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As in each part, the problem is always the same in each example. It is:

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

Here are the two substances:

Sn(OH)2	Ksp = 5.45 x 10¯27
Ag2CrO4	Ksp = 1.12 x 10¯12

Tin(II) hydroxide

Here is the equation for dissociation:

Sn(OH)₂ <===> Sn²⁺ + 2 OH¯

and here is the K_sp expression:

K_sp = [Sn²⁺] [OH¯]²

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

As in right now!!

The ratio between Sn²⁺ and OH¯ is one-to-two. That means that however much Sn²⁺ dissolves, we get DOUBLE that amount of OH¯. This is important, so go slow and think it through.

One Sn²⁺ makes two OH¯. That means that if 'x' Sn²⁺ dissolves, then '2x' of the OH¯ had to have dissolved.

Boy, I hope you got that!! Let's move on.

We now write an equation:

5.45 x 10¯²⁷ = (x) (2x)²

Wait, Mr. ChemTeam person, why did you DOUBLE the concentration of hydroxide?

I didn't. The concentration of OH¯ is equal to double the concentration of Sn²⁺. Let's move on.

Oops! Important point. Don't write 2x². That's wrong! It's (2x)²; two x the quantity squared.

So we have:

4x³ = 5.45 x 10¯²⁷

and solving that, we get:

x = 1.11 x 10¯⁹ M

Now, this is the answer because there is a one-to-one ratio between Sn²⁺ and Sn(OH)₂

Silver chromate

Here is the usual info:

Ag₂CrO₄ <===> 2 Ag⁺ + CrO₄²¯
K_sp = [Ag⁺]² [CrO₄²¯] = 1.12 x 10¯¹²

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

1.12 x 10¯¹² = (2x)² (x)

In other words, what I'm doing is allowing x to equal the concentration of whatever ion has a one in front of it. That ion will always be in a one-to-one ratio with the solid which is dissolving. So solving for x gives me the molar solubility of the substance.

We now have:

4x³ = 1.12 x 10¯¹²

which is solved to give the answer:

x = 6.54 x 10¯⁵ M

Do you want to see more problems like these?

Go to Solving K_sp Problems - Part One
Go to Solving K_sp Problems - Part Three
Go to Solving K_sp Problems - Part Four

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