Thallium (I) iodate is only slightly soluble in water. Its Ksp at 25°C is
3.07 x 10-6. Estimate the solubility of thallium iodate in units of grams per
100.0 ml of water.
TlIO3(s) <==> Tl^+ + IO3^-
Let x = molar solubility of TlIO3.
The x is molar solubility of Tl^+ and x is molar solubility of IO3^-
Ksp = (Tl^+)(IO3^-) = 3.07E-6
Plug in and solve for x.
Post your work if you get stuck.
A small correction here.
x is in mols/L. You want to change that to grams/100.
0 mL.
x = 3.07 x 10-6
1 mol TlIO3 = 474.3 g
x = 3.07 x 10-6 mol/L
x = 474.3 g/L x 3.07 x 10-6 mol/L
x = 1.45 x 10-3 g/100.0 mL
Well, it seems like you're trying to calculate the solubility of thallium iodate in grams per 100.0 ml of water. Let me jump right in and help you with that!
First, you correctly expressed the dissolution of thallium iodate as:
TlIO3(s) <==> Tl^+ + IO3^-
Now, you've defined x as the molar solubility of TlIO3. Since we're looking for grams per 100.0 ml of water, we need to convert the molar solubility to grams.
To do this, we need to know the molar mass of TlIO3. Do you happen to have that information?
To convert the molar solubility from moles per liter (mol/L) to grams per 100.0 ml, you can use the molar mass of thallium iodate.
1. Find the molar mass of thallium iodate (TlIO3):
- Thallium (Tl) atomic mass = 204.38 g/mol
- Iodine (I) atomic mass = 126.90 g/mol
- Oxygen (O) atomic mass = 16.00 g/mol
- Molar mass of TlIO3 = (204.38 g/mol) + (126.90 g/mol) + (3 * 16.00 g/mol) = 397.28 g/mol
2. Convert the molar solubility from mol/L to grams/100.0 ml:
- Given that the molar solubility of TlIO3 is x mol/L.
- 1 L = 1000 ml, so the conversion factor is (100/1000) ml/L = 0.1 ml/L.
- The molar solubility in grams/100.0 ml is x (mol/L) * 397.28 (g/mol) * 0.1 (ml/L).
Therefore, the solubility of thallium iodate in grams per 100.0 ml of water is approximately 39.728x grams.
To estimate the solubility of thallium iodate (TlIO3) in grams per 100.0 ml of water, you can follow these steps:
1. Write the balanced equation for the dissolution of TlIO3:
TlIO3(s) ⇌ Tl^+ + IO3^-
2. Let x represent the molar solubility of TlIO3 (in moles per liter).
3. Since TlIO3 dissociates completely to Tl^+ and IO3^-, the concentration of Tl^+ and IO3^- ions will also be x.
4. Use the given Ksp (solubility product constant) equation:
Ksp = [Tl^+][IO3^-] = 3.07 x 10^-6
5. Substitute the value of x into the Ksp equation:
(x)(x) = 3.07 x 10^-6
x^2 = 3.07 x 10^-6
6. Solve for x by taking the square root of both sides:
x ≈ √(3.07 x 10^-6)
7. The value of x is the molar solubility of TlIO3 in moles per liter.
8. To convert it to grams per 100.0 ml of water, you need to consider the density of water.
9. The density of water at 25°C is approximately 0.997 g/ml. Therefore, 100.0 ml of water is approximately 99.7 grams.
10. Convert the molar solubility (x) to grams per 100.0 ml by multiplying it with the molar mass of TlIO3 (which is 389.26 g/mol) and dividing by the volume of water in grams (99.7 grams):
Solubility (grams/100.0 ml) = (x * molar mass of TlIO3) / volume of water in grams
Remember to use the approximate value of x obtained in step 6.
By following these steps, you can estimate the solubility of thallium iodate in grams per 100.0 ml of water.