- Dissolution and precipitation
- Common polyatomic ions
- Introduction to solubility equilibria
- Worked example: Calculating solubility from Kₛₚ
- 2015 AP Chemistry free response 4
- The common-ion effect
- pH and solubility
- Solubility and complex ion formation
pH and solubility
For ionic compounds containing basic anions, solubility increases as the pH of the solution is decreased. For ionic compounds containing anions of negligible basicity (such as the conjugate bases of strong acids), solubility is unaffected by changes in pH. Created by Jay.
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- Is Cl- a much weaker base than F- because Cl is less electronegative than F?(15 votes)
- Yes, the order of basicity is F⁻ ≫ Cl⁻ > Br⁻ > I⁻. The corresponding pKb values are 11, 21, 23, and 25.
Yes, Cl is less electronegative than F, but Cl⁻ is a much larger ion than F⁻. The higher negative charge density on F⁻ outweighs its greater electronegativity, so F⁻ is the stronger base.(26 votes)
- Wouldn't adding HCl lead to formation of Cl- ions which would drive the reaction in reverse to form more AgCl (s) according to Le Chatlier Principle?(5 votes)
- Yes, adding HCl would form Cl⁻ ions, which would drive the reaction to the left and decrease the solubility of AgCl.
He picked the wrong acid. He should have picked nitric acid as his example.
The NO₃⁻ ion is not a common in, so it will have no effect.
And Cl⁻ is not basic, so H⁺ will have no effect.
Any acid that does not have a common ion or an ion that will form a precipitate with Ag⁺ will have no effect on the solubility of AgCl.(17 votes)
- I know it's not good practice to ask questions on Khan, but I hope this is well received.
I've been running through MCAT questions, and in Examcrackers 1001 MCAT questions for chemistry, I came across this very scenario as taught by Jay.
The question reads: A saturated solution of CaF2 is created by adding excess CaF2 to water. The excess CaF2 precipitate is filtered out of solution. 0.1 M HF is added to the solution. Does a precipitate still form?
A) Yes, because fluoride ions from the dissociation of HF act as a common ion causing precipitation
B) Yes, because HF is a strong acid
C) No, because acid increases the solubility of CaF2
D) No, because protons push the equilibrium of the overall reaction to the right balancing out the leftward shift caused by HF
I answered C based on what was taught in this tutorial but that was incorrect, and the answer is actually A.
I can see both sides of the coin, and Examcrackers justify their answer by saying that the overall reaction CaF2 (s) + 2H+ (aq) <-> Ca2+ (aq) + 2HF (aq) proves that any addition of HF will shift the equilibrium to the left as per Le Chatelier's principle.
I am unsure which is correct and would very much appreciate any help given :)(7 votes)
- The previous video "Solubility and the common ion effect" explains why answer A is correct. This video confirms it. In this video F- is reacting with acid to form HF, but in your example, the opposite reaction is occurring. HF is dissociating to F-. Increasing the F- concentration is an increase in a product of the CaF2 reaction. Therefore, the reaction shifts to the left, creating more of the reactant, CaF2. Since the solution is already saturated, excess CaF2 forms the precipitate.(14 votes)
- "decreasing pH increases solubility." Is that correct to say? Or would it be better to say decreasing pH caused more CaF2 to dissolve since Ksp does not change....(6 votes)
- Both statements are correct.
Yours gives an explanation that is not present in the first statement.(4 votes)
- Would there be a similar effect if I add NaOH to the solution? 2 OH- ions bond with Ca2+ to form Ca(OH)2, equilibrium shifts to right, more CaF2 dissolves?(3 votes)
- The OH⁻ won't have much effect, because Ca(OH)₂ is much more soluble than CaF₂.(2 votes)
- How would the solubility of CaF be affected by adding a base to the solution? Would its solubility decrease?(2 votes)
- I would say it would decrease solubility of CaF2 (CaF2 <---> Ca2+ + 2F-). In solution F- ions react with water bay this reaction: F- + H2O <--> HF + OH- adding a base, this reaction is moved to the right, increasing the concentration of F- ions which moves upper reaction to right, favoring precipitate formation(2 votes)
- Is the HF in solution as a molecule?(2 votes)
- No, it is not. HF exists in solution mainly as a hydrogen bonded ion pair, H₃O⁺·F⁻. There are essentially no HF molecules.
Two equilibria are involved.
H₂O + HF ⇌ H₃O⁺·F⁻; K₁ ≫ 1
H₃O⁺·F⁻ ⇌ H₃O⁺ + F⁻; K₂ ≪ 1
The first equilibrium lies far to the right. The second equilibrium lies to the left, so HF is a weak acid.
