(a) The molar solubility of PbBr2 at 25^∘C is 1.0 ×10^-2 mol / L . Calculate Ks p

step 1 We're going to be doing KSP calculations. These are a little tougher than previous calculations,

(a) The molar solubility of PbBr2 at 25^∘C is ...

$\begin{array}{llll} \text { (a) The molar solubility of } \mathrm{PbBr}_{2} & \text { at } 25^{\circ} \mathrm{C} & \text { is }\end{array}$ $1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .$ Calculate $K_{s p} .(\mathbf{b})$ If $0.0490 \mathrm{~g}$ of $\mathrm{AgIO}_{3}$ dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate $K_{s p}$ value from Appendix $\mathrm{D},$ calculate the $\mathrm{pH}$ of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_{2}$

$\begin{array}{llll} \text { (a) The molar solubility of } \mathrm{PbBr}_{2} & \text { at } 25^{\circ} \mathrm{C} & \text { is }\end{array}$
$1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .$ Calculate $K_{s p} .(\mathbf{b})$ If $0.0490 \mathrm{~g}$ of $\mathrm{AgIO}_{3}$ dis-
solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate $K_{s p}$ value from Appendix $\mathrm{D},$ calculate the $\mathrm{pH}$ of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_{2}$

Key Concepts

pH and Basic Solutions Calculation

The pH of a solution is related to its hydrogen ion concentration, and in the case of a basic solution, the hydroxide ion concentration is used to determine pOH and subsequently pH. For sparingly soluble bases that release OH- ions, calculating the pH involves using the known or calculated solubility to determine the concentration of OH- in solution, and then applying the relationships pOH = -log[OH-] and pH + pOH = 14.

Mass-to-Mole Conversion

Solubility is sometimes given in mass per liter rather than moles per liter. Converting from grams per liter to moles per liter is essential for using the solubility in equilibrium calculations. This conversion is performed by dividing the mass by the molar mass of the compound, linking quantitative lab measurements to theoretical predictions.

Dissolution Equilibrium and Stoichiometry

When a sparingly soluble salt dissolves, it reaches a dissolution equilibrium where the salt and its ions coexist in solution. The balanced chemical equation for the salt’s dissolution provides the stoichiometric relationships between the undissolved salt and the ions formed, which is essential for setting up the Ksp expression and performing solubility calculations.

Molar Solubility

Molar solubility is defined as the number of moles of a substance that can dissolve per liter of solution to form a saturated solution. This value is important because it allows for the determination of the concentrations of individual ions in solution, using the stoichiometry of the dissolution reaction.

Solubility Product Constant (Ksp)

Ksp is the equilibrium constant that applies to the dissolution of a sparingly soluble salt into its constituent ions in a saturated solution. It is expressed as the product of the ionic concentrations, each raised to the power of its stoichiometric coefficient, and serves as a measure of the salt’s solubility. A smaller Ksp value indicates lower solubility, while a larger value indicates higher solubility.

Transcript

00:01 We're going to be doing ksp calculations.

00:03 These are a little tougher than previous calculations, but not too bad.

00:07 We have three to do.

00:09 In our first substance, we're given that the molar solubility of lead to bromide, which is pbbr2, at 25 degrees celsius, is 1 .0 times 10 to the minus 2 moles per liter or molarity.

00:34 And we're asked to find ksp.

00:39 Remember, ksp is a solubility product.

00:43 Do any equilibrium type calculation, we need to write the chemical equation.

00:49 Pvbr2 solid is in equilibrium with its ions, the cadion, it's pv2 plus, get aqueous solution, and 2br minus two bromides in aqueous solution.

01:10 We are given that the molar solubility is 1 .0 times 10 to the minus 2 for this substance, which means it'll be the same for this because it's got a 1 -1 -mole ratio and two times that concentration for the bromide because that's got a one to two more ratio.

01:35 If you recall the ksp for any substance that breaks into three ions is four x, where x is the concentration, cubed.

01:46 Substituting in our given concentration, we get four times 1 .0 times 10 to the minus two cubed.

01:58 So our ksp is for this one, 4 .0 times 10 to the minus 6th.

02:11 That is our ksp for problem a.

02:17 Problem b gives us the following information.

02:23 We have 0 .0490 grams of silver iodate per liter of solution.

02:38 Find ksp.

02:43 Since we're given grams, we're going to have to convert that to moles.

02:47 Let's do that first.

02:48 0 .0490 grams of agi -03.

02:55 I just cheated and looked this up and got 282 .77 grams per mole, which gave me, i didn't even figure that out.

03:11 Let me go ahead and figure that out for us, because i just put that into our equation and squared it right away.