Calculate the molar solubility of BaSO4 in a solution in which [H3O+] is 0.50 M. Ksp(BaSO4) = 1.1×10-10,
Ka(HSO4-)=1.02×10-2
.......BaSO4 ==> Ba^2+ + SO4^2-
I......solid......0.......0
C......solid......x.......x
E......solid......x.......x
But with [HO^+] added, that combines with the SO4^2- to give HSO4^- as this:
......SO4^2- + H^+ ==> HSO4^- and increases the solubility of BaSO4.
In other words part of the SO4^2- is missing.
alpha1 is the fraction of SO4^2- actually there. That is
alpha 1 = k2/(k2+H^+) =
0.0102/(0.5102) = very close to 0.02.
So Ksp = (Ba^2+)(SO4^2-)
If we call (Ba^2+) = solubility = S, the Ksp = (S)(S*0.02).
Plug in values for Ksp and solve for S. I get approx 7E-5M but that's just a close estimate. To show the added solubility due to the H^+, note that in pure water the solubility of BaSO4 is sqrt(1.1E-10) = 1.05E-5M so the solubility has been increased by about 7x by adding acid to make 0.5M H^+.
Well, it seems like we have a solubility problem here, but don't worry, I'm here to clown around and help you out!
To calculate the molar solubility of BaSO4, we first need to find the concentration of Ba2+ and SO42- ions in the solution. Since barium sulfate (BaSO4) dissociates into Ba2+ and SO42- ions, we can assume that the concentration of Ba2+ is equal to the concentration of SO42-.
Let's denote the molar solubility of BaSO4 as "x" (in mol/L). Therefore, the concentration of Ba2+ ions in the solution is also "x" (in mol/L).
The equation for the dissociation of BaSO4 is:
BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)
Now, let's write an expression for the solubility product constant (Ksp):
Ksp = [Ba2+][SO42-]
Since the concentration of Ba2+ is "x" and the concentration of SO42- is also "x", we can substitute these values into the expression for Ksp:
Ksp = x * x = x^2
We are given that the Ksp value for BaSO4 is 1.1×10^-10. We can then set up the equation:
1.1×10^-10 = x^2
Solving for "x," we find that the molar solubility (or concentration) of BaSO4 is approximately 1.05×10^-5 M.
So, the molar solubility of BaSO4 in a solution with [H3O+] of 0.50 M is around 1.05×10^-5 M.
To calculate the molar solubility of BaSO4 in a solution with a known [H3O+], we need to consider the reaction between BaSO4 and H3O+.
The dissolution of BaSO4 in water can be described by the following equation:
BaSO4 (s) ⇌ Ba2+ (aq) + SO4^2- (aq)
For each mole of BaSO4 that dissolves, one mole of Ba2+ and one mole of SO4^2- ions are formed.
For the reaction to occur, the HSO4- ion (formed from the dissociation of H2SO4, assuming H2SO4 is the strong acid donating the H3O+ ions) can react with the Ba2+ ion to form an insoluble compound, BaSO4. This reaction can be described as follows:
Ba2+ (aq) + HSO4- (aq) ⇌ BaSO4 (s) + H+ (aq)
Given that the Ka(HSO4-) is 1.02×10^-2, we can use the equilibrium expression for the reaction above:
Ka = [H+][SO4^2-] / [HSO4-]
Since the concentration of [HSO4-] can be assumed to be equal to the concentration of [H3O+], we can write:
[H+][SO4^2-] / [H3O+] = 1.02×10^-2
We are given that [H3O+] = 0.50 M, so we can substitute that value into the equation:
[H+][SO4^2-] / 0.50 = 1.02×10^-2
Solving for [H+][SO4^2-], we get:
[H+][SO4^2-] = 1.02×10^-2 × 0.50
[H+][SO4^2-] = 5.1×10^-3
Next, we can use the equilibrium constant expression for Ksp of BaSO4:
Ksp = [Ba2+][SO4^2-]
Given that Ksp(BaSO4) is 1.1×10^-10, we can substitute the molar solubility (x) for [Ba2+] and [SO4^2-]:
Ksp = x^2
1.1×10^-10 = x^2
Taking the square root of both sides, we get:
x = √(1.1×10^-10)
x ≈ 1.05×10^-5 M
Therefore, the molar solubility of BaSO4 in a solution with [H3O+] of 0.50 M is approximately 1.05×10^-5 M.
To calculate the molar solubility of BaSO4 in a solution with a given [H3O+], we can use the solubility product constant (Ksp) expression for BaSO4 and the acid dissociation constant (Ka) expression for HSO4- to determine the concentrations of the ions involved. Let's break down the steps to find the molar solubility:
Step 1: Write the balanced equation for the dissociation of BaSO4:
BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)
Step 2: Write the solubility product constant expression (Ksp) for BaSO4:
Ksp = [Ba2+][SO42-]
Step 3: Write the equilibrium expression for the dissociation of HSO4-:
HSO4-(aq) ↔ H+(aq) + SO42-(aq)
Step 4: Write the expression for the acid dissociation constant (Ka) for HSO4-:
Ka = [H+][SO42-]/[HSO4-]
Step 5: Determine the concentrations of H+ and SO42- in terms of [H3O+]:
[H+] = [H3O+]
[SO42-] = [H3O+] (since HSO4- is in equilibrium with H+ and SO42-)
Step 6: Substitute the concentrations from step 5 into the Ka expression:
Ka = [H3O+][H3O+]/[HSO4-] = [H3O+]^2/[HSO4-]
Step 7: Solve for [HSO4-]:
[HSO4-] = [H3O+]^2/Ka
Step 8: Substitute the concentration of [HSO4-] from step 7 into the Ksp expression from step 2:
Ksp = [Ba2+][SO42-] = [Ba2+][H3O+]
Step 9: Substitute the known values into the Ksp expression:
1.1×10^-10 = [Ba2+][H3O+]
Step 10: Rearrange the equation to solve for [Ba2+]:
[Ba2+] = 1.1×10^-10 /[H3O+]
Therefore, the molar solubility of BaSO4 in a solution with [H3O+] of 0.50 M is given by [Ba2+], which can be calculated using the equation:
[Ba2+] = 1.1×10^-10 /0.50
Calculate the value to find the molar solubility of BaSO4 in the given solution.