Lecture 7 - Acids and Bases (Part 4), Solvation Equilibrium

Lecture Recording:

Lecture Summary/Notes:

Henderson-Hasselbalch Equation

This equation is used when you have a solution of a weak acid and its strong conjugate base. We shall derive it, just for fun :)

We start with both the acid and its conjugate base in the solution. Hence, in the final conditions, we have the acid minus a certain amount (x), and the conjugate base minus a certain amount (x), and we gain x amount of H+. We write out the equilibrium equation. In this case, we are able to use the simplifying assumption in both the numerator and the denominator since we are assuming that the acid is a weak acid, and thus we are going to assume that the initial concentrations of acid and conjugate base are going to be 100 times greater than the Ka. Then we take the log of that, and then rearrange it a bit to end up with the HENDERSON-HASSELBALCH EQUATION! YAYYYY!!! Now we can use it to solve lots of problems :).

There are two conditions to meet before we can use the H-H equation:

The conjugate base (A-) concentration to acid concentration ratio must be less than 10 and greater than 0.1, and both acid and base have to have an initial concentration greater than 100 times the Ka.

If these two conditions are met, the H-H equation can be used, and it also produces a BUFFER solution.

Buffers

A buffer solution is a solution that stabilizes pH at a certain point. Adding more acid/base doesn't change the pH by very much. Diluting the solution does not change the pH at all.

Textbook Example - P.E.A. and P.E.B. (page 755)

P.E.A. - We simply used the H-H equation and subbed in all the givens, and isolated for x. Once we found the concentration of NH3, we multiplied it by the volume to get the moles, then divided by the molar mass to get the mass :).

P.E.B. In this example, we found the moles of the conjugate base, and the moles of the acid. OAc- is the conjugate base, where Ac is CH3CO; this will make more sense if you look at the previous problem on page 755. (In this case, we can use moles instead of concentrations for the initial and final conditions since the volume is the same~). We are given the Ka in the previous example (1.8 x 10^-5), and thus were given the pKa (-log Ka). We substitute the pKa into the H-H equation as well as the moles of the base and the acid to get the pH. (Remember that if we found the molarity of the base and acid, it would be n/V, and the V's would cancel out anyways). 

Textbook Example - P.E.A. and P.E.B. (page 758)

P.E.A. - a) We primarily found the concentrations of both the acid and the base (by dividing it by the volume), and got the pKa from the table of pKa values on page 710, and substituted the values into the H-H equation to find the pH. EASY! b) If we add 0.005 moles of HCl, we are adding 0.05 moles (since we are adding protons - H+) to the acid, and subtracting 0.005 moles from the conjugate base in the final conditions. (The final conditions in this case is the initial concentrations of the acid and base, and thus we plug in these values into the H-H equation to find the pH). c) We are doing the same thing as in b, but adding 0.005 moles to the conjugate base instead of the acid).

P.E.B. - From example 17-6 we gathered the moles of base and acid when the pH = 5.o9 (by multiplying the concentrations by 0.3L), and the pKa (4.74). We are adding to the acid an x # of moles, and subtracting x # of moles from the base to get the desired pH of 5.03. So we plug those values into the H-H equation and then solve for x! We divide the found x value (# of moles that we add to the acid) by the molarity of the acid (6.0M) to find the volume of HNO3 we must add to get the desired pH, and presto! :)

Making a buffer!

• Choose compounds that do not interfere with the reaction
• Phosphate, borate are often used
• You wouldn't use HCN where hemoglobin is involved
• Choose compounds that stabilize the pH where you want it to

Titration

A titration is a controlled reaction of an acid with a base.

Titrations are often used to find the unknown concentration of a known reactant (You know what the substance is, but don't know the concentration). If you are using a pH indicator (as in the diagram), the point where the pH of the solution strikes up is the equivalence point, which is where the # of moles of base = # of moles of acid. You would multiply the concentration of the titrant (known concentration of base) by the volume of the base at the equivalence point to find the # of moles of base, and then divide that by your initial volume of the acid to find the concentration of the acid! :D:D:D:D. If you don't have a pH indicator, you can use the color change as an indicator of the equivalence point. When the color of the solution changes, you use the volume of base at that point to do the same calculations.

