1. Acids and Bases

2. 15.5 - Autoionization of Water & pH

3. Water is Amphoteric!
Amphoteric: a compound able to react both as an acid and as a base.
Even when in a pure form, water acts as an acid and a base with itself. This process is called autoionization.
A very small amount of ions will form in pure water. Some molecules of H2O will act as acids, each donating a proton to a corresponding H2O molecule that acts as a base. The proton-donating molecule becomes a hydroxide ion (OH-) while the proton-accepting molecule becomes a hydronium ion (H3O+).

4. The equilibrium constant is called:
the ion product constant for water (kw)
(sometimes called the dissociation constant for water)
In pure water, since H2O is the only source of these ions, the concentration of H3O+ ions and OH- are equal, and the solution is neutral. Since the concentrations are equal, we easily calculate them from Kw.
[H3O+] = [OH-] = √ Kw = 1.0 x 10^-7
(in pure water at 25 degrees C)

5. Kw = [H3O+] [OH-] = 1.0 x 10^-14 at 25C
Formula (on reference table):
Kw = [H3O+] [OH-] = 1.0 x 10^-14 at 25C
The concentration of H3O+ times the concentration of OH-
is always equal to 1.0 x 10^-14 at 25C.

6. Therefore…
If [H3O+] increases,
Then [OH-] must decrease in order for the ion product constant (that’s nothing fancy, just kw the equilibrium constant) to remain 1.0 x 10^-14.

7. In an acidic solution [H3O+] > [OH-]
This change would make an acidic solution.
An acidic solution contains an acid that creates additional H3O+ ion causing the [H3O-] to increase and the [OH-] to decrease.
In an acidic solution [H3O+] > [OH-]

8. If [H3O+] = 1.0 x 10^-3, what is the concentration of OH-?
Walk Through:
If [H3O+] = 1.0 x 10^-3, what is the concentration of OH-?
Solve the ion product constant expression for