Ksp of PbI2 at 25 degrees celcius and 15 degrees celcius is 7.5×10^-8 and 7.5×10^-9 respectively. (Pb=207 I=127)

1) write an expression for the equilibrium of PbI2

2)write an equation for the Ksp of PbI2

3)calculate the solubility of PbI2 at 25 degrees celcius in gdm^-3

4)what is the mass of PbI2 precipitated when temperature of 1l of PbI2 solution is reduced from 25 degrees celcius to 15 degrees celcius

5)state whether the solubility of PbI2 is endothermic/exothermic

Please tell me the difference between 1st and 2nd questions?

Also I don't need answers for any of these questions...but a little GUIDANCE on each question would be much appreciated :) Thanks.

First things first. Learn how to spell celsius.

1. The equilibrium equation is just the solubility in water.
PbI2(s) ==> Pb^2+(aq) + 2I^-(aq)

2. Surely you know how to write the Ksp.

3. From the equation is (1), solve for the solubility.
.....PbI2 ==> Pb^2+ + 2I^-
I...solid.....0........0
C...solid-x...x........x
E...solid-x...x........x

Ksp = (Pb^2+)(I^-)^2
Substitute the E line into Ksp expression and solve for x = solubility. Be sure and use the correct value of Ksp for the temperature in the problem. The solubility you get is in mol/dm3. Convert to grams/dm3

4.Redo the solubility which you calculated in 3 and calculate solubility for 15 C. Convert solubility to g/dm3. Then subtract the two to find how much will ppt at the lower temperature.

5. First time I've seen a question like this but you just reason it out.
PbI2 --> Pb^2+ + 2I^-
Look at your data for solubility at the two T values. It is MORE solubility at the higher T so your equation could be written as
PbI2 + heat ==> Pb^2+ + 2I^-
Since you know the solubility is increased (adding heat drives the equation equilibrium to the right---from Le Chatelier's Principle) with heat, that makes it endothermic. Right?

.....PbI2 ==> Pb^2+ + 2I^-

I...solid.....0........0
C...solid-x...x........2x
E...solid-x...x........2x

I put 2x considering stoichiometry. Am I wrong here?
Thanks.

1) To write an expression for the equilibrium of PbI2, we need to consider the dissociation of PbI2 in water. PbI2 dissociates into Pb2+ and 2I- ions. Therefore, the expression for the equilibrium can be written as:

PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)

2) The equation for the Ksp (solubility product constant) of PbI2 is derived from the equilibrium expression:

Ksp = [Pb2+][I-]^2

Where [Pb2+] represents the concentration of lead ions and [I-] represents the concentration of iodide ions in the solution at equilibrium.

3) To calculate the solubility of PbI2 at 25 degrees Celsius in g/dm^-3, you need to use the Ksp value provided at that temperature. The solubility (S) can be calculated using the formula:

S = √(Ksp)

Substitute the Ksp value (7.5×10^-8) into the equation to find the solubility.

4) The difference between the first and second questions lies in the level of detail asked. The first question only requests the equilibrium expression, which is a qualitative representation of the system. On the other hand, the second question asks for the quantitative equation for the Ksp, which involves the concentration of the ions.

5) To determine whether the solubility of PbI2 is endothermic or exothermic, you need to consider the temperature change when the compound dissolves. If the solubility increases as the temperature increases, it indicates an endothermic process. If the solubility decreases as the temperature increases, it suggests an exothermic process. In this case, you can compare the Ksp values at 25 degrees Celsius and 15 degrees Celsius. It appears that the solubility decreases with a decrease in temperature, which suggests that the solubility of PbI2 is exothermic.