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# Solution: 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

###### 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

###### 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:

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