Use Appendix D to calculate the Ksp of AgCl.

Name: Problem 142, Chapter 19: Chemistry: The Molecular Nature of Matter and Change
Uploaded: 2020-11-20T01:13:32-08:00
Duration: 6 min 14 s
Channel: David Collins
Description: Use Appendix D to calculate the Ksp of AgCl.

Key Concepts

Use of Thermodynamic Data for Equilibrium Calculations

In many chemistry contexts, thermodynamic data such as standard free energies, enthalpies, or standard electrode potentials are used to calculate equilibrium constants. These data, often compiled in appendices of textbooks, allow for the determination of constants like K_sp by linking experimentally measurable properties to theoretical equilibrium expressions.

Application of the Law of Mass Action

The law of mass action underpins the calculation of equilibrium constants by stating that, at equilibrium, the concentration of products and reactants is related in a fixed manner. For a sparingly soluble salt, the law of mass action is used to derive the expression for K_sp by multiplying the concentrations of the resulting ions, which provides a quantitative measure of the salt's solubility.

Solubility Product Constant (K_sp)

The solubility product constant is an equilibrium constant that applies to the dissolution of a sparingly soluble salt. It represents the product of the molar concentrations of the ions produced in solution at equilibrium, each raised to the power of its stoichiometric coefficient. This constant provides insight into how much of the solid salt can dissolve in a solvent under equilibrium conditions.

Solubility Equilibrium

Solubility equilibrium refers to the dynamic balance between a sparingly soluble solid and its dissolved ions in a saturated solution. At equilibrium, the rate at which the solid dissolves equals the rate at which ions reassemble into the solid, and the concentrations of the dissolved ions are fixed by the solubility product constant.

Transcript

00:02 Question 142 is a weak acid calculation, where the weak acid is carbonic acid that is formed due to the absorption of carbon dioxide into water.

00:15 They want to know what the ph of rainwater is, given the amount of carbon dioxide that can be absorbed into rainwater and the dissociation values of carbonic acid.

00:28 We know that carbon dioxide will absorb into water, producing carbon dioxide aqueous, according to this expression here.

00:41 This was given in a previous problem, 3 .1 times 10 to the negative 2, will then be the calcium concentration in water, divided by the pressure of the carbon dioxide concentration in water, divided by the carbon dioxide pressure above water.

00:58 Then once the carbon dioxide aqueous is in water, it can dissociate into h plus and h.

01:08 Co3 minus, and it has its own equilibrium constant that was provided in this question, 4 .5 times 10 to the negative 7.

01:17 It may not have been obvious that the k -a value they gave you was for this reaction, but if you look it up online, it is obvious that this k value is for carbon dioxide reacting with water, producing h plus and bicarbonate.

01:33 They also tell us that if we have one atmosphere of carbon dioxide above a solution then we can get 88 mil liters of carbon dioxide dissolved in 100 mil liters of water but in one atmosphere we don't have one atmosphere of pressure of carbon dioxide because it's only 0 .04 percent carbon dioxide so that means we'd only have 0 .04 % of 88 in 100 milliliters, which would be 0 .0352 milliliters per 100 milliliters of water.

02:16 So now we go back to this equation here and we see that a rearrangement will allow us to solve for the carbon dioxide concentration in water.

02:31 It will simply be equal to, actually let me back up, instead of using this equation now, what we want to do is convert the concentration of carbon dioxide in water in these funny units to molarity.

02:49 So if this is the amount that we have in mill liters per 100 milliliters of solvent, we can convert the milliliters of co2 into liters of co2, which is kind of strange now that it's dissolved in water.

03:07 And this is where things seem to go awry a little bit.

03:11 And then we can convert the liters into moles, recognizing it's at, wait, it's not at stp, it's at 25 degrees celsius instead of 0 degrees celsius.

03:27 So a 25 degrees celsius, molar volume is you can calculate this using pv equals nrt 24 .5 liters rather than 22 .4 .4.

03:40 So this is how many liters at 25 degrees celsius, one atmosphere corresponds to one mole of gas.

03:49 Then we want to convert our mill liters of h2o into liters of h2o, and this gives us 1 .44 times 10 to the negative 5 molar...

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