In concentrated solutions, the additional HF causes the ion pair to dissociate and form the hydrogen-bonded HF₂⁻ ion:
H₃O⁺·F⁻ + HF ⇌ H₃O⁺ + HF₂⁻
At the limit of pure liquid HF, there is self-ionization:
3HF ⇌ H₂F⁺ + HF₂⁻
Liquid HF is an extremely acidic substance.(3 votes)
- Isn't the formula for fluorine Fl?(2 votes)
- It's just F. Sal gets it wrong in a few videos and puts "Fl"(3 votes)
I am struggling calculating the solubility of BaF2 at pH of 4,
whats given is the following:
somehow i ended up with some crazy S^5 kind of equation and it doesnt seem right.... thanks a lot for the help(3 votes)
- I didn't understand why cl didn't react with H? Even if it is a weaker conjugate base, why didn't this happen?(1 vote)
- Because HCl is a strong acid that means the reaction HCl -> H+ + Cl- is one way, it does not go backwards.(3 votes)
- [Instructor] Changing the pH of a solution can affect the solubility of a slightly soluble salt. For example, if we took some solid lead two fluoride, which is a white solid, and we put it in some distilled water, the solid is going to reach an equilibrium with the ions in solution. Lead two fluoride forms lead two plus ions and fluoride anions in a one to two mole ratio. So if we have two lead two plus ions in this diagram, we need four fluoride anions. At equilibrium, the rate of this solution is equal to the rate of precipitation, and therefore, the amount of solid and the concentration of ions in solution remains constant. And this forms a saturated solution of lead two fluoride. To the system at equilibrium, we're gonna add some H plus ions. So by increasing the concentration of H plus ions in solution, we're decreasing the pH of the solution. When the H plus ions are added to the solution, most of them react with the fluoride anions that are present. So H plus plus F minus forms HF. Comparing the first diagram to the second diagram, I just happened to add three H plus ions, which will react with three of the fluoride anions that are present to produce three HF. Notice how the concentration of fluoride anions in solution has decreased from the first diagram to the second diagram, because of the addition of the H plus ions. So the system was at equilibrium and the concentration of fluoride anions was decreased. According to Le Chatelier's Principle, the system will shift in the direction that decreases the stress. So if the stress is decreased concentration of fluoride anions, the system will shift to the right to make more fluoride anions. And when the system shifts to the right, more lead two fluoride dissolves to increase the concentration of Pb two plus and fluoride anion. We can see that comparing the second diagram to the third diagram, so the amount of solid has gotten smaller, since some of that lead two fluoride dissolved, and we've increased the concentration of Pb two plus and F minus in solution. The solid keeps dissolving and the concentration of ions keeps increasing in solution until the system reaches equilibrium. So for a saturated solution of lead two fluoride at equilibrium, decreasing the pH or making the solution more acidic by increasing the concentration of H plus ions, increases the solubility of lead two fluoride, which is why we saw more of the solid dissolve when the H plus ions were added. This effect of decreasing the pH and increasing the solubility of a slightly soluble salt happens whenever the slightly soluble salt contains a basic anion. For this example, the basic anion is the fluoride anion, which reacts with the added H plus ions. And when the basic anion reacts, that decreases the concentration of that basic anion, which caused the equilibrium to shift to the right. And there are many other examples of basic anions, two more would be the hydroxide anion and the carbonate anion. And if a compound contains a basic anion, such as the hydroxide anion, hydroxide functions as a base and reacts with H plus ions to form H2O. So therefore, the solubility of a compound containing a hydroxide ion would increase as H plus ions are added to the solution. It's also important to note for this lead two fluoride problem, if the pH is decreased at a constant temperature, the Ksp value for PbF two remains constant. So the molar solubility does increase, but the Ksp value remains the same. This time, instead of lead two fluoride, let's look at lead two chloride. Lead two chloride is also a white solid. So if we dissolve some in solution, eventually, we would reach an equilibrium between the solid and the ions in solution. So this diagram here shows a saturated solution of lead two chloride and the system is at equilibrium. And to the system at equilibrium, we decrease the pH by adding H plus ions to the solution. In this case, the chloride anions aren't basic enough to react with the H plus ions. Therefore, we do not form HCl and the concentration of chloride anions remains the same as it was in the original diagram. So if the concentration of chloride ions remains the same and the concentration of lead two plus ions would remain the same, the system is still at equilibrium and decrease in the pH had no effect on the solubility of the solids. So we have the same amount of lead two chloride solid on the bottom of the beaker in both diagrams. So whenever an anion has an extremely weak base, like the chloride anion, we say that this is an anion of negligible basicity and the solubility of salts with anions of negligible basicity is unaffected by changes in pH.