precipitation equilibrium precipitation reactions reach a position of equilibrium – even the most...
TRANSCRIPT
![Page 1: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/1.jpg)
Precipitation Equilibrium
Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight extent, establishing an equilibrium with it’s ions in solution.
For example, a solution of
lead II chloride.
![Page 2: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/2.jpg)
Ion-Product Equilibrium Systems There are 2 types of precipitation equilibria:
(1) between a precipitate and its ions.A precipitate (ppt.) forms when a cation from one solution combines with another forming an insoluble ionic solid.
For example:
Sr(NO3)2 (aq) + K2CrO4 (aq) ⇌ 2KNO3 (aq) + SrCrO4 (s)
(2) between a precipitate and the species used to dissolve it.
AgCl (s) + 2NH3 (aq) Ag(NH⇌ 3)2+ (aq) + Cl- (aq)
![Page 3: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/3.jpg)
The Solubility Product Constant The Ksp expression – this is the equilibrium
constant expression for the dissolving of a solid.
The smaller the Ksp value the less soluble the precipitate.
And like the any equilibrium constant value, the Ksp value is at a fixed temperature.
![Page 4: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/4.jpg)
Writing Ksp Expression:
Write the ion-product expression for each compound: (a) magnesium carbonate (b) iron(II) hydroxide (c) calcium phosphate (d) silver chromate (e) silver sulfide*(be careful with sulfide ions, they are so reactive with water that they will produce HS -
and OH-)
![Page 5: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/5.jpg)
Calculating Ion Concentration Calcium phosphate is a water-insoluble
mineral, large quantities of which are used to make commercial fertilizers. Taking it’s equilibrium constant value of 1 x 10-33, calculate: The concentration of the phosphate ion in
equilibrium with the solid if the [Ca2+] = 1 x 10-9 M Ans = 1 x 10-3 M
The concentration of the calcium ions in equilibrium with the solid if the [PO4
3-] = 1 x 10-5 M Ans = 2 x 10-8 M
![Page 6: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/6.jpg)
Determining Precipitate Formation: Ksp values can be used to predict when a
precipitate will form or not. To do this we work with the reaction quotient once more: If Q > Ksp - the reaction will shift to the left so
ppt. forms If Q < Ksp - the reaction will shift to the right so
no ppt. forms If Q = Ksp - then the solution is saturated with
ions at the point of ppt.
![Page 7: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/7.jpg)
Determining if a ppt. will form… Sodium chromate is added to a solution in which the
original concentration of Sr2+ is 0.0060 M (a) assuming the [Sr2+] stays constant, will a precipitate of
strontium chromate form when the chromate ion concentration becomes 0.0030 M? ans: no precipitate will form, Q < K
(b) will a precipitate of strontium chromate form if 0.200 L of 0.0060 M of strontium nitrate solution is mixed with 0.800 L of 0.040 M potassium chromate? ans: a precipitate will just barely form, Q > K (just slightly)
![Page 8: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/8.jpg)
Ksp & Water Solubility:
One way to establish equilibrium between a slightly soluble solid is to stir the solid with water to form a saturated solution.
Solubility of the solid, in moles per liter, is related to the solubility product constant. For example, determine the solubility of barium
sulfate. ans: s = 1.0 x 10-5 M
![Page 9: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/9.jpg)
Example: Determine Solubility Calculate the solubility of barium fluoride in
moles/liter and grams/liter. ans: 3.6 x 10-3 M & 0.63 g/L
![Page 10: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/10.jpg)
Ksp & The Common Ion Effect The solubility of an insoluble substance
decreases when adding a solution with a common ion in comparison to adding water (similar to Le Chatelier’s Principle).
Which will have the higher solubility: barium sulfate in water OR barium sulfate in a
0.1M solution of sodium sulfate Why?
![Page 11: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/11.jpg)
Applying the Common-Ion Effect Taking the equilibrium constant value of
barium sulfate into account, estimate its solubility in a 0.10 M solution of sodium sulfate (hint: you will need an equilibrium table). ans: 1.1 x 10-9 M
![Page 12: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/12.jpg)
Selective Precipitation
A way to separate 2 cations in water solution is to add an ion that precipitates only one of the cations.
For example: A flask contains a solution 0.10 M Cl- and 0.010 M CrO42-.
When AgNO3 is added: (a) which anion, chloride or chromate, precipitates first?
ans: The chloride ion will ppt. first since it requires the lowest silver ion concentration to ppt.
(b) what percentage of the first anion has been precipitated when the second anion starts to precipitate?
ans: 99.98% of chloride ions precipitated.
![Page 13: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/13.jpg)
Dissolving Precipitates
Water insoluble ionic solids can be brought into solution by adding a reagent to react with either the anion or the cation. The 2 most useful reagents for this are: 1. Strong Acids – the H+ will react with the basic
anions 2. NH3 or OH- to react with the metal cations
![Page 14: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/14.jpg)
Dissolving Zinc Hydroxide
Write the equation that represents the dissolving of zinc hydroxide by a strong acid (hint: two-step process).
Determine the equilibrium constant for this reaction (hint: rule of multiple equilibria). The Kw for 1 mole of water is 1 x 10-14.
![Page 15: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/15.jpg)
Strong Acids
Strong acids can be used to dissolve water-insoluble salts in which the anion is a weak base: Almost all carbonates Many sulfides
Example: Write balanced equations to explain why each of the following precipitates dissolve in a strong acid (assume the acid is in excess): Aluminum hydroxide Calcium carbonate Cobalt II sulfide
![Page 16: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/16.jpg)
Complex Ion Formation
Ammonia and hydroxides are commonly used to dissolve precipitates containing a cation that forms a stable complex with NH3 or OH- (reference table 16.2 – page 434).
Write the reaction by which zinc hydroxide dissolves in ammoina and solve for the equilibrium constant of this reaction using: Step 1: Zn(OH)2 (s) ⇌ Zn2+ + 2OH- Ksp = 4.0
x 10-17 Step 2: Zn2+ + 4NH3 (aq) ⇌ Zn(NH3)4
2+ + 2OH- Kf = 3.6x108
Kf = equilibrium constant for the formation of complex ions (page 416)
![Page 17: Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight](https://reader035.vdocuments.site/reader035/viewer/2022072006/56649cff5503460f949cfeb0/html5/thumbnails/17.jpg)
Dissolving AgCl with Ammonia Consider the reaction by which silver chloride
dissolves in ammonia: Taking the Ksp AgCl = 1.8 x 10-10 and the Kf
Ag(NH3)2+ = 1.7 x 107, calculate the K for this
reaction. ans: 3.1 x 10-3
Calculate the number of moles of AgCl that dissolves in one liter of 6.0 M ammonia. ans: 0.30 moles/liter