1. Chapter 16 Acids and Bases

2. Some Definitions Arrhenius
Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions.
Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.

3. Some Definitions Brønsted–Lowry Acid: Proton donor
Base: Proton acceptor

4. A Brønsted–Lowry acid… …must have a removable (acidic) proton
A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons.

5. If it can be either…
...it is amphiprotic. HCO3− HSO4− H2O

6. What Happens When an Acid Dissolves in Water?
Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid.
As a result, the conjugate base of the acid and a hydronium ion are formed.

7. Conjugate Acids and Bases:
From the Latin word conjugare, meaning “to join together.”
Reactions between acids and bases always yield their conjugate bases and acids.

8. Acid and Base Strength
Strong acids are completely dissociated in water.
Their conjugate bases are quite weak.
Weak acids only dissociate partially in water.
Their conjugate bases are weak bases.

9. Acid and Base Strength
Substances with negligible acidity do not dissociate in water.
Their conjugate bases are exceedingly strong.

10. Acid and Base Strength HCl(aq) + H2O(l)  H3O+(aq) + Cl−(aq)
In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.
HCl(aq) + H2O(l)  H3O+(aq) + Cl−(aq)
H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right K is not measured (K>>1).

11. Acid and Base Strength C2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1).

12. Autoionization of Water
As we have seen, water is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.
This is referred to as autoionization.
H2O(l) + H2O(l)
H3O+(aq) + OH−(aq)

13. Ion-Product Constant The equilibrium expression for this process is
Kc = [H3O+] [OH−]
This special equilibrium constant is referred to as the ion-product constant for water, Kw.
At 25°C, Kw = 1.0  10−14

14. pH
pH is defined as the negative base-10 logarithm of the hydronium ion concentration.
pH = −log [H3O+]

15. pH In pure water, Kw = [H3O+] [OH−] = 1.0  10−14
Because in pure water [H3O+] = [OH−],
[H3O+] = (1.0  10−14)1/2 = 1.0  10−7

16. pH Therefore, in pure water, pH = −log (1.0  10−7) = 7.00
An acid has a higher [H3O+] than pure water, so its pH is <7
A base has a lower [H3O+] than pure water, so its pH is >7.

17. pH
These are the pH values for several common substances.

18. Other “p” Scales
The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).
Some similar examples are
pOH −log [OH−]
pKw −log Kw

19. Watch This!
Because [H3O+] [OH−] = Kw = 1.0  10−14, we know that −log [H3O+] + −log [OH−] = −log Kw = or, in other words, pH + pOH = pKw = 14.00

20. How Do We Measure pH? For less accurate measurements, one can use
Litmus paper
“Red” paper turns blue above ~pH = 8
“Blue” paper turns red below ~pH = 5
An indicator

21. How Do We Measure pH?
For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

22. Strong Acids
You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.
These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.
For the monoprotic strong acids,
[H3O+] = [acid].

23. Strong Bases
Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).
Again, these substances dissociate completely in aqueous solution.

24. Dissociation Constants
For a generalized acid dissociation,
the equilibrium expression would be
This equilibrium constant is called the acid-dissociation constant, Ka.
HA(aq) + H2O(l)
A−(aq) + H3O+(aq)
[H3O+] [A−]
[HA]
Kc =

25. Dissociation Constants
The greater the value of Ka, the stronger the acid.

26. Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is Calculate Ka for formic acid at this temperature.
We know that
[H3O+] [COO−]
[HCOOH]
Ka =

27. Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is Calculate Ka for formic acid at this temperature.
To calculate Ka, we need the equilibrium concentrations of all three things.
We can find [H3O+], which is the same as [HCOO−], from the pH.

28. Calculating Ka from the pH
pH = −log [H3O+] 2.38 = −log [H3O+] −2.38 = log [H3O+] 10−2.38 = 10log [H3O+] = [H3O+] 4.2  10−3 = [H3O+] = [HCOO−]

29. Calculating Ka from pH Now we can set up a table… [HCOOH], M [H3O+], M
[HCOO−], M
Initially
0.10
Change
−4.2  10-3
+4.2  10-3
+4.2  10−3
At Equilibrium
0.10 − 4.2  10−3
= = 0.10
4.2  10−3

30. Calculating Ka from pH [4.2  10−3] [4.2  10−3] Ka = [0.10]
= 1.8  10−4

31. Buffers Definition: a solution that resists change in pH
Typically a mixture of the acid and base form of a chemical
Can be adjusted to a particular pH value
Blood: pH =
Too acidic? Increase respiration rate expelling CO2, driving reaction to the left and reducing H+ concentration.
Excretory system – excrete more or less bicarbonate

32. Buffers Definition: a solution that resists change in pH
Typically a mixture of the acid and base form of a chemical
Can be adjusted to a particular pH value
pH below 7.4 in rats – CaCO3 in BONE dissociates, carbonates soak up extra H+ to buffer blood. But bones weakened.

