16.7: Acid-Base Titration Curves

Acid-Base Titrations:

An acid-base titration is a procedure for determining the amount of acid (or base) in a solution by determining the volume of base (or acid) of known concentration that will completely react with it.

Can titrate:
1) A strong acid with a strong base
2) A strong acid with a weak base
3) A weak acid with a strong base
4) A weak acid with a weak base → this kind of titration will not be on the exam.

An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Here is one for a strong acid titrated with a strong base:

The change in pH as the base is added will be slow at first, then it will reach a point where the pH increases drastically before leveling out again.

The middle of this rapid incline is called the equivalence point: the point in a titration when the smallest amount of titrant has been added that is sufficient to fully neutralize or react with the acid or base (the point when all the acid and base have reacted).

Example:

The case of a strong acid and a strong base:

Note: The pH at the equivalence point for a strong acid and a strong base will always be 7.

What is the pH when 27.0mL of 0.450 M HCl is titrated (read as mixed!) with 45.0mL of 0.500 M NaOH?

Remember with mixing/titrating, the first thing we do is convert to moles:

mols HCl = (0.0270 L)(0.450 mols/L) = 0.0122 mol HCl
mols NaOH = (0.0450 L)(0.500 mols/L) = 0.0225 mol NaOH

The acid and base react to form a salt:

HCl (aq) + NaOH (aq) → H₂O (l) + NaCl (aq)



Convert to concentration by dividing by new volume:

NaOH = (0.0103 mol) / (0.072 L) = 0.143 M NaOH = 0.143 M OH^-
pOH = -log(0.143) = 0.844
pH = 14 – pOH = 13.2

Example:

The case of a strong acid and a weak base:

What is the pH at the equivalence point when 0.150 M of HCl titrates 20 mL of 0.200 M NH₃?

Remember that at the equivalence point, mols of HCl = mols of NH₃.

Since we're dealing with a titration, first convert to mols:

mols NH₃= (0.200 mols/L)(0.0200 L) = 4.00 x 10^-3 mols NH₃
and mols NH₃ = mols HCl = 4.00 x 10^-3 mols

To get the final volume after titration, we'll need to know the volume of HCl. Since we know the mols and molarity, we can use what we know to find the volume.

Concentration = Mols x Volume,
Volume = Mols / Concentration

Volume HCl = (4.00 x 10^-3 mols) / (0.150 M) = 0.0267 L HCl
So total volume of titration is 0.0267L + 0.0200 L = 0.0467 L
Concentrations of HCl and NH₃ = (4.00 x 10^-3 mols) / (0.0467 L) = 0.0857 M

So 0.0857 M of the acid and 0.0857 M of the base react to form 0.0857 M of a salt:

HCl + NH₃ → NH₄Cl

NH₄Cl dissociates in water and leaves NH₄⁺ and Cl^- ions. Cl^- ions form a strong acid which dissociates 100% in water, so no reaction there. When NH₄⁺ ions react with water:

NH₄⁺ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + NH₃ (aq)



K_a = (x²) / (0.0857 – x) → approximation holds
K_a= (x²) / (0.0857)

We need to find the K_a for NH₄⁺, so first find K_b of the conjugate base, NH₃, and plug it in:

K_a = K_w / K_b

K_b for NH₃ = 1.8 x 10^-5
K_a for NH₄⁺= (1 x 10^-14) / (1.8 x 10^-5) = 5.56 x 10^-10

5.56 x 10^-10 = (x²) / (0.0857)
x = 6.9 x 10^-6 = [H₃O⁺]
pH = -log(6.9 x 10^-6) = 5.16

Example:

The case of a weak acid and a strong base:

What is the pH when 35.0mL of 0.215 M HClO is titrated by 0.400 M LiOH to the equivalence point?

Remember that at equivalence, mols HClO = mols LiOH.

And since we're titrating, remember the first thing to do is convert to mols:

mols HClO = (0.350 L)(0.215 mols/L) = 0.00753 mols HClO
and mols HClO = mols LiOH = 0.00753 mols

To get the final volume after titration, we'll need to know the volume of LiOH. Since we know the mols and molarity, we can use what we know to find the volume.

