The solubility product of AgC2 O4 at 25^∘ C is 2.3 ×10^-1 i M^3. A solution of K2 C2 O4 containi

The solubility product of AgC2 O4 at 25^∘ C is 2.3 ×10^-1 i...

The solubility product of $\mathrm{AgC}_{2} \mathrm{O}_{4}$ at $25^{\circ} \mathrm{C}$ is $2.3 \times 10^{-1 i} \mathrm{M}^{3}$. A solution of $\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ containing $0.15$ moles in $500 \mathrm{ml}$ water is shaken at $25^{\circ} \mathrm{C}$ with excess of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ till the following equilibrium is reached: $\mathrm{Ag}_{2} \mathrm{CO}_{3}+\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{K}_{2} \mathrm{CO}_{3}$ At equilibrium, the solution contains $0.035$ mole of $\mathrm{K}_{2} \mathrm{CO}_{3}$. Assuming the degree of dissociation of $\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ and $\mathrm{K}_{2} \mathrm{CO}_{3}$ to be equal, calculate the solubility product of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ (a) $2.3 \times 10^{-11} \mathrm{M}^{3}$ (b) $7.0 \times 10^{-10} \mathrm{M}^{3}$ (c) $3.0 \times 10^{-13} \mathrm{M}^{3}$ (d) $7.0 \times 10^{-12} \mathrm{M}^{3}$

The solubility product of $\mathrm{AgC}_{2} \mathrm{O}_{4}$ at $25^{\circ} \mathrm{C}$ is $2.3 \times 10^{-1 i} \mathrm{M}^{3}$. A solution of $\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
containing $0.15$ moles in $500 \mathrm{ml}$ water is shaken at $25^{\circ} \mathrm{C}$ with excess of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ till the following equilibrium is reached:
$\mathrm{Ag}_{2} \mathrm{CO}_{3}+\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{K}_{2} \mathrm{CO}_{3}$
At equilibrium, the solution contains $0.035$ mole of $\mathrm{K}_{2} \mathrm{CO}_{3}$. Assuming the degree of dissociation of $\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ and $\mathrm{K}_{2} \mathrm{CO}_{3}$ to be equal, calculate the solubility product of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$
(a) $2.3 \times 10^{-11} \mathrm{M}^{3}$
(b) $7.0 \times 10^{-10} \mathrm{M}^{3}$
(c) $3.0 \times 10^{-13} \mathrm{M}^{3}$
(d) $7.0 \times 10^{-12} \mathrm{M}^{3}$

Key Concepts

Stoichiometry in Equilibrium Reactions

This concept involves the quantitative relationships between reactants and products in a chemical reaction. In equilibrium systems, stoichiometry is used to relate the moles of each species involved, ensure mass balance, and develop expressions for equilibrium constants like the solubility product. It is crucial in determining how changes in one component affect the concentrations of others at equilibrium.

Solubility Product (Ksp)

This concept refers to the equilibrium constant for the dissolution of a sparingly soluble ionic compound. It is defined as the product of the molar concentrations of the constituent ions each raised to the power of their stoichiometric coefficients in the dissolution equation. The Ksp quantifies the limit at which the compound will dissolve in water, providing important information about precipitation and solubility equilibria in aqueous solutions.

Chemical Equilibrium

Chemical equilibrium is the state in which the rates of the forward and reverse reactions in a chemical system are equal, so there is no net change in the concentrations of reactants and products. In the context of solubility and salt precipitation, understanding equilibrium allows one to predict the extent to which a salt will dissolve or precipitate when mixed with other reagents.

Degree of Dissociation

The degree of dissociation refers to the fraction of a solute that separates into its constituent ions in solution. When compounds, especially salts that fully dissociate, are involved in equilibrium, assuming equal degrees of dissociation can simplify calculations by allowing for a common relationship between ion concentrations. This is particularly useful when comparing the behavior of similar salts in a mixed system.

Transcript

00:01 So for this problem, we're given two different substances, and we're given their solubility, and we're asked to calculate the ksp value for each of the substances.

00:11 Now, the solubility that we're given is the unit or the amount of substance in grams per 100 grams of water.

00:21 But in order to use the solubility in our ksp equation, the first thing we have to do is convert this into, moles per liter.

00:33 So we're going to start by doing some dimensional analysis.

00:38 And so what we have to start is our 2 .4 times 10 to negative 4 grams of barium sulfate.

00:51 And that is per 100 grams of water.

00:59 So i'm going to start by first getting the grams of water.

01:02 Into liters.

01:04 And the first thing we know is that one gram of water is equal to one milliliter of water.

01:16 If i'm trying to get to liters, i know that there are 1 ,000 milliliters in one liter of water.

01:27 Okay, so going through, this will cancel my grams, grams, milk, mills, mills, so i'm left with liters on the bottom.

01:39 Perfect.

01:41 Now i have to get the top, the grams of barium sulfate, into moles.

01:48 So i'm going to use the gram formula mass, or the molar mass of barium sulfate.

01:53 The molar mass of barium sulfate is 233 .38 grams per mole.

02:09 We'll be able to cancel our units.

02:12 The grams of barium sulfate are going to become moles of barium sulfate.

02:18 So in order to plug this into my calculator, i'm going to multiply straight across the top, straight across the bottom.

02:24 We get 2 .4 times 10 to the negative 4 multiplied by a thousand, and then divide that by 233 .38 multiplied by 100.

02:37 When i do that, i plug it into my calculator and i get, so without rounding too much first, i'm going to get 1 .0 to 84 times 10 to the negative 5th molar or moles per liter.

02:59 Now for the ksp expression, ksp is the product of the ion concentrations.

03:10 Now for barium sulfate, there's one barium and there's one sulfate polyatomic ion.

03:17 So the ksp is going to be equal to the concentration of the barium 2 plus ion and the concentration of the sulfate ion.

03:32 So for this, if this is my solubility, this is what i'm going to plug in for each of these ions.

03:41 So remember, in solution, barium sulfate, one mole of this.

03:45 Solid would give you two moles of ions, one barium and one sulfate.

03:52 So we're going to take 1 .084 times 10 to the minus 5, and the concentration of sulfate would be the same because it's a 1 to 1 ratio.

04:17 So 1 .02 .84 times 10 to the minus 5 squared.

04:25 That's going to give me 1 .0576, but we only need two sighings, right? so our ksp value should only be reported to two significant figures.

04:36 So we're going to take 1 .1, and that is multiplied by 10 to the negative 10th power.

04:44 So this is my ksp value for the barium sulfate.

04:50 We're asked to do the same thing for calcium hydroxide.

04:54 And again, i need to take this form of solubility and put it into the format moles per liter.

05:01 So we'll do the same thing.

05:05 We start out knowing that we can dissolve...

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