Calculate the molar solubility of BaSO4 in a solution
in which [H3O+] is 3.5 M. Ksp(BaSO4) = 1.1 x 10^-10, Ka(HSO4)=1.02 x 10^-2
BaSO4 ==> Ba^2+ + SO4^2- (eqn 1) and
SO4^2- + H^+ ==> HSO4^- (eqn 2)eqn 2
Let S = solubility so
(Ba^2+) = S and (SO4^2-) +(HSO4^-) = S
You want to find S. You know k2 for HSO4^- which is
0.0102 = (H^+)(SO4^2-)/(HSO4^-). You know(H^+) and k2 from the problem so you calculate (HSO4^-) = (H^+)(SO4^2-)/k2 = ? and plug that into eqn 2, solve for S and plug that into Ksp expression for BaSO4. Solve for S and that's it. Post your work if you get stuck.
Well, I haven't seen a BaSO4 trying to solve a puzzle, but I'll give it a shot!
To find the molar solubility of BaSO4, we first need to set up the solubility product expression:
Ksp = [Ba2+][SO42-]
Now, let's assume x represents the molar solubility of BaSO4. Since BaSO4 dissociates into one Ba2+ ion and one SO42- ion, we can express the concentrations as follows:
[Ba2+] = x
[SO42-] = x
Now, let's use the Ka of HSO4- to find the concentration of H3O+:
Ka = [H3O+][SO42-]/[HSO4-]
[H3O+] = Ka*[HSO4-]/[SO42-]
[H3O+] = (1.02 x 10^-2)*(x)/x
[H3O+] = 1.02 x 10^-2
Now let's plug [Ba2+] and [SO42-] into the solubility product expression:
Ksp = (x)(x)
1.1 x 10^-10 = x^2
Solving for x, we find:
x = √(1.1 x 10^-10)
So, the molar solubility of BaSO4 in the given solution is approximately √(1.1 x 10^-10) moles per liter.
But hey, don't be too disappointed if BaSO4 can't solve this puzzle... it's not the sharpest compound in the flask!
To calculate the molar solubility of BaSO4 in a solution, we need to consider the solubility product constant (Ksp) and the effect of [H3O+] on the solubility.
Step 1: Write the balanced equation for the dissociation of BaSO4:
BaSO4(s) <--> Ba2+(aq) + SO4^2-(aq)
Step 2: Use the stoichiometry of the balanced equation to determine the molar solubility of BaSO4. Let's assume x represents the molar solubility of BaSO4.
BaSO4(s) <--> Ba2+(aq) + SO4^2-(aq)
Initial: 0 0 0
Change: -x +x +x
Equilibrium: -x x x
Step 3: Write the solubility product expression, Ksp, for BaSO4:
Ksp = [Ba2+][SO4^2-]
Step 4: Substitute the equilibrium concentrations into the Ksp expression:
Ksp = x * x
Step 5: Solve for x:
1.1 x 10^-10 = x^2
Step 6: Take the square root of both sides to solve for x:
x = sqrt(1.1 x 10^-10)
Step 7: Calculate the value of x using a calculator:
x ≈ 1.05 x 10^-5
Therefore, the molar solubility of BaSO4 in a solution with [H3O+] of 3.5 M is approximately 1.05 x 10^-5 M.
To calculate the molar solubility of BaSO4 in a solution with a known [H3O+], we need to consider the reaction that occurs between BaSO4 and water. The balanced equation for the dissociation of BaSO4 is:
BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)
Since BaSO4 is insoluble, it will not dissociate in water to a significant extent. However, we can determine the concentration of the ions in equilibrium by using the solubility product constant (Ksp) of BaSO4, which is given as 1.1 x 10^-10.
The Ksp expression for this dissociation can be written as:
Ksp = [Ba2+][SO42-]
We can assume that the concentration of SO42- is equal to the concentration of Ba2+ because they have a 1:1 stoichiometric ratio.
Let's denote the molar solubility of BaSO4 as "s". The concentration of Ba2+ and SO42- in terms of "s" will be [Ba2+] = s and [SO42-] = s.
Substituting these values into the Ksp expression, we get:
Ksp = (s)(s) = s^2 = 1.1 x 10^-10
Now, we need to find the value of "s" (molar solubility).
To solve for "s", we will use the known value of Ksp and calculate the square root of the Ksp value:
s = √(1.1 x 10^-10)
Using a calculator, you will find that the molar solubility of BaSO4 is approximately 1.05 x 10^-5 M.
Please note that the given value of [H3O+] = 3.5 M and the value of Ka(HSO4) = 1.02 x 10^-2 are not directly related to the calculation of the molar solubility of BaSO4. If you have any further questions or need assistance with a related topic, feel free to ask!