(a) La solubilidad molar del PbBr2 a 25^∘ C es de 1.0 ×10^-2 mol / L. Calcule Kp s. (b) Si se di

step 1 We are asked to complete three KSP calculations. I'll do them each separately. For the first cal

(a) La solubilidad molar del PbBr2 a 25^∘ C es de 1.0 ×10^-2...

(a) La solubilidad molar del $\mathrm{PbBr}_2$ a $25^{\circ} \mathrm{C}$ es de $1.0 \times 10^{-2}$ $\mathrm{mol} / \mathrm{L}$. Calcule $K_{p s}$. (b) $\mathrm{Si}$ se disuelven $0.0490 \mathrm{~g}$ de $\mathrm{AgIO}_3$ por litro de disolución, calcule la constante del producto de solubilidad. (c) Con base en el valor apropiado de $K_{p s}$ tomado del apéndice $D$, calcule la solubilidad del $\mathrm{Cu}(\mathrm{OH})_2$ en gramos por litro de disolución.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant is an equilibrium constant specific to sparingly soluble salts and is defined as the product of the molar concentrations of the ions produced in the dissolution, each raised to the power of its coefficient in the balanced dissolution equation. It provides a quantitative measure of a salt's solubility under a given set of conditions.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve per liter of solution before the solution becomes saturated. It is a crucial parameter when relating the amount of dissolved salt to the concentration of its constituent ions in establishing the equilibrium expression for the solubility product.

Ionic Equilibrium and Stoichiometry

This key concept involves writing the dissolution equation for a sparingly soluble salt and using the stoichiometry to relate the molar solubility to the concentrations of the resulting ions. Correctly setting up these relationships is essential for forming the Ksp expression and solving for unknown concentrations or solubility limits.

Mass-to-Mole Conversion

Mass-to-mole conversion is a fundamental step in many solubility problems where the amount of solute is given in grams per liter but the equilibrium calculations require molar concentrations. This conversion uses the molar mass of the solute to translate a measured mass into the number of moles, facilitating accurate determination of solubility and Ksp values.

Transcript

00:01 We are asked to complete three ksp calculations.

00:05 I'll do them each separately.

00:07 For the first calculation, we are given the molar solubility of lead to bromide at 25 degrees celsius.

00:21 And we're told it's 1 .0 times 10 to the minus 2 moles per liter.

00:32 From this, we're asked to find ksp.

00:38 As always, we write our chemical equation, and we can also write our ksp expression.

01:16 We note that our molar solubility is given here, so that is the concentration of the lead.

01:24 It'll be 2x for this because there's two bromings for every one of these.

01:30 And let us set up our chemical equation.

01:45 There's our lead, there's our iodate, and we end up with 4 .0 times 10 to the minus 6th is our ksp.

02:16 Problem one is done.

02:24 For problem b, we're considering, i believe it's silver iodate, and we are given that we have 0 .0 490 grams per liter of solution.

02:55 Find ksp.

03:01 So for this, we're going to have to convert grams to moles and then do a calculation similar to our first one.

03:08 Actually, a little bit easier because each of our, our cation ion and ion each have a charge of one.

03:16 Step 1.

03:17 Grams to moles.

03:27 Well, my last mass of the silver iodate is 282 .8 grams per mole.

03:38 So we get 1 .733 times 10 to the minus 4th moles per liter.

03:51 My ksp expression for this one is pretty simple.

03:55 So all we do is multiply our given value by two.

04:23 I'll just put it in for each expression.

04:26 1 .733 times 10 to the minus 4th times 1 .733 times 10 to the minus 4th.

04:42 And we end up with 3 .00 times 10 to the negative eighth is our ksp for our second one.

04:57 For our third and final problem, we are looking at, i think it was calcium hydroxide.

05:12 Yep, calcium hydroxide.

05:24 And we're going to find the solubility.

05:31 And this time, in grams per liter.

05:37 We're going to have to look up our ksp for this one.

05:41 So we look up our ksp in our book, we get that it is 6 .5 times 10 to the, minus sixth...

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