(a) The molar solubility of PbBr2 at 25^∘ C is 1.0 ×10^-2 mol / L Calculate Ks p-(𝐛) If 0.0490 g

step 1 So now we'll work on problem 54 from chapter 17. In this problem, we have three parts. In the fi

(a) The molar solubility of PbBr2 at 25^∘ C is 1.0 ×10^-2...

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

Key Concepts

pH Calculation in Saturated Solutions

Calculating the pH of a saturated solution, especially for substances that form basic or acidic solutions, involves determining the concentration of hydrogen ions or hydroxide ions generated from the dissolution. For a base like calcium hydroxide, converting the molar solubility to hydroxide ion concentration and then using the definitions of pOH and pH allows for the determination of the solution’s pH.

Unit Conversion in Solubility Calculations

This concept involves converting a given solubility, often presented in grams per liter, into molar solubility by utilizing the molar mass of the compound. It is a necessary step in determining the concentrations of ions for the Ksp calculation, ensuring consistency in the units used in equilibrium expressions.

Equilibrium Expressions and Stoichiometry

Understanding the stoichiometric relationships in the dissolution reaction of a salt is critical for setting up the equilibrium expression of the solubility product. Each ion produced by the dissolution reaction appears in the expression for Ksp with an exponent equal to its coefficient, linking the molar solubility to the equilibrium concentrations of all species in solution.

Molar Solubility

Molar solubility is the number of moles of a salt that can dissolve in one liter of solution to reach saturation. It is an essential parameter for determining the ion concentrations in a saturated solution and is directly used to calculate the solubility product constant when combined with the stoichiometric relationships inherent to the salt’s dissolution reaction.

Solubility Product Constant (Ksp)

This concept refers to the equilibrium constant for the dissolution of a sparingly soluble salt in water. It is defined as the product of the molar concentrations of the ions in solution, each raised to the power of its stoichiometric coefficient according to the balanced dissolution equation. Ksp provides insight into the extent of a salt's solubility and is pivotal in predicting precipitation in solutions.

Transcript

00:01 So now we'll work on problem 54 from chapter 17.

00:08 In this problem, we have three parts.

00:11 In the first part, we're told that the molar solubility of lead bromide at 25 degrees celsius is 1 .0 times 10 to the minus 2 moles per liter.

00:20 Calculate ksp.

00:24 So we're dealing with lead bromide, pb, br2, and we're told that the solubility is 1 times 10 to the minus 2 molar.

00:36 So we know that ksp for this compound will be equal to the concentration of lead times the concentration of bromine squared.

01:00 And so the concentration of lead, if we have a saturated solution or based on the molar solubility, is 1 times 10 to the minus 2.

01:13 And the concentration of bromine would be two times that because of the stochialometry of the compound.

01:18 And we square that value, and that gives us a value of 4 .0 times 10 to minus 6, as our ksp value for lead bromide.

01:35 Now in part b, we're asked if 0 .0 490 grams of silver iodate dissolves per liter of solution, calculate the solubility product constant.

01:49 So they're giving us the solubility in grams per liter.

01:58 So what we need to go and go ahead and do is convult that to moles per liter.

02:02 So we have 0 .0 490 grams times one mole divided by the molar mass, which is 282 .77.

02:18 And that's equal to 1 .73 times 10 to the minus 4 moles of silver iodate...

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