the Ksp of Bi2s3 at room temmperature is 1.60x10-72, what is its solubility in g/100 ml? Please help me with the solution. Our teacher haven't discussed this yet so I have no idea how to solve this.. and if i ever got the answer with your solution. could you please check my answer?
............Bi2S3 --> 2Bi^2+ + 3S^2-
Initial.....solid......0........0
Change......solid......2x.......3x
equilibrium.solid......2x........3x
Ksp = (Bi^3+)^2(S^2-)^3
Substitute the E line into the Ksp equation and solve for x.
x will give you the solubility in mols/L. Convert to g/L by g = mols x molar mass. Then convert to g/100 mL.
To find the solubility of Bi2S3 in g/100 ml using its Ksp, we need to follow these steps:
Step 1: Write the balanced equation for the dissolution of Bi2S3
Bi2S3(s) ⇌ 2Bi3+(aq) + 3S2-(aq)
Step 2: Write the expression for the solubility product constant (Ksp)
Ksp = [Bi3+]^2 [S2-]^3
Step 3: Let's determine the molar solubility of Bi2S3
Let 's' be the molar solubility of Bi2S3 (in mol/L)
From the balanced equation, we can see that:
[Bi3+] = 2s (since the coefficient in front of Bi3+ is 2)
[S2-] = 3s (since the coefficient in front of S2- is 3)
Substituting these values into the Ksp expression:
Ksp = (2s)^2 (3s)^3
1.60 x 10^-72 = 36s^5
Step 4: Solve for s (molar solubility)
s^5 = (1.60 x 10^-72) / 36
s^5 = 4.44 x 10^-74
Using a calculator to find the fifth root of both sides:
s ≈ 1.23 x 10^-15 mol/L
Step 5: Convert the molar solubility to g/100 ml
To convert from mol/L to g/100 ml, we need to multiply by the molar mass of Bi2S3 and then by the conversion factor 100 ml / 1 L.
The molar mass of Bi2S3 = (2 x 208.98 g/mol) + (3 x 32.07 g/mol) = 514.35 g/mol
s = 1.23 x 10^-15 mol/L * 514.35 g/mol * 100 ml/1 L
s ≈ 6.34 x 10^-12 g/100 ml
So, the solubility of Bi2S3 at room temperature is approximately 6.34 x 10^-12 g/100 ml.
Please double-check your calculations using the provided steps and let me know if you have any further questions!
To find the solubility of Bi2S3 in g/100 ml, you need to use the solubility product constant (Ksp) and the molar mass of Bi2S3.
1. Start by writing the balanced equation for the equilibrium reaction:
Bi2S3 (s) ↔ 2Bi3+ (aq) + 3S2- (aq)
2. The equilibrium expression for Ksp is:
Ksp = [Bi3+]^2 * [S2-]^3
3. Let's assume that x g of Bi2S3 dissolves per 100 ml of water. This means that:
[Bi3+] = 2x
[S2-] = 3x
4. Substitute the expressions for [Bi3+] and [S2-] into the Ksp expression:
Ksp = (2x)^2 * (3x)^3 = 108x^5
5. Rearrange the equation to isolate x:
x = (Ksp/108)^(1/5)
6. Plug in the given Ksp value and solve for x:
x = (1.60x10^-72 / 108)^(1/5) ≈ 1.59x10^-15
7. Convert the solubility to g/100 ml:
Since 1 mol of Bi2S3 has a molar mass of (2 x 208.98) + (3 x 32.06) g/mol = 514.42 g/mol
Therefore, solubility = (1.59x10^-15 mol/100 ml) x (514.42 g/mol) ≈ 8.18x10^-13 g/100 ml
Now, you can check your answer by comparing it to the solution I provided.