what is the solubility of barium sulfate in a solution containing 0.050 M sodium selfate? The Ksp value for barium sulfate is 1.1E-10.

An insoluble salt with formula MX3 has a solubility product constant written in terms of solubility,s, in the form
a) Ksp=s2
b)Ksp=4s3
c) Ksp=27s4
d)Ksp=256s5

A) Solubility of BaSO4 is worked similar to the CaCO3 problem. The principle difference is that the solution has Na2SO4 in it. Note Na2SO4 is a common ion (the sulfate ion is the common ion). What does that do. Sulfate comes from two sources; i.e., from BaSO4 (where it will be x) and from Na2SO4 (where it will be 0.05M). So the BaSO4 problem will be

Ksp = (Ba^2+)(SO4^2-)
Ksp = (x)(x+0.05) and solve for x.

B)I assume X is a halogen which then means M must be +3 element.
MX3(s) ==> M^3+ + 3X^-
Ksp = (M^3+)(X^-)^3
So set up the ICE chart in terms of s (follow the examples before where we let solubility be x) and plug that into Ksp expression. The correct answer is listed in the choices.

Oh, look who's into all these solubility problems! I hope you're not drowning in all these numbers!

Now, to tackle your questions. For the solubility of barium sulfate, we need to consider the common-ion effect since we have sodium sulfate in the solution. This means that the concentration of sulfate ions will affect the solubility of barium sulfate.

Let's assume that x is the solubility of barium sulfate. Then, the concentration of sulfate ions in the solution will be 0.050 + x.

Using the Ksp value, we can set up the equation:
1.1E-10 = x * (0.050 + x)

Now, here's the punchline... The solubility of barium sulfate is so low that it's practically insoluble! So, we can assume that x is negligible compared to 0.050. That means we can approximate the equation to:

1.1E-10 = x * (0.050)

Now, you can solve for x, and that will give you the solubility of barium sulfate in your solution. Good luck with the math! Remember, laughter is the best solution for any problem, even solubility ones!

To determine the solubility of barium sulfate in a solution containing 0.050 M sodium sulfate, we need to calculate the ion concentrations and compare them to the solubility product constant (Ksp) for barium sulfate.

Step 1: Write the balanced equation for the dissociation of barium sulfate in water:
BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)

Step 2: In the solution containing sodium sulfate, it dissociates as follows:
Na2SO4(s) ⇌ 2Na+(aq) + SO42-(aq)

Step 3: We know the concentration of sodium sulfate is 0.050 M. Using the dissociation equation, we can determine the concentration of sulfate ions (SO42-) in the solution, which is also 0.050 M.

Step 4: Since barium sulfate is an insoluble salt, it will dissociate only slightly in water. Therefore, we can assume that the concentration of barium ions (Ba2+) in the solution will be negligible compared to the initial concentration of sodium sulfate.

Step 5: Based on the solubility product constant (Ksp) for barium sulfate, we can use the equation:

Ksp = [Ba2+][SO42-]

Since the concentration of Ba2+ can be assumed to be negligible, we can substitute the concentration of SO42- (0.050 M) into the equation:

Ksp = [Ba2+][0.050 M]

Step 6: Rearrange the equation to solve for the solubility (s):

Ksp = [Ba2+][SO42-]
1.1E-10 = s * 0.050 M

Step 7: Solve for the solubility (s):

s = 1.1E-10 / 0.050 M

Thus, the solubility of barium sulfate in the solution containing 0.050 M sodium sulfate is equal to 2.2E-9 M.

Regarding the second question about the solubility product constant (Ksp) for an insoluble salt with the formula MX3, the correct answer is:

c) Ksp = 27s4

To determine the solubility of barium sulfate in a solution containing 0.050 M sodium sulfate and using the given Ksp value, we can apply the concept of common ion effect and the solubility product constant.

Barium sulfate (BaSO4) dissociates into barium ions (Ba2+) and sulfate ions (SO42-) in water, according to the balanced chemical equation:

BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)

The solubility product constant expression (Ksp) is derived from the balanced chemical equation and represents the product of the concentrations of the dissociated ions raised to the power of their respective stoichiometric coefficients:

Ksp = [Ba2+][SO42-]

Given the Ksp value for barium sulfate as 1.1 x 10^(-10), we can assume that the initial concentrations of Ba2+ and SO42- ions are x M each (since they dissociate in a 1:1 ratio).

Due to the presence of sodium sulfate in the solution, there is already a concentration of 0.050 M SO42- ions. Therefore, the concentration of SO42- ions from the dissociation of barium sulfate will be (x + 0.050) M.

Using the Ksp expression, we can set up the equation:

Ksp = [Ba2+][SO42-]
1.1 x 10^(-10) = x * (x + 0.050)

Simplifying the equation:

x^2 + 0.050x - 1.1 x 10^(-10) = 0

Now, you can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a)

For the given quadratic equation, the values of a, b, and c are as follows:

a = 1
b = 0.050
c = -1.1 x 10^(-10)

Plug these values into the quadratic formula and calculate the two possible values of x (solubility) by considering both the positive and negative square root. Select the positive value as it represents the concentration.

Once you find the value of x, you can substitute it into the given answer choices to determine which expression represents the solubility product constant (Ksp) in terms of solubility (s) for an insoluble salt with formula MX3.