Calculate the Kf of Ag(NH3)2+from Ag^+(a q)+e^- ⇌Ag(s) E^∘=0.80 V Ag(NH3)2^+(a q)+e^- ⇌Ag(s)+2

Calculate the Kf of Ag(NH3)2+from Ag^+(a q)+e^- ⇌Ag(s) ...

Key Concepts

Nernst Equation and Its Relation to Equilibrium

The Nernst equation relates the electrode potential of a redox reaction to the concentrations of the reactants and products and, importantly, to the reaction's equilibrium constant. Specifically, the difference in standard potentials between related half?reactions can be linked via the relationship ?E° = (RT/nF) ln K, where K is the equilibrium constant. This connection is crucial when calculating formation constants from electrochemical data.

Standard Electrode Potential

This concept refers to the voltage associated with a reduction half?reaction under standard conditions. It is a measure of the tendency of a chemical species to be reduced, and it is key to comparing the reactivity of different half?reactions. In problems like this, standard electrode potentials allow the determination of equilibrium constants when reactions are coupled.

Formation Constant of Complex Ions

The formation constant (Kf) quantifies the stability of a complex ion formed between a metal ion and one or more ligands. It represents the equilibrium constant for the reaction in which the complex is formed from its constituents. In the context of this problem, the formation constant is calculated by comparing the standard electrode potentials of related redox reactions, thereby providing insight into the thermodynamic stability of the complex.

Transcript

00:01 Question number 144 is a complex ion exchange question where the calcium is exchanged for lead with the ligand in the complex ion being edta in each case.

00:21 This is the equilibrium that we're considering.

00:24 The calcium here gets substituted for lead in order to form these two products.

00:30 And the equilibrium constant for this is 2 .5 times 10 to the 7, which would be expressed as a ratio of the product concentrations over the reactant concentrations.

00:44 So to solve for the concentration of the lead, edta complex, we are going to assume that this reaction is 100 % efficient, meaning that every mole of lead, that is present ends up being converted into this lead edta complex.

01:12 So we can calculate the concentration of the lead edta complex at equilibrium, assuming that every mole of lead that was present has now become this complex.

01:24 So to calculate this concentration, we need to know the moles divided by the volume.

01:29 Well, we know the micrograms of lead that is.

01:34 Present per mill liter.

01:36 This will now correspond to the amount of lead complex that is present per 100 milliliters.

01:44 First thing that we need to do is convert the micrograms of lead into grams of lead.

01:52 Then we can go from grams of, i'm sorry, then what i'm doing here is i'm converting the mill liters in the denominator to liters so i can ultimately end up with moles per liter, which is molarity.

02:05 Then i want to convert the grams of lead to moles of lead, and then recognize that every one mole of lead has now become moles of the lead edta complex.

02:19 So my units are now moles, lead, edta complex, divided by liter solution.

02:26 This then gives me the concentration of the lead complex that has been formed in the blood.

02:36 To calculate the calcium concentration, i recognize that it is going to be equal to the concentration of the lead complex that's formed.

02:47 Why? because i will assume that there was no calcium in the blood to begin with, and every one of these that is created, i also create a calcium.

02:57 So these two concentrations must be equal.

03:03 But what about the concentration of the calcium edta complex that was added in order to push this equilibrium to the right.

03:13 Well, not all of it was consumed.

03:15 I added 0 .1 liters, 100 mil liters, at 0 .1 molar.

03:21 That gives me the total moles of this that i added.

03:24 The amount that was consumed is going to correspond to the volume of blood that was present multiplied by the concentration of the lead that was present, the 120 micrograms per 100 milliliters.

03:44 I then recognize that the moles, so this will be the moles of lead that was added, which is now also the moles of calcium, edta that was consumed.

03:59 So knowing now that i have, let's see here, so i've got the grams of, okay let me step back.

04:09 So this is the grams of lead...

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