Solving Ksp Problems: Part Two - 4s3

Solving Ksp Problems:
Part One - s2
Solving Ksp Problems:
Part Three - 27s4
Solving Ksp Problems:
Part Four - 108s5
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The general problem is this:

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.

However, there is additional explaining to do when compared to the Part One section where I used AgCl and s2.


Example #1: Calculate the molar solubility of tin(II) hydroxide in pure water. Ksp = 5.45 x 10¯27

Solution:

1) Here is the equation for dissociation:

Sn(OH)2(s) ⇌ Sn2+(aq) + 2OH¯(aq)

2) Here is the Ksp expression:

Ksp = [Sn2+] [OH¯]2

So far, nothing out of the ordinary. However, that two in front of the hydroxide is important and will come into play real soon.

3) As in right now!!

Let us assign the variable 's' to be the molar solubility of Sn(OH)2.

From the chemical equation, we see that the amount of Sn(OH)2 that dissolves and the amount of Sn2+ that results are in a 1:1 molar ratio. Therefore:

[Sn2+] = s

When we look at the hydroxide, a different picture emerges. For every one Sn(OH)2 that dissolves, two OH¯ are produced. This means:

[OH¯] = 2s

This is an important point. Make sure you understand what happened. That s and 2s are rather important.

4) We now write an equation:

5.45 x 10¯27 = (s) (2s)2

Wait, Mr. ChemTeam person, why did you DOUBLE the concentration of hydroxide? (This is a common question and it shows the person does not understand the reasoning behind arriving at the 2s variable.)

I didn't double the concentration of the hydroxide. The moles of OH¯ that dissolved are double the moles of Sn(OH)2 that dissolved. For every one mole of tin(II) hydroxide that dissolves, two moles of hydroxide are present in solution. This is important. I didn't double the hydroxide concentration. I wrote it in terms of the amount of Sn(OH)2 that dissolved.

Oops! Important point. Don't write 2s2. That's wrong! It's (2s)2; two s the quantity squared. Not two s squared.

5) So we have:

4s3 = 5.45 x 10¯27

6) and solving that (divide by 4, then cube root), we get:

s = 1.11 x 10¯9 M

Example #2: Calculate the molar solubility of silver chromate in pure water. Ksp = 1.12 x 10¯12

Solution:

1) Here's the usual info:

Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42¯(aq)
Ksp = [Ag+]2 [CrO42¯]

2) In this problem the cofficient of 2 is on the first ion, not the second. No problem! Same solving technique

1.12 x 10¯12 = (2s)2 (s)

Keep in mind: what I'm doing is allowing s to equal the molar amount of the solid that dissolves. That substance is usually (not always) in a one-to-one ratio with the solid which is dissolving. So solving for s gives me the molar solubility of the substance.

3) We now have:

4s3 = 1.12 x 10¯12

4) which is solved to give the answer:

s = 6.54 x 10¯5 M

Example #3: Calcium fluoride, Ksp = 1.46 x 10¯10

Solution:

1) Here is the dissociation equation:

CaF2(s) ⇌ Ca2+(aq) + 2F¯(aq)

2) Here is the Ksp expression:

Ksp = [Ca2+] [F¯]2

3) The Ksp expression leads to this equation:

1.46 x 10¯10 = (s) (2s)2

4) Solving gives:

s = 3.32 x 10¯4 M

Example #4: Copper(I) sulfide, Ksp = 2.26 x 10¯48

Solution:

1) The dissociation equation and the Ksp expression:

Cu2S(s) ⇌ 2Cu+(aq) + S2¯(aq)
Ksp = [Cu+]2 [S2¯]

2) This gives:

2.26 x 10¯48 = (2s)2 (s)

3) when solved, gives:

s = 8.27 x 10¯17 M

Example #5: Fe(OH)2, Ksp = 4.87 x 10¯17

The molar solubility of Fe(OH)2 is 2.30 x 10¯6 M.

Bonus Example #1: Lead(II) azide, Pb(N3)2, Ksp = 2.5 x 10¯9

This substance ionizes as follows:

Pb(N3)2 ⇌ Pb2+(aq) + 2N3¯(aq)


Bonus Example #2: Copper(II) ferrocyanide, Cu2[Fe(CN)6], Ksp =1.3 x 10¯16

Cu2[Fe(CN)6](s) ⇌ 2Cu2+(aq) + [Fe(CN)6]4¯(aq)

Solving Ksp Problems:
Part One - s2
Solving Ksp Problems:
Part Three - 27s4
Solving Ksp Problems:
Part Four - 108s5
Back to Equilibrium Menu