Concentration = Mols x Volume,
Volume = Mols / Concentration

Volume LiOH = ( 0.00753 mols) / (0.400 M) = 0.0188 L LiOH
So total volume of titration is 0.0350L + 0.0188 L = 0.0538 L
Concentrations of HClO and LiOH = (0.00753 mols) / (0.0538 L) = 0.140 M

So 0.140 M of the acid and 0.140 M of the base react to form 0.140 M of a salt:

HClO + LiOH → LiClO

LiClO dissociates in water and leaves Li⁺ and ClO^- ions. Li⁺ ions form a strong base which dissociates 100% in water, so no reaction there. When ClO^- ions react with water:

ClO^- (aq) + H₂O (l) ⇌ HClO (aq) + OH^- (aq)



K_b = (x²) / (0.140 – x) → approximation holds
K_b = (x²) / (0.140)

We need to find the K_b for ClO^-, so first find K_a of the acid, HClO, and plug it in:

K_b = K_w / K_a

K_a for HClO = 3.5 x 10^-8
K_b for ClO^- = (1 x 10^-14) / (3.5 x 10^-8) = 2.86 x 10^-7

2.86 x 10^-7 = (x²) / (0.140)
x = 2.00 x 10^-4 = [OH^-]
pOH = -log(2.00 x 10^-4) = 3.7
pH = 14 – 3.7 = 10.3

16.6: Henderson-Hasselbalch Equation

Henderson-Hasselbalch Equation:

An equation that relates the pH of a buffer for different concentrations of conjugate acid and base:

or


And thank you wikipedia for making it easy to see how it relates to things we've already learned:

(Note: I know wikipedia is shunned for it's user contributed content, but every wikipedia article has a detailed list of linked references at the bottom of the page that can be checked to verify that an article's content isn't just made up bullshit. When in doubt, check references.)

Also note that this equation only works when the approximation method holds, when C_a / K_a is greater than 100.

16.5-16.6: The Common Ion Effect-Buffers

Common Ion Effect: the shift in an ionic equilibrium caused by the addition of a solute that provides an ion that takes part in the equilibrium.

Example:

The problem we did in class, at least from what I wrote down, does not follow what the book says. According to Le'Chatelier's principle, adding a strong acid to a weak acid should decrease the degree of ionization for the weak acid, since adding more product (H₃O⁺) will cause the reaction to shift to the left. For the problem we did in class, what I wrote down has the reaction shifting to the right. So instead, see the example from the book, pages 672-673.

Example:

What is the pH for a solution of 0.5 M HClO and 1.5 M HBr?

The common ion here is H⁺ (H₃O⁺). Since we know HBr is a strong acid, we also know it will dissociate 100% in water, adding 1.5M H₃O⁺ to the solution.

HClO (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + ClO^- (aq)



K_a = x(1.50 + x) / (0.50 – x) → approximation holds
3.5 x 10^-8 = (1.50x) / (0.50)
x = 1.17 x 10^-8
[H₃O⁺] = 1.50 + 1.17 x 10^-8 = 1.50 M
pH = -log(1.50) = -0.18

Note: It is possible to have a pH greater than 14 and less than 0. For example, if you have greater than 1M of a strong acid, the pH will be less than zero.

Also, the pH of a solution containing a weak acid and a strong acid will generally be equal to the pH of the strong acid alone.

Buffers:
(Basically the same as a common ion problem.)

Buffer: a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

Buffers contain either a weak acid and its conjugate base in a salt or a weak base and its conjugate acid in a salt. Consider a buffer that contains approximately equal concentrations of a weak acid HA and it's conjugate base A^-. When a strong acid is added to the buffer, it supplies hydronium ions that react with the base A^- :

H₃O⁺ (aq) + A^- (aq) → HA (aq) + H₂O (l)

When a strong base is added to the buffer it supplies hydroxide ions which react with the acid HA:

OH^- (aq) + HA (aq) ⇌ H₂O (l) + A^- (aq)

A buffer solution resists changes in pH through its ability to combine with both H₃O⁺ and OH^- ions.

Examples of buffers:

HCN and NaCN (notice the common ion CN^-)
NH₃ and NH₄Cl

Example:

What is the pH of a solution of 0.400 M HC₂H₃O₂ and 0.200 M NaC₂H₃O₂?

HC₂H₃O₂ → weak acid → HA
NaC₂H₃O₂ → salt containing conjugate base → C₂H₃O₂^- → A^-
Note: Na⁺ doesn't do much but show up. Na⁺ is like the person C₂H₃O₂^- brought to the party then didn't hang out with once they got there. Which is just mean. So C₂H₃O₂^- is kind of an asshole (A^-). Maybe that's an easy way to remember it.

