Find the molar solubility of BaCrO4(Ksp=2.1 ×10^-10) in (a) pure water and (b) 1.5 ×10^-3 M Na2

step 1 A KSP of BACR -O -4 is 2 .1 times 10 to the negative 10th. In order to find the solubility, we h

Find the molar solubility of BaCrO4(Ksp=2.1 ×10^-10) in (a)...

Key Concepts

Solubility Product (Ksp)

The solubility product constant is an equilibrium constant specific to the dissolution of a sparingly soluble ionic compound. It is calculated from the concentrations of the ions at equilibrium, each raised to the power of their stoichiometric coefficients in the dissolution reaction. Ksp provides a quantitative measure of a compound's solubility in water.

Molar Solubility

Molar solubility refers to the number of moles of a substance that can dissolve per liter of solution to reach saturation. It is determined by setting up the equilibrium expression using the Ksp and solving for the concentration of ions formed from the dissociation of the compound in water.

Common Ion Effect

The common ion effect describes the decrease in solubility of a sparingly soluble salt when a soluble salt containing one of the ions is added to the solution. The additional presence of the common ion shifts the dissolution equilibrium towards the undissolved salt, thereby reducing its molar solubility.

Transcript

00:01 A ksp of bacr -o -4 is 2 .1 times 10 to the negative 10th.

00:07 In order to find the solubility, we have to start by writing out the equation for a saturated solution of bacr -o -4.

00:16 B -a -c -r -o -4 solid partially dissociates to b -a -2 plus, equates plus cr 04 2 minus aqueous.

00:36 This means our ksp expression is the concentration of ba2 plus times the concentration of cr 04 2 minus.

00:50 Now let's set up a nice table.

00:56 We don't use solids in our calculations, so we can just ignore bacr 04 here.

01:03 But for our aqueous compounds, our initial concentrations are just zero.

01:13 Then for our change, we have some unknown amount dissociating into ba2 plus and cro4 -2 minus, and we can just call that x.

01:26 So plus x for each of these, since there's a one -to -one ratio between them and also between ba, cro4 in each of them.

01:36 Then our equilibrium is just x for each since it's the initial plus the change.

01:46 So plugging in our equilibrium concentrations into our ksp expression, we get ksp is equal to x squared.

01:58 And then we can take our value of ksp and solve for x.

02:02 So x squared is our ksp value, which is 2 .1 times 10 to the negative 10.

02:10 Solving for x, we get 1 .4 times 10 to the negative 5th.

02:18 That means that our final concentration of our products here, ba2 plus and cro42 -2 minus, are each 1 .4 times 10 to the negative 5 molar...

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