Lecture 4 - Chemical Equilibrium (Part 2), Acids and Bases

Lecture Recording:

Lecture Summary/Notes:

We finished off chapter 15 today, and started Chapter 16 (Acids and Bases). We did A LOT of practice examples from the textbook today.

Reaction Quotient (Qc)

If we have a generic equation:

aA + bB <-> cC + dD

the reaction quotient for this equation would be:

Looks a lot like the equilibrium constant! What's the difference?
The  reaction quotient can be found for any set of concentrations (does not have to be at equilibrium). 
If a reaction has not yet reached equilibrium, Qc (reaction quotient) will be a different number from Kc (equilibrium constant). The comparison between the two constants will be useful in determining the 'movement' of the reaction. 

Tips for determining the movement of an equilibrium reaction.

We went through a couple of examples to apply this concept.


Textbook Example - P.E.A. and P.E.B. (page 672)

P.E.A - We first found the moles of each reactant in the equation (# of moles = mass/molar mass), and then found molarity (recall that molarity is equal to moles/volume). Since the mass and the volume is the same for each reactant, they cancel out with each other when put into the reaction quotient/equilibrium equation, leaving just the reciprocals of their molar masses. Qc turned out to be 5.7, which is greater than the given equilibrium constant (from the previous example) 1.00. Since Qc > Kc, the reaction will move to the left, showing an increase in CO and H2O.

 P.E.B. - This reaction involves gases, so we first found the  pressure-based reaction quotient (using the given partial pressures), and then compared it to the pressure-based equilibrium constant (found by converting the given concentration-based equilibrium constant). Qp was found to be less than Kp, indicating that the net change will occur to the right. 

Le Chatelier's Principle

When an equilibrium system changes in either temperature, pressure or concentration of a species, the system responds by setting a new equilibrium that sort of makes up for the change. We went through many examples to illustrate.


Textbook Example - P.E.A. and P.E.B. (page 674)

P.E.A. - Adding more moles of O2 to the reaction will increase the pressure of O2. To relax this change, the equilibrium will shift to the right.

P.E.B. - a) No effect - the activity of a pure solid is always 1.
             b) The reaction will shift to the left (because of the increased pressure)
             c) No effect - the activity of a pure solid is always 1.


Textbook Example - P.E.A. and P.E.B. (page 676)

P.E.A. - The pressure of both has doubled, and the total pressure has increased, thus the reaction will move to the left (because of the greater number of moles of gas on the left).

P.E.B. - A change in volume will result in a change in the pressure. In this case, the equilibrium would not be able to relax a change in the pressure, thus volume would have no effect on the equilibrium.

Textbook Example - P.E.A. and P.E.B. (page 678)

P.E.A. - We are given an enthalpy that is positive. From this we know that the reaction is endothermic, meaning that it absorbs heat as it goes to the right. Thus, raising the temperature would make this reaction move to the right (since absorption of heat would make it colder, compensating for the increase in temperature) and decreasing the temperature would make this reaction move to the left. Thus, the amount of NO2 would be greater at high temperatures.

P.E.B. - This is an exothermic reaction (since the given enthalpy is negative). The reaction will try to go to the left if the temperature is raised, thus the concentration of NH3 will be greater at 100C


Textbook Example - P.E.A. and P.E.B. (page 680)

P.E.A. - We first found the concentrations of each of the reactants by dividing the # of moles by the volume, and then found the equilibrium constant... too easy?

P.E.B. - (in the right corner are the given values). We found the equilibrium equation for  the reaction, and then isolated and solved for N2O4. We then multiplied it by the volume to get the number of moles, and then multiplied its molar mass by the found # of moles to get the mass.

Textbook Example - P.E.A. (page 681)

P.E.A. - For this reaction, we are given the initial conditions of the reaction (1.86 moles of  NOBr) and the final conditions, which is the conditions of the reaction when it comes to an equilibrium (0.082 moles of Br2). Using stoichiometry, we found the number of moles of NO and NOBr for the final conditions. We then found the concentrations using the moles, and then subbed them into the equilibrium equation to find Kc and Kp.

