Problem: What is the solubility of M(OH) 2 in a 0.202 M solution of M(NO3)2? Ksp = 6.65 × 10−18 for M(OH)2.Calculate the molar solubility of lead (II) thiocyanate in 0.600 M KSCN.? K sp=2.00 x 10-5

FREE Expert Solution
89% (298 ratings)
FREE Expert Solution

$M(OH)_2$ has a $K_{sp}$, which stands for the solubility product constant. That means that $M(OH)_2$ is a solid, and it will dissociate as follows:

$M(OH)_{2(s)} \rightleftharpoons M^{2+}_{(aq)} + 2\ OH^-_{(aq)}$

Anytime we deal with $K_{sp}$, we are dealing with an ICE chart. Remember, in ICE charts we ignore anything that is a solid or a liquid, meaning we will ignore $M(OH)_{2(s)}$.

Now if this were dissolving in pure water, we would have 0 for the initial amounts of both products. But in this case, the compound is dissolving in 0.202 M of $M(NO_3)_2$. If we dissociate this, it becomes:

$M(NO_3)_2\rightleftharpoons M^{2+} + 2\ NO_3^{-}$

Notice that $M^{2+}$ is present in both compounds $M(OH)_2$ and $M(NO_3)_2$. That means that we are dealing with the common ion effect. Initially, we will have 0.202 M of $M^{2+}$ for our ICE chart. Initially, we have 0 $2\ OH^- _{(aq)}$. Then we complete the ICE chart as follows:

89% (298 ratings)
View Complete Written Solution
Problem Details

What is the solubility of M(OH) 2 in a 0.202 M solution of M(NO3)2? Ksp = 6.65 × 10−18 for M(OH)2.

Calculate the molar solubility of lead (II) thiocyanate in 0.600 M KSCN.? K sp=2.00 x 10-5

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Ksp concept. You can view video lessons to learn Ksp. Or if you need more Ksp practice, you can also practice Ksp practice problems.

How long does this problem take to solve?

Our expert Chemistry tutor, Jules took 8 minutes and 15 seconds to solve this problem. You can follow their steps in the video explanation above.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Ricciardo's class at OSU.