But we know 0.200 M NaC₂H₃O₂ will dissociate to yield 0.200 M C₂H₃O₂^-.

HC₂H₃O₂ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + C₂H₃O₂^- (aq)



Notice that there is a nonzero initial concentration of A^-. This is from the salt.

K_a = x(0.200 + x) / (0.400 – x) → approximation holds
1.7 x 10^-5 = x(0.200) / (0.400)
x = 3.4 x 10^-5 = [H₃O⁺]
pH = -log(3.4 x 10^-5) = 4.47

Example:

What is the pH of a solution that is 0.45 M HF and 0.115 M NaF?

HF (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + F^- (aq)



K_a = x(0.115 + x) / (0.45 – x) → approximation holds
6.8 x 10^-4 = x(0.115) / (0.45)
x = 0.00266 = [H₃O⁺]
pH = -log(0.00266) = 2.57

Example:

For a buffer containing HC₂H₃O₂ and NaC₂H₃O₂, which will react with NaOH?

Consider NaOH to be OH^-.

HC₂H₃O₂ + OH- → H₂O + C₂H₃O₂^-

Which will react with HCl?
Consider HCl to be H⁺ (H₃O⁺).

C₂H₃O₂^- + H₃O⁺ → HC₂H₃O₂ + H₂O
or A^- + H₃O⁺ → HA + H₂O

Note: When acids and bases are added to a buffer, the pH will change, but only slightly. The pH does not stay exactly the same.

Example:

THIS IS THAT HUGE 3 PART PROBLEM REGARDING BUFFERS
(Note: K_a for HC₃H₅O₂ = 1.3 x 10^-5)

What is the pH for a buffer made from MIXING 47.0 mL of 0.678 M HC₃H₅O₂ and 33.0 mL of 0.533 M NaC₃H₅O₂?

Note: In this problem, we are making a buffer.

HC₃H₅O₂ → weak acid → HA
NaC₃H₅O₂ → salt with conjugate base → NaA → A^-

Whenever you see the words MIXING or TITRATING it is a clue that the volume is going to change, so molarity will change. The first step in these problems is always to convert to moles:
In order to do that we need to convert mL to L:
(47.0 mL) x (1 L / 1000mL) = 0.047L
(33.0 mL) x (1L / 1000mL) = 0.033L

(0.047L) x (0.678 mols / L HA) = 0.0319 mols HA
(0.033L) x (0.533 mols / L A^-) = 0.0176 mols A^-

Now divide by the new volume to get the new molarity:

New volume = 0.047L + 0.033 L = 0.080 L
0.0319 mols HA / 0.080 L = 0.399 M HA
0.0176 mols A^- / 0.080 L = 0.220 M A^-

HA + H₂O ⇌ H₃O⁺ + A^-




K_a = x(0.220 + x) / (0.399 – x) → approximation holds
1.3 x 10^-5 = x(0.220) / (0.399)
x = 2.36 x 10^-5 = [H₃O⁺]
pH = -log(2.63 x 10^-5) = 4.63

Now find the pH when 95.0 mL of the buffer we've just created is mixed with 5.00 mL of 0.100 M HCl.

We know that it is the conjugate base in our buffer that will react with the HCl (the A^-).

We also know that HCl is a strong acid that dissociates 100 % in water.

When mixing, convert to moles:
(0.399 M HA / L)(0.095 L) = 0.379 mols HA
(0.220 M A^- / L)(0.095 L) = 0.0209 mols A^-
(0.100 M HCl / L)(0.005 L) = 5 x 10^-4 mols HCl

Now we need to look at how the conjugate base of the buffer and the HCl are going to react and how much HA they are going to produce:

HCl + A^- → HA + Cl^-



We subtract the mols of HCl from the left side of the reaction because the HCl will be used up before the A^- is used up, there is less of it (limiting). And since this is what we are subtracting from the left side, it is also what we are adding to the right side. So now:

A^- = 0.0204 mols
HA = 0.0384 mols

Remember we're still in mols, so we need to divide by our new total volume to get molarity:

A^- = 0.0204 mols / 0.1 L = 0.204 M
HA = 0.0384 mols / 0.1 L = 0.384 M

0.384 M is our new acid concentration. What happens when it dissociates in water? We get our final hydronium ion concentration from which we can calculate pH:

HA + H₂O ⇌ H₃O⁺ + A^-




K_a = x(0.204 + x) / (0.384 – x) → approximation holds
1.3 x 10^-5 = x(2.04) = (0.384)
x = 2.45 x 10^-5 = [H₃O⁺]
pH = -log(2.45 x 10^-5) = 4.61


Now find the pH when 95.0 mL of the buffer we initially created is mixed with 5.00 mL of 0.350 M NaOH:

Since we're dealing with a changing volume, first thing to do is convert to moles:
(0.399 mols / 1 L HA)(0.0950L) = 0.0379 mols HA
(0.220 mols/ 1 L A^-)(0.0950L) = 0.0209 mols A^-
(0.350 mols/ 1 L NaOH)(0.005L) = 0.00175 mols OH^-

Which component of the buffer will the NaOH react with? The HA.

OH^- + HA → H₂O + A^-




We know to subtract 0.00175 mols because there is less of it and everything is 1:1. Now take the new values for HA and A^-, which are in mols, and divide them by the new volume to get the concentration in molars:

HA = 0.0362 mols / 0.1 L = 0.362 M HA
A^- = 0.0227 mols / 0.1 L = 0.227 M A^-

Now this acid will dissociate in water to yield H₃O⁺, from which we can find the new pH:

HA (aq) + H₂O (l) ⇌ A^- (aq) + H₃O⁺ (aq)



K_a = x(0.227 + x) / (0.362 – x) → approximation holds
1.3 x 10^-5 = x(0.227) / (0.362)
x = 2.07 x 10^-5 = [H₃O⁺]
pH = -log(2.07 x 10^-5) = 4.68



Notice that the pH of the original buffer was 4.63. When we added a strong acid, the pH only dropped to 4.61, and when we added a strong base, the pH went up to 4.68. So although the pH didn't stay exactly the same, it stayed pretty stable.

16.2: Polyprotic Acids

Polyprotic Acids:

-Polyprotic Acid: an acid that yields two or more H₃O⁺ per molecule.

Example:

H₃PO₄ is a triprotic acid, it can lose three hydrogens to yield three H₃O⁺ in aqueous solution. But in that solution, H₃PO₄, H₂PO₄^-, HPO₄^2-, and PO₄^3- are all present.

1) H₃PO₄ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + H₂PO₄^- (aq)
K_a1 = [H₃O⁺][H₂PO₄^-] / [H₃PO₄]

2) H₂PO₄^- (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HPO₄^2- (aq)
K_a2 = [H₃O⁺][HPO₄^2-] / [H₂PO₄^-]

3) HPO₄^2- (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + PO₄^3- (aq)
K_a3 = [H₃O⁺][PO₄^3-] / [HPO₄^2-]

K_a1, the first ionization constant, is much larger than the second one. For a triprotic acid, K_a2, the second ionization constant is much larger than the third one.

Example:

For 1.25 M of H₃PO₄ at 25^oC, find pH, [H₃PO₄], [H₂PO₄^-], [HPO₄^2-], [PO₄^3-], and [H₃O⁺].

(K_a for H₃PO₄ = 6.9 x 10^-3, K_a for H₂PO₄^- = 6.2 x 10^-8, K_a for HPO₄^2- = 4.8 x 10^-13)

Since this reaction takes place in parts, lets look at the first part and see what information we get from it. The first reaction is:

H₃PO₄ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + H₂PO₄^- (aq)



(x²) / (1.25 - x) = 6.9 x 10^-3

Approximation method holds, so:

(x²) / (1.25) = 6.9 x 10^-3
x = 0.0929

According to the table, [H₃O⁺] = x = 0.0929
and pH = -log[H₃O⁺] = 1.03
[H₂PO₄^-] = x = 0.0929, but it is used in the next reaction, so we don't know the final concentration yet.

The second reaction is:

H₂PO₄^- (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HPO₄^2- (aq)




Notice that the initial concentration of H₂PO₄^- is the 0.0929M we had left over from the first reaction. And the initial concentration of H₃O⁺ is also 0.0929M, which was yielded in the first reaction also.