Textbook Example - P.E.A. (page 682)

P.E.A. - Pretty self explanatory... I think

Textbook Example - P.E.A. (page 684)

P.E.A. - Here, we are given the initial conditions, but we don't know at what point the reaction comes to an equilibrium, so we say that we use up x moles of reactants and gain x moles (2x because of the coefficient 2 in 2HI). So we find set the concentration of each part, carrying the x's through. The volumes cancel out in the equation. We set that equation to equal the given equilibrium constant and then rearrange the equation into quadratic form. We then used the quadratic formula to find the two solutions for x. The one of the left was not possible since we only had 0.15 moles of reactant to begin with (can't use more than what we have!). We subbed the right x back into 2x to find the # of moles of HI when the reaction comes to an equilibrium.

Textbook Example - P.E.A. and P.E.B. (page 685)

P.E.A. - We were given the initial conditions. In this case, since we don't have any reactants on the left to begin with, we are forced to start the reverse reaction. We do not know when the reaction comes to an equilibrium, so we say that we use up x moles of Fe3+ (resulting in 1.20M-x) and gain x moles of Ag+ and Fe2+. We write out the equilibrium equation for this reaction, and set it to equal 2.98 (given constant). We rearrange the equation to quadratic form and then use the quadratic formula to find possible x values. We cannot have a negative value. Ag+ and Fe2+ are both equal to just x mol/L, and Fe3+ is equal to 1.20M-x.  We checked our work by subbing the found concentrations back into Kc equation, to see if we can get the given value.

P.E.B. - Same process. Just in this case, you can make the math easier by finding the square roots of both sides (where shown). Isolate and solve for x, sub it back into the final conditions to find the molarity of each reactant/product.

Holy moly, that was a lot of practice problems....
That's the end of chapter 15! Onto acids and bases!~

Acids and Bases

There are two classical definitions of acids and bases that we covered.
1. Arrhenius Definition

• Acid: Any substance which increases Hydrogen ion concentrations (when added to water)
• i.e. HCl(aq) -> H+(aq) +Cl-(aq)
• In this case HCl would be the acid (generated an H+ ion)
• Base: Any substance which increases OH- concentrations (when added to water)
• i.e. NaOH(aq) -> Na+(aq) + OH-(aq)
• In this case the NaOH is the base (generated OH-)
• By the way:

2. Bronsted-Lawry Definition

• Acid: Any substance that donates a proton
• Base: Any substance that accepts a proton
• i.e. NaOH(aq) -> Na+(aq) + OH-(aq)
• For this definition, the OH would be the base where
• OH- + H+ -> H2O
• More 'complete' than the Arrhenius definition - it applies to non-aqueous solutions as well (even gases)

Strong Acids/Bases Vs. Weak Acids/Bases

Strong Acid: reaction (acid dissociation) goes through all the way (solid arrow pointing right). The acid dissociates,leaving no amount of undissociated acid left in the solution.

• i.e. AH(aq) -> A-(aq) + H+(aq)

Strong Base: dissociates completely in the solution

• i.e. NaOH(aq) -> Na+(aq) + OH-(aq)

Weak Acid - measurable amount of acid left over.

• i.e. AH(aq) <-> A-(aq) + H+(aq)   
• The <-> is the equilibrium arrows... :P

Weak Base - measurable amount of base left over.

• i.e. B(aq) + H2O(l) <-> BH+(aq) + OH-(aq)

Conjugate Acids/Bases

For the generic acid dissociation reaction AH(aq) -> A-(aq) + H+(aq), the AH would be the acid, and the A- would be the conjugate base. For the reaction B(aq) + H2O(l) <-> BH+(aq) + OH-(aq), the B would be the base, and the BH+ would be the conjugate acid.


In general...
Weak acid -> Strong conjugate base
Strong acid -> Weak conjugate base
Strong base -> Weak conjugate acid
Weak base -> Strong conjugate acid

See figure 16.1 for a table of acids and bases and their relative strength :)

Acid/Base Dissociation Constant (Ka/Kb)

For weak acids/bases you will need to write an equilibrium equation.

They're written the same way as any other equilibrium equation.

