Question

If only 0.225 g of Ca(OH)2 dissolves in enough water to give 0.100 L of aqueous solution at a given temperature, what is the Ksp value for calcium hydroxide at this temperature

Answer 1

Answer: The
`K_(sp)` value for calcium hydroxide at this temperature is
`1.08 * 10^(-4)`.

Step-by-step explanation:

Given: Mass of
`Ca(OH)_(2)` = 0.225 g

Volume = 0.100 L

As moles is the mass of substance divided by its molar mass.

So, moles of
`Ca(OH)_(2)` (molar mass = 74 g/mol) is calculated as follows.

`Moles = (mass)/(molar mass)\\= (0.225 g)/(74 g/mol)\\= 0.003 mol`

Molarity is the number of moles of substance present in a liter of solution.

Hence, molarity of given solution will be as follows.

`Molarity = (moles)/(Volume (in L))\\= (0.003 mol)/(0.1 L)\\= 0.03 M`

The equation for dissociation of
`Ca(OH)_(2)` is as follows.

`Ca(OH)_(2) \rightarrow Ca^(2+) + 2OH^(-)`

This means that
`[Ca^(2+)] = 0.03` and
`[OH^(-)] = 2 * 0.03 = 0.06`. Hence,
`K_(sp)` value for this reaction is calculated as follows.

`K_(sp) = [Ca^(2+)][OH^(-)]^(2)\\= (0.03) * (0.06)^(2)\\= 1.08 * 10^(-4)`

Thus, we can conclude that the
`K_(sp)` value for calcium hydroxide at this temperature is
`1.08 * 10^(-4)`.