Textbook Example - P.E.A. (page 764)

P.E.A. In b), we got the volume of the NaOH by dividing the  found # of moles of OH by the concentration of NaOH. We then subtracted the volume of the HCl from the volume of the NaOH at 50% neutralization. For c), it should be 0.250M, but 0.250L in the 4th line of the problem. For d) The # of moles of NaOH in 1mL is 2.50 x 10^-4mol. If we find the concentration of OH- in 40mL of solution, the pOH is 2.2, leaving a pH of 11.8. The pH of the solution jumps up from 7 (at equivalence point) to 11.8 with the addition of 1mL!

Buffer Point

This is the point where the concentration of conjugate base [B] = concentration of acid [A]

This is the point where the buffer has the LEAST amount of change in pH

A buffer at the buffer point is the best buffer you can make!

It has the greatest capacity and the best stability at this point.

We talked about pH indicators at this point - sometimes colours act as natural acid-base indicators! In an acid form, the substance may be colourless or a certain color, but when deprotonated, it becomes coloured, or another colour! See below:

HIn <=> H+ + In-

Where In stands for indicator. HIn is colourless, but In- is coloured!

**p.761 has a table of indicators**

An example discussed in class was red cabbage! Search 'red cabbage pH indicator' on YouTube - lots of fun :P.

Weak Acid + Strong Base Titration

If we plot out the pH and the volume of a titration where we are mixing a weak acid with a strong base, it would look like this:

When we have a strong base reacting with a weak acid, we have a buffer! (Remember the Henderson-Hasselbalch conditions), and thus we have a buffer region - this is the area where the pH is stable - the buffer point (see previous section) is the point in the middle of this region. For this case, the equivalence point has a pH greater than 7. In the case of a weak base reacting with a strong acid (not a buffer), the equivalence point will be at a pH less than 7.

Textbook Example - P.E.A. (page 767)

P.E.A. In a) SORRY - it should be 0.150-x, not 0.115-x in the final conditions and in the equilibrium equation! We found the initial pH by finding the initial H+ concentration. In b), if the neutralization is 25% complete, then there the solution will be 25% base and 75% acid, so we multiply the moles of the acid and base by 0.75 and 0.25, reciprocally. We then sub those values into the H-H equation to find the resulting pH. In c), there will be 100% base and about 0% acid. We find the volume of the base at this point, and add it to the initial volume of the acid (total volume). Thus, the concentration of the conjugate base will be the # of moles/total volume. Once we find the concentration of the base, we want to know what the pH of that solution is. So we say that Kb=Kw/Ka to find the Kb, and make another equilibrium equation for F-. We use the simplifying assumption to find x (the concentration of OH-), and then use that to find the pOH, and then subtract it from 14 to get the pH. HUH? Yeah, I don't really get it yet.

Weak Polyprotic Acid + Strong Base Titration

In this case, we will have a graph that looks like this:

There will be a separate buffer point and equivalence point for every deprotonation.  In this case, there are 3 dissociations, so it could be an acid like H3PO4.

Solubility Product (18.1)

If you have a sparingly soluble salt XY (sparingly soluble meaning that it won't completely dissociate), and it dissociates into X and Y, you will have a reaction like this:

And then if you write out the equilibrium equation, you will notice that the activity of the reactant (the salt) is 1 (since the salt is solid). Then you will end up with an equilibrium constant made of the product of the two products! This has a special name -> Solubility product, YAY! **p.785 presents a table of Ksp values of some sparingly soluble salts**

Textbook Example - P.E.A. & P.E.B. (page 786)

VERYY easy questions. Just setting up the equilibrium equations for finding the solubility product :)