[kensher]

I will present the assignment and my calculations. Fair to say, it's completely different from the answer in the book.

a) Write the equilibrium for Ag₂SO₄(s) dissolved in water and the expression for the solubility product.

My answer

Ag₂SO₄(s)  2Ag²⁺ (aq) + SO₄^2- (aq)
K_sp = [Ag+]² * [SO₄^2-]

So far so good.

b) K_sp (Ag₂SO₄) = 1,6 * 10^-5
Calculate the solubility of Ag₂SO₄(s) in moles/L and in mg/100 mL.

My answer

Solubility of Ag₂SO₄ at equilibrium:

1 mole of Ag₂SO₄  2 moles of Ag⁺  1 mole of So₄^2-

x * (2x)² = 1,6 * 10^-5 (moles/L)³

4x³ = 1,6 * 10^{-5
           
3}√4x = ³√1,6 * 10^-5 (moles/L)³

4x = 4 * 10^-6

x =        4 * 10^-6 moles/L
           ^{______________}
                   4

x = 1 * 10^-6 moles/L

The answer from my book shows: 1,6 * 10^-2 moles/L and 0,5g/L

Where do I go wrong?

c) Calculate the solubility in moles/L of Ag₂SO₄(s) in 0,22 moles/L AgNo₃

My answer

1,6 * 10^-5 (moles/L)³= (0,22 moles/L + 2x)² * x

We assume that 2x is so small compared to 0,22 moles/L, that we remove it from the equation.

0,0484x = 1,6 * 10^-5 (moles/L)³

x = 3,3 * 10^-4 (moles/L)³

My books answer is the same, but in moles/L and not in (moles/L)³ (3,3 * 10^-4

What's the difference here?

[Corribus]

4x = 4 * 10^-6

You appear to have just made some algebra error at this step. I'm not sure exactly what you did, though.

4x³ = 1.6E-5

Divide both sides by 4 to get x³ = 4E-6

Cube root of 4E-6 gives x = 0.0158.

Good news is that you set the problem up right, just be careful following through on the math.

[kensher]

Thanks for the reply!

[kensher]

How about c) btw? I will repost it because I see that i didn't complete it.

c) Calculate the solubility in moles/L of Ag₂SO₄(s) in 0,22 moles/L AgNo₃

My answer

1,6 * 10-5 (moles/L)³= (0,22 moles/L + 2x)² * x

We assume that 2x is so small compared to 0,22 moles/L, that we remove it from the equation.

0,0484x = 1,6 * 10-5 (moles/L)³

x = 3,3 * 10-4 (moles/L)³

What I don't understand

My book's answer is the same, but in moles/L (3,3 * 10-4 moles/L) and not in (moles/L)³ (3,3 * 10-4 moles/L)³

Does anybody know why this is?

[Borek]

1,6 * 10-5 (moles/L)³= (0,22 moles/L + 2x)² * x

We assume that 2x is so small compared to 0,22 moles/L, that we remove it from the equation.

0,0484x = 1,6 * 10-5 (moles/L)³

0.0484x - what happened to the units?

[kensher]

Corrected, but I still don't understand the last steps; unless you are allowed to divide (moles/L)³ with (moles/L)² and get moles/L¹ = moles/L.

My answer

1,6 * 10^-5 (moles/L)³= (0,22 moles/L + 2x)² * x

We assume that 2x is so small compared to 0,22 moles/L, that we remove it from the equation.

0,0484x (moles/L)² = 1,6 * 10-5 (moles/L)³

x = 3,3 * 10^-4 (moles/L)³

[Borek]

Corrected, but I still don't understand the last steps; unless you are allowed to divide (moles/L)³ with (moles/L)² and get moles/L¹ = moles/L.

Sure you are, that's how you deal with units. You can add/subtract only values that have same units, you can cancel units when multiplying/dividing, powers (including fractional ones) apply as well. However, functions like sin/log/exp require dimensionless arguments.

[kensher]

Thanks for the reply!