The solubility of AgCl(s) in water at 25^∘ C is 1.33 ×10^-5 mol / L and its ΔH^∘ of solution is

The solubility of AgCl(s) in water at 25^∘ C is 1.33 ×10^-5...

The solubility of $\mathrm{AgCl}(s)$ in water at $25^{\circ} \mathrm{C}$ is $1.33 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$
and its $\Delta H^{\circ}$ of solution is $65.7 \mathrm{~kJ} / \mathrm{mol}$. What is its solubility at $50.0^{\circ} \mathrm{C} ?$

Key Concepts

Solubility Equilibria

This concept refers to the equilibrium established between a solid phase and its constituent ions in a saturated solution. It typically involves the solubility product constant (Ksp), which quantifies the extent to which a compound dissolves in water. Understanding solubility equilibria is crucial to predicting how much of a compound can dissolve under various conditions.

Enthalpy of Solution

The enthalpy of solution represents the heat change when a substance dissolves in a solvent. It can be either endothermic or exothermic. In endothermic dissolution, increasing the temperature generally increases solubility because additional heat favors the dissolution process, while in exothermic processes, higher temperatures might reduce solubility.

Temperature Dependence of Equilibrium

This concept involves the way temperature changes affect chemical equilibria. According to Le Chatelier's principle, if a process is endothermic, increasing the temperature shifts the equilibrium toward the products, thus increasing solubility. Conversely, for exothermic processes, higher temperatures can shift the equilibrium toward the reactants, decreasing solubility.

van't Hoff Equation

The van't Hoff equation is a mathematical expression that relates the change in the equilibrium constant of a chemical reaction to the change in temperature and the enthalpy change of the reaction. It provides a framework for quantifying how solubility (through Ksp as an equilibrium constant) varies with temperature, making it a key tool in understanding solubility trends.

Transcript

00:01 In the given question, we have been provided solubility of agcl at 25 degrees celsius is equal to 1 .33 into 10 to the power minus 5 mole per liter.

00:13 Okay, that is the solubility at 25 degrees celsius.

00:19 And we have to find the solability at 50 degrees celsius.

00:24 That is the question.

00:26 Now, we have been further provided that the delta h of this reaction is 65 .7 kilojoule per mole.

00:36 Okay, so how we can do this question? first, we will find the equilibrium constant at 25 degrees celsius.

00:43 Okay, so in this case, since our agcl solubility is given, so agcl dissociate into ag positive, that is the ketan, and cl minus, that is the an ion.

00:56 And equilibrium constant will be equals to ksp for this type of compounds or ina compound and ksp will be equal to solubility square.

01:06 Okay, so solubility is 1 .33 into 10 to the power minus 5.

01:11 So when we do its square, our ksp will come out to be 1 .77 into 10 to the power minus 10 and that is the value of k1 at 25 degrees celsius.

01:27 Now we can calculate k2 at a different temperature.

01:31 Temperature which is 50 degrees celsius by utilizing the equation that ln k2 over k1 is equal to minus delta h0 of the reaction divided by universal gas constant 1 upon t2 minus 1 upon t1.

01:48 But always remember the temperature should be in kelvin.

01:51 So 25 degrees celsius become 298 kelvin.

01:55 We add 273 to convert it in kelvin.

01:58 Okay.

01:59 And similarly over 50 degrees celsius.

02:01 Will be equal to 323 kelvin.

02:05 Okay.

02:06 So now we can find the value of equilibrium constant at 50 degrees celsius.

02:10 So we light it as k2 only.

02:14 K1 is 1 .7.

02:16 7 into 10 to the power minus 10...