(0.0929 + x)(x) / (0.0929 - x) = 6.2 x 10^-8

Approximation method holds, so:

(0.0929)(x) / (0.0929) = 6.2 x 10^-8
x = 6.2 x 10^-8

According to the table, [H₂PO₄^-] = 0.0929M - 6.2 x 10^-8 = 0.0929M
[HPO₄^2-] = x = 6.2 x 10^-8
[H₃O⁺] = 0.0929M - 6.2 x 10^-8 = 0.0929M

We know that K_a2 is supposed to be much smaller than K_a1. When we subtract this very small value from 0.0929, the difference is nearly negligible, and it rounds back to 0.0929. So really, we found these values after the first reaction.

The third reaction is:

HPO₄^2- (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + PO₄^3- (aq)



4.8 x 10^-13 = (0.0929 + x)(x) / (6.2 x 10^-8 - x)

Approximation method holds, so:

4.8 x 10^-13 = (0.0929)(x) / (6.2 x 10^-8)
x = 3.20 x 10^-19 M
According to the table, [PO₄^3-] = x = 3.20 x 10^-19 M

(Note that there is no K_a for strong acids, because they ionize 100% in water).

16.1: Acid-Ionization Equilibria

Recap:

In pure water, [H₃0⁺] = [OH^-] = 1 x 10^-7 M and pH 7 is neutral.

Acidic Solution:
[H₃0⁺] is greater than [OH^-]
[H₃0⁺] is greater than 1 x 10^-7 M
[OH^-] is less than 1 x 10^-7 M
pH is less than 7

Basic Solution:
[OH^-] is greater than [H₃0⁺]
[H₃0⁺] is less than 1 x 10^-7 M
[OH^-] is greater than 1 x 10^-7 M
pH is greater than 7

Weak Acids/Bases:

Example:

A strong acid, HCl dissociates in water 100%, so a concentration of 0.0100 M HCl would yield +0.0100 M H₃O⁺, with a pH of -log(0.0100) = 2.
A weak acid dissociates less than 100%. So 0.0100 M of a weak acid would yield less than 0.0100 M H₃O⁺. For a weak acid, the concentrations of ions in solution are determined from the acid-ionization constant.

Acid-ionization constant: the equilibrium constant for the ionization of a weak acid. Denoted K_a.

Shorthand for a weak acid is HA (where H is the hydrogen and A is the rest) for a monoprotic acid,
H₂A for a diprotic acid, and H₃A for a triprotic acid. So:

HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq)




where [H₃0⁺] and [H⁺] are interchangeable.

Example:

HC₆H₄NO₂ (aq) + H₂0 (l) ⇌ H₃O⁺ (aq) + C₆H₄NO₂^- (aq)
is the same as
HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq)

[HA] = 0.012 M
pH = 3.39
What is the K_a and what is the degree of ionization?

Degree of Ionization: the fraction of molecules that react with water to give ions. Given as a percentage it is called the percent ionization.




Since we know the pH is 3.39, we can find [H₃O⁺].

[H₃O⁺] = 10^-pH
[H₃O⁺] = 4.07 x 10^-4

So, x = 4.07 x 10^-4

K_a = (4.07 x 10^-4)(4.07 x 10^-4) / (0.0120)
K_a = 1.38 x 10^-5

To obtain the degree of ionization, notice that 4.07 x 10^-4 mol out of 0.0120 mol of acid ionizes, so

degree of ionization = (4.07 x 10^-4) / (0.0120) = 0.0339
percent ionization = 0.0339 x 100 = 3.39 %

Approximation Method:

Let C_a represent the initial concentration of the weak acid.

There will be less than a 5% error with the approximation method if C_a / K_a is greater than 100.

Example:

What is the pH for 0.500 M of acetic acid? K_a = 1.7 x 10^-5




K_a = 1.7 x 10^-5 = (x²) / (0.500 - x)

To solve this, we need to use the quadratic equation. Or, we can test to see if we are able to apply the approximation method and use that instead, since it is much quicker and easier.

Is C_a / K_a greater than 100?

(0.500) / (1.7 x 10^-5) = 29411.8

Yes.

Here is the Approximation: in the term (0.500 - x), the x will be so small that subtracting it from 0.500 will be negligible, and the term is roughly the same as simply 0.500. This approximation greatly simplifies the equation:

1.7 x 10^-5 = (x²) / (0.500)
x = 0.00292
pH = -log(0.00292) = 2.54

Remember, this substitution only works if C_a / K_a is greater than 100. If it's not, then the quadratic equation must be used.