33. Buffers Definition: a solution that resists change in pH Why use them?
Typically a mixture of the acid and base form of a chemical
Can be adjusted to a particular pH value
Why use them?
Enzyme reactions and cell functions have optimum pH’s for performance
Important anytime the structure and/or activity of a biological material must be maintained

34. How buffers work Equilibrium between acid and base.
Example: Acetate buffer
CH3COOH  CH3COO- + H+
If more H+ is added to this solution, it simply shifts the equilibrium to the left, absorbing H+, so the [H+] remains unchanged.
If H+ is removed (e.g. by adding OH-) then the equilibrium shifts to the right, releasing H+ to keep the pH constant

35. Limits to the working range of a buffer
Consider the previous example:
CH3COOH  CH3COO- + H+
If too much H+ is added, the equilibrium is shifted all the way to the left, and there is no longer any more CH3COO- to “absorb” H+.
At that point the solution no longer resists change in pH; it is useless as a buffer.
A similar argument applies to the upper end of the working range.

36. Chemistry of buffers
Lets look at a titration curve

37. Titration is used to determine the concentration of an acid or base by adding the OTHER and finding an equivalency point…

38. Titration is used to determine the concentration of an acid or base by adding the OTHER and finding an equivalency point…
Suppose you have a KOH solution, and you want to know its concentration (molarity).
Slowly add an acid (HCl) with a known concentration (0.1 M) and find the equivalency point…in this case it will be at pH = 7… and we use an indicator that changes color at that pH determine when that point has been reached.
So, suppose it takes 10ml of 0.1 M HCl to buffer 50 ml of the KOH.

39. Titration is used to determine the concentration of an acid or base by adding the OTHER and finding an equivalency point…
So, suppose it takes 10ml of 0.1 M HCl to buffer 50 ml of the KOH.
The original concentration of the base =
Vol Acid x conc. Of acid
Volume of Base
10 ml x 0.1 M
50 ml = 0.02 M

40. Chemistry of buffers
Ka = equilibrium constant for H+ transfer… also described as the dissociation constant…the tendancy of an acid to dissociate. AH  A- (base conjugant) + H+
Ka = [A-] [H+]/ [AH] = [base] [H+] / [acid]
Weak acids have low values… contribute few H+ ions…
Because we are usually dealing with very small concentrations, log values are used…
The log constant =

41. Chemistry of buffers Ka = [A-] [H+]/ [AH] = [base] [H+] / [acid]
Weak acids have low values… contribute few H+ ions…
Because we are usually dealing with very small concentrations, log values are used…
The log constant =
SO! Since pK is the negative log of K, weak acids have high values … (-2 – 12).
HCl = -9.3 – very low ~complete dissociation

42. Chemistry of buffers -pKa -pH
First rearrange the first equation and solve for [H+]
[H+] = Ka x [acid]/[base]
Then take the log of both sides
log10[H+] = log10Ka + log10 [acid]/[base]
-pH
-pKa

43. Chemistry of buffers -pH = -pKa + log10 [acid]/[base]
Multiply both sides by –1 to get the Henderson- Hasselbach equation
pH = pKa - log10 [acid]/[base]

44. Chemistry of buffers
What happens when the concentration of the acid and base are equal?
Example: Prepare a buffer with 0.10M acetic acid and 0.10M acetate
pH = pKa - log10 [acid]/[base]
pH = pKa - log10 [0.10]/[0.10]
pH=pKa
Thus, the pH where equal concentrations of acid and base are present is defined as the pKa
A buffer works most effectively at pH values that are + 1 pH unit from the pKa (the buffer range)

45. equilibrium
pKa value
H3PO4 H2PO4− + H+
pKa1 = 2.15
H2PO4− HPO42− + H+
pKa2 = 7.20
HPO42− PO43− + H+
pKa3 = 12.37

46. Drives equilibria and reversible states of compounds
Carbonic Acid Bicarbonate Carbonate

47. Factors in choosing a buffer
Be sure it covers the pH range you need
Generally: pKa of acid ± 1 pH unit
Consult tables for ranges or pKa values
Be sure it is not toxic to the cells or organisms you are working with.
Be sure it would not confound the experiment (e.g. avoid phosphate buffers in experiments on plant mineral nutrition).

48. What to report when writing about a buffer:
The identity of the buffer (name or chemicals)
The molarity of the buffer
The pH of the buffer
Examples:
“We used a 0.5M Tris buffer, pH 8.0.”
“The reaction was carried out in a 0.1M boric acid – sodium hydroxide buffer adjusted to pH 9.2.”

49. Three basic strategies for making a buffer
1. Guesswork – mix acid and base at the pH meter until you get the desired pH.
Wasteful on its own, but should be used for final adjustments after (2) or (3).
2. Calculation using the Henderson-Hasselbach equation.
3. Looking up recipe in a published table.

50. Calculating buffer recipes
Henderson-Hasselbach equation
pH = pKa - log10 [acid]/[base]
Rearrange the equation to get
10(pKa-pH) = [acid]/[base]
Look up pKa for acid in a table. Substitute this and the desired pH into equation above, and calculate the approximate ratio of acid to base.
Because of the log, you want to pick a buffer with a pKa close to the pH you want.

51. Example
You want to make about 500 mL of 0.2 M acetate buffer (acetic acid + sodium acetate), pH 4.0.
Look up pKa and find it is 4.8.
10( ) = = 6.3 = [acid]/[base]
If you use 70 mL of base, you will need 6.3X that amount of acid, or 441 mL. Mix those together and you have 511 mL (close enough).