq)
Ksp = [Ni2+]e[OH–]e2 = 2.0×10–15
Initial0.250
Change+x+2x
Equilibrium0.25 + x2x
Again, x is expected to be small so 0.25 + x ~ 0.25
2.0×10–15 = [0.25][2x]2
x = 4.5×10–8
So [OH–] = 2(4.5×10–8) = 9.0×10–8 M just as precipitation of nickel(II) hydroxide starts.
pOH = –log(9.0×10–8) = 7.05 so pH = 6.95
Question 3: What is the iron(III) concentration when the nickel(II) just starts to precipitate?
Fe(OH)3(s)
→ ← Fe3+(aq) + 3 OH–(aq)
Ksp = [Fe3+]e[OH– ]e3 = 4×10–38
Initialx9.0×10–8
Change+ 0+ 0
Equilibriumx9.0×10–8
4×10–38 = [x][9.0×10–8]3
x = 5×10–17
[Fe3+] = 5×10–17 M just as precipitation of nickel(II) hydroxide starts.
Question 4: What percentage of iron(III) has precipitated when the nickel(II) just starts to precipitate?
The percentage of iron ion remaining in solution as the nickel just starts to precipitate is:
% Fe3+ in solution = (5×10–17/0.25)×100% = 2×10–14 %
Essentially all of the iron(III) has precipitated, but the nickel ion is still in solution.
This is known as fractional precipitation.
This same process allows for geological ore deposition.
In the above example, a base was used to cause precipitation.
Acids and bases can also increase solubility.
Most sparingly soluble salts are composed of either an acidic cation or a basic anion (or both). This means that the acid/base chemistry of the ions can be used to change the solubility properties of the salt.
What are the relative solubilities of calcium hydrogen phosphate under neutral, acid, and basic conditions?
Under neutral pH conditions:
CaHPO4(s) → ← Ca2+(aq) + HPO42–(aq)
Under acidic conditions, say with addition of hydrochloric acid:
CaHPO4(s) → ← Ca2+(aq) + HPO42–(aq)
HCl(aq) + HPO42–(aq) → H2PO4–(aq) + Cl–(aq)
Net:
CaHPO4(s) + HCl(aq) → Ca2+(aq) + H2PO4–(aq) + Cl–(aq)
The salt readily dissolves in the acidic solution.
The dihydrogen phosphate ion can also react with the hydrochloric acid, so the complete reaction is given by:
CaHPO4(s) + 2 HCl(aq) → Ca2+(aq) + H3PO4(aq) + 2 Cl–(aq)
Under basic conditions, say with the addition of sodium hydroxide:
CaHPO4(s) → ← Ca2+(aq) + HPO42–(aq)
NaOH(aq) + HPO42–(aq) → PO43–(aq) + Na+(aq) + H2O(l)
3 Ca2+(aq) + 2 PO43–(aq) → Ca3(PO4)2(s)
Net:
3 CaHPO4(s) + 2 NaOH(aq) → Ca3(PO4)2(s) + HPO42–(aq) + 2 Na+(aq) + 2 H2O(l)
Here, the solubility decreases (calcium phosphate is less soluble than calcium hydrogen phosphate).
Of course, in real life the chemistry is even more complicated because the following precipitation can occur when the hydroxide ion concentration is high enough:
5 Ca2+(aq) + 3 PO43–(aq) + NaOH(aq) → Ca5(PO4)3(OH)(s) + Na+(aq)
Ca5(PO4)3(OH) is called hydroxyapatite and is the mineral component of bone and teeth.