the solubility product of PbSO4 at 298K is 1.6x10^-8 mol^2dm^-6. Calculate the solubility of PbSO4 in 0.10mol dm^-3 sulphuric acid.Thanks!

PbSO4==> Pb^2+ + SO4^2-

...x.......x......x

Ksp = 1.6E-8 = (Pb^2+)(SO4^2-)

How advanced is this class? This is quite a difficult problem.
The solubility is increased due to the presence of H^+ that reacts with the SO4^2- to produce HSO4^-. To counter that, the SO4^2- from the H2SO4 decreases the solubility because of the common ion effect. Finally, the SO4^2- from the H2SO4 is not as concentrated as one might think. For example, it is closer to 0.00985M in 0.1M H2SO4 and not 0.1M.
My guess is that your prof expects you to use 0.1M for (SO4^2-) and to ignore the effect of the increased H^+. If that is the case, then
Ksp = (Pb^2+)(SO4^2-) = 1.6E-8
For (Pb^2+) we substitute x.
For (SO4^2-) we substitute x+0.1 and solve for x. That's the simple way of doing it but I'm guessing that is what your prof wants.

Im a first year student in tertiary education,started this course for less than 3 months,thanks for helping,it really helps me to better understand the method to use for the common ion effect.

To calculate the solubility of PbSO4 in 0.10 mol dm^-3 sulphuric acid, we can use the concept of common ion effect.

Step 1: Write the dissolution equation for PbSO4 in water:
PbSO4 (s) -> Pb2+ (aq) + SO4^2- (aq)

Step 2: Write the dissolution equation for H2SO4 in water:
H2SO4 (aq) -> 2H+ (aq) + SO4^2- (aq)

Step 3: According to the common ion effect, the presence of the common ion (SO4^2-) from H2SO4 will suppress the solubility of PbSO4. We need to determine the concentration of SO4^2- in the solution.

Step 4: The solubility product expression for PbSO4 is given as:
Ksp = [Pb2+][SO4^2-]

Step 5: Given that Ksp = 1.6x10^-8 mol^2 dm^-6, and assuming the solubility of PbSO4 is 's' mol dm^-3, we have:
1.6x10^-8 = s * s

Step 6: Solve for 's':
s^2 = 1.6x10^-8
s = √(1.6x10^-8)
s ≈ 1.3x10^-4 mol dm^-3

Step 7: The concentration of SO4^2- in the solution is the same as the concentration of H2SO4, which is 0.10 mol dm^-3.

Step 8: To find the solubility of PbSO4 in the presence of 0.10 mol dm^-3 sulphuric acid, we need to calculate the common ion effect.

Step 9: The concentration of SO4^2- ions due to H2SO4 is 0.10 mol dm^-3.

Step 10: The final concentration of SO4^2- ions is the sum of the concentration due to the solubility of PbSO4 and the concentration due to H2SO4, which is:
0.10 mol dm^-3 (from H2SO4) + 1.3x10^-4 mol dm^-3 (from PbSO4) = 0.10 mol dm^-3

Therefore, the solubility of PbSO4 in 0.10 mol dm^-3 sulphuric acid is approximately 1.3x10^-4 mol dm^-3.

To calculate the solubility of PbSO4 in 0.10 mol dm^-3 sulphuric acid, we need to consider the common ion effect. Since sulphuric acid has a common ion (SO4^2-) with PbSO4, it will decrease the solubility of PbSO4 in the solution.

The solubility product constant, Ksp, for PbSO4 is given as 1.6x10^-8 mol^2 dm^-6. This constant can be used to determine the solubility of PbSO4 in pure water (without the presence of sulphuric acid). However, to calculate the solubility in the presence of sulphuric acid, we need to take into account the common ion effect.

Let's set up the solubility equilibrium expression for PbSO4:

PbSO4 ⇌ Pb^2+ + SO4^2-

The concentration of Pb^2+ and SO4^2- ions can be calculated using the solubility (s) of PbSO4:

[Pb^2+] = s
[SO4^2-] = s

Now, let's consider the reaction between PbSO4 and H2SO4:

PbSO4 + H2SO4 ⇌ Pb^2+ + SO4^2- + 2H+

We know that the concentration of H2SO4 is 0.10 mol dm^-3. This means that the concentration of H+ ions is also 0.10 mol dm^-3 since H2SO4 dissociates to produce two H+ ions.

Since the stoichiometry of H+ with respect to PbSO4 is 2:1, the concentration of H+ ions will be twice the concentration of Pb^2+ ions:

[H+] = 2[Pb^2+] = 2s

Using the solubility equilibrium expression, we can write:

Ksp = [Pb^2+][SO4^2-]
1.6x10^-8 = (s)(s)
1.6x10^-8 = s^2

Solving this equation gives us:
s = sqrt(1.6x10^-8) = 1.26x10^-4 mol dm^-3

This is the solubility of PbSO4 in pure water. However, due to the presence of sulphuric acid, we need to consider the common ion effect.

Since the concentration of H+ ions is 0.10 mol dm^-3, and the solubility of PbSO4 is 1.26x10^-4 mol dm^-3, the actual solubility of PbSO4 in 0.10 mol dm^-3 sulphuric acid will be reduced.

To calculate the reduced solubility, we need to consider the amount of PbSO4 that reacts with H+ ions from H2SO4. Since the stoichiometry of H+ ions with respect to PbSO4 is 2:1, half of the PbSO4 will react with H+ ions.

Therefore, the solubility of PbSO4 in 0.10 mol dm^-3 sulphuric acid is:
s reduced = s - (0.5[H+]) = 1.26x10^-4 - (0.5*0.10) = 1.21x10^-4 mol dm^-3

Hence, the solubility of PbSO4 in 0.10 mol dm^-3 sulphuric acid is 1.21x10^-4 mol dm^-3.