Hey guys! Stuck on some! My maybe answers are written next to the problem (1 problem-5 parts). Thanks for any help:
1. Lead (II) chromate, PbCrO4, is a slightly soluble salt. Write a net ionic equation for the
dissolution of solid PbCrO4 in water.
My thoughts: PbCrO4+2H+-->2Pb2++CrO72-+H2O
Or is it simply PbCrO4->Pb2++CrO42-
(is this an equilibrium equation???)
Write an equilibrium constant expression for this reaction.
K =
My thoughts: Ksp=[Pb2+][CrO42-]
What might one add to increase the solubility of the PbCrO4? ________________________
My thoughts: A strong acid like HCl
Write equations to describe how this added reagent works to increase the PbCrO4 solubility.
My thoughts: I have no idea, please help!!!!
What reagent could be added to decrease the concentration of Pb2+ in solution?
My thoughts: Ba(OH)2 (?) because the Pb2+ will react with the OH- to form Pb(OH)2, which will leave less Pb2+ ions?
1. The correct net ionic equation for the dissolution of solid PbCrO4 in water is:
PbCrO4(s) -> Pb2+(aq) + CrO4^2-(aq)
It is an equilibrium equation.
2. The equilibrium constant expression for this reaction is:
Ksp = [Pb2+][CrO4^2-]
Your thoughts are correct.
3. To increase the solubility of the PbCrO4, you can add a strong acid like HCl.
Your thoughts are correct.
4. The equations to describe how the added reagent (HCl) works to increase the PbCrO4 solubility are:
HCl(aq) + CrO4^2-(aq) -> HCrO4^-(aq) + Cl^-(aq)
PbCrO4(s) + 2HCl(aq) -> Pb2+(aq) + 2Cl^-(aq) + H2CrO4(aq)
5. To decrease the concentration of Pb2+ in solution, you can add Ba(OH)2, just as you thought. This is because the Pb2+ will react with the OH- to form Pb(OH)2, which will precipitate:
Pb2+(aq) + 2OH^-(aq) -> Pb(OH)2(s)
1. The net ionic equation for the dissolution of solid PbCrO4 in water can be written as follows:
PbCrO4(s) ⇌ Pb2+(aq) + CrO42-(aq)
The first equation you wrote is incorrect because it includes H+ ions, which are not a part of the reaction. The second equation you wrote is correct, but it is a simplified form of the reaction without indicating the dissociated ions in the solution.
2. The equilibrium constant expression (Ksp) for this reaction is indeed:
Ksp = [Pb2+][CrO42-]
You correctly identified the concentrations of the dissolved ions.
3. To increase the solubility of PbCrO4, one can add a common ion. In this case, adding a strong acid like HCl would increase the solubility. This is because the common ion effect reduces the solubility of a slightly soluble salt by adding an ion that is already present in the equilibrium expression. When more Pb2+ ions are present, it shifts the equilibrium to the right, increasing the solubility of PbCrO4.
4. When HCl is added, it dissociates into H+ and Cl- ions. The H+ ions then combine with the CrO42- ions to form HCrO4- ions, according to the balanced chemical equation:
CrO42-(aq) + 2H+(aq) → HCrO4-(aq)
This reaction helps in increasing the solubility of PbCrO4 by removing some of the CrO42- ions from the solution, reducing their concentration and shifting the equilibrium to the right.
5. To decrease the concentration of Pb2+ ions in the solution, a reagent that can form an insoluble compound with Pb2+ can be added. In this case, adding Ba(OH)2 would be suitable. The Ba2+ ions react with Pb2+ ions to form an insoluble compound, Pb(OH)2, according to the balanced chemical equation:
Pb2+(aq) + 2OH-(aq) → Pb(OH)2(s)
By forming Pb(OH)2, the concentration of Pb2+ in solution decreases, resulting in a decrease in the concentration of Pb2+ ions.
To answer your questions:
1. The net ionic equation for the dissolution of solid PbCrO4 in water is:
PbCrO4(s) -> Pb2+(aq) + CrO42-(aq)
This equation represents the dissociation of the solid PbCrO4 into its constituent ions (Pb2+ and CrO42-) in water. The arrow indicates the direction of the reaction, from left to right, which means the solid is dissolving.
2. The equilibrium constant expression for this reaction is:
Ksp = [Pb2+][CrO42-]
Ksp represents the solubility product constant, which is the equilibrium constant for the dissolution of a sparingly soluble salt. In this case, Ksp is equal to the concentrations of the Pb2+ and CrO42- ions raised to their stoichiometric coefficients.
3. To increase the solubility of PbCrO4, you can add a strong acid like HCl. The presence of excess H+ ions from the strong acid can shift the equilibrium position of the dissolution reaction towards the formation of more Pb2+ and CrO42- ions.
4. The equation to describe how the added acid (HCl) increases the solubility of PbCrO4 is:
PbCrO4(s) + 2H+(aq) -> Pb2+(aq) + CrO42-(aq) + H2O
In this equation, the H+ ions react with the solid PbCrO4, causing it to dissociate into Pb2+ and CrO42- ions. The equilibrium is shifted to the right, resulting in increased solubility of PbCrO4.
5. To decrease the concentration of Pb2+ in solution, you can add a reagent like Ba(OH)2. The Ba2+ ions from Ba(OH)2 will react with Pb2+ ions to form insoluble Pb(OH)2 precipitate, effectively reducing the concentration of Pb2+ ions in solution.
Overall equation for the reaction between Ba(OH)2 and Pb2+:
Ba(OH)2(aq) + Pb2+(aq) -> Pb(OH)2(s) + Ba2+(aq)
The formation of the insoluble Pb(OH)2 shifts the equilibrium towards the precipitation of Pb2+ ions, leading to a decreased concentration in solution.