Textbook Example - P.E.A. (page 701)

a) HF is the (conjugate) acid, H2O is the (conjugate) base, F- is the conjugate (base) of HF and H3O+ is the conjugate (acid) of H2O. Forward reaction, (Reverse reaction)
b) HSO4- is the (conjugate) acid, NH3 is the (conjugate) base, SO42- is the conjugate (base) of HSO4- and NH4+ is the conjugate (acid) of NH3. Forward reaction, (Reverse reaction)

c) HCl is the (conjugate) acid, CH3COO- is the (conjugate) base, Cl- is the conjugate (base) of HCl and CH3COOH is the conjugate (acid) of CH3COO-. Forward reaction, (Reverse reaction)

Note: kSanchez wz here. Carry on.

Self-Ionizing Property of Water

Water, all by itself can dissociate. This is the reaction:

H2O(l) <-> H+(aq) + OH-(aq)

The equilibrium constant for this reaction has its own name - the water dissociation constant - and it is represented by Kw.

Kw = 10^-14  at 25ºC

Kw = [H+][OH-] (always constant)

These statements are always true, no matter if there is a change in temperature

In pure, distilled, de-gassed water (25C), [H+] = [OH-]= 10^-7

   > pH=pOH=7 (neutral, pure)

You can also convert from Ka to Kb and vice-versa by saying that Ka*Kb = Kw. So Ka = Kw/Kb, and Kb = Kw/Ka.

pH and pOH

When you have the p- of something, you take the -log of that quantity.

• pH = -log [H+]
• pOH = -log [OH-]
• pKw = -log Kw
• pKw = pH + pOH
• pH+pOH = 14

• pH > 7  - BASE
• pH = 7  - NEUTRAL
• pH < 7  - ACID

Textbook Example - P.E.A. and P.E.B. (page 706)

P.E.A. - Here we're given the pH. We rearrange the equation pH=-log[H+] to isolate and solve for [H+]. We then use another known equation to solve for OH. 

P.E.B. - We find the H+ concentration that we have originally, and multiply it by the initial volume to see how many moles of H we have (this does not change). We then found the molarity of the desired pH. We set that molarity as equal to the # of moles over the desired volume and then solved for V2.

Textbook Example - P.E.A. and P.E.B. (page 707)

P.E.A. - We were given the initial condition, and the concentration of H+. By stoichiometry, we know that [I-] will equal [H+]. We then find the pH using the known equation, and [OH-] using the given info and a known equation.

P.E.B. - Here, we are dissolving a set amount of HCl gas in order to make 625mL of solution. First, we have to find out the number of moles of H+ we have in the first place. We do this by using the PV=nRT equation, solving for n. Once we found n, we were able to find the concentration of H+ in the solution easily, by dividing the # of moles by the volume of the solution, and then of course we were able to find the pH using our handy dandy pH = -log [H+]

Lecture 5 - Acids and Bases (Part 2)

Lecture Recording:

 

Lecture Summary/Notes:

Finally got these up! Enjoy! 

Ka, Kb, pH, pOH

If we  have a general equilibrium reaction:

Where Ka is the acid dissociation constant, we can use that value to solve for [H+], which in turn can be used to find pH. **p.710 in the textbook has a table of Ka, Kb, pKa, and pKb values of some weak acids and bases**


NOTE: Kw/Ka = Kb. Remember Kw = 1.0 x 10^-14. Here is a derivation of that equation:

Textbook Example - P.E.A. and P.E.B. (page 712)

P.E.A. - We are given the initial value of HOCl and the pH. We find the final conditions, saying that we use an x amount of HOCl and form x amount of H+ and OCl-. We write out the equilibrium equation in terms of x. We then find the x value (pink cloud) by solving for [H+] ( x = [H+] ). We then substitute the x value back into the equilibrium equation to solve for Ka.

P.E.B. - In this case, we are solving for the equilibrium constant of a base (Kb). We find the final conditions of the equilibrium reaction, and write out the equilibrium equation in terms of x. We then solve for x (in this case, [OH-] = x). We are given the pH, but need the pOH, so we subtract the pH from 14 to get pOH. We then isolate [OH-] to get x, and then substitute it back into the equilibrium equation to solve for Kb.

Textbook Example - P.E.B. (page 713)

P.E.B. - We are given the mass (1.0 g), we find the molar mass (180.2 g/mol) of Acetylsalicylic acid so that we can find the # of moles (5.55 x 10^-3). We then divide it by the given volume (0.325 L) to find the concentration. We then use this value as the initial concentration of the acid, and then find our final conditions. We then, again, form the equilibrium equation. In this case we are given the Ka, so we set the equation equal to the value, and then solve for x using the quadratic formula. We end up with 2 x values, and eliminate the negative value (cannot have negative concentration), leaving us with the x value, which is = to [H+]. We then use that value to find pH. (Thank you, anonymous!)

Simplifying Assumption

This is super useful for when we are solving for x, given the Ka/Kb value. The assumption is that if the   equilibrium constant is very very small compared to the initial concentration of the acid/base, then the reaction will go to such a small extent that it will hardly change the initial concentration of the acid/base. See the following example:

The red says: This reaction will go to such a small extent, that the x value is going to be a number significantly less than 0.1. We can therefore make a simplifying assumption. Thus, we can just ignore the x in the denominator. Think about it this way: if x is 0.001, and the initial concentration is 0.1, 0.1-0.001=0.099, which we will make 0.1 anyways (rounded for proper sig figs). So we can -assume- that it will hardly affect it, and ignore it in the denominator, so that we can solve for x in the numerator more easily. Get it? :) 

Condition for using the simplifying assumption:
The initial concentration is at least 100 times greater than the equilibrium constant.
(So [HA]0 >100Ka, or [Hb]0>100Kb)

Percent Ionization

Percent ionization is the percentage of the acid that actually dissociates.

Textbook Example - P.E.A. and P.E.B. (page 717)

P.E.A. - If we write out the dissociation reaction of HF, and then find the final conditions of the equilibrium reaction (in terms of x). We then write out the equilibrium equation, setting it equal to the given Ka. In this case, we are able to use the simplifying assumption since the initial concentration (0.20M) is 100X greater than the Ka. So we then solve for x very easily, and set it equal to both the H+ and F- concentrations. F- is the conjugate base of HF. We then found the percent ionization of HF by setting the [F-]/[HF]0 and multiplying it by 100%. We get a final answer of 5.6%. This means (green text) that only 5.6% of the HF molecules will dissociate and make F- molecules. The rest stay as HF molecules. This is a very WEAK acid! For 0.020M, we would go through the whole process again, solving for [H+] where in the denominator, instead of 0.2, we would have 0.02. Solving for x, we would get 0.0036M, and then we would have percent ionization = (0.0033/0.020) x 100% to give us 18.2% (thank you, anonymous!).

P.E.B. - For this question we are given the percent ionization and the initial concentration of the acid. We first solved for the concentration of the conjugate base. (on the left is the percent ionization equation). We know that x amount of conjugate base is formed, so the found concentration of the conjugate base is = x. We then substitute x into the equilibrium equation [Ka = (x^2)/([HA]-x)] to find the Ka value.

Polyprotic Acids

Polyprotic acids are acids with more than one proton (H). Two examples of polyprotic acids are H2SO4 (sulphuric acid) and H3PO4 (Phosphoric acid). With these acids, the number of dissociations possible is equal to the number of H's in the acid. See below:

There are two possible dissociations for sulphuric acid

There are three possible dissociations for phosphoric acid. For each dissociation, there is a separate dissociation constant. (If you can't read it) Ka1 = 7.1 x 10^-3, Ka2 = 6.3 x 10^-8, Ka3 = 4.2 x 10^-13. After every dissociation, the Ka value drops considerably. WHY? It's because if you deprotonate the acid over and over, you have to lose an ion, which drops the acidity by lots!

**p. 718 in the textbook has a table of ionization constants of some polyprotic acids**
**If you dissolve a polyprotic acid in water, you only need to consider the FIRST dissociation. If you want to know the concentration of the second dissociation anion, it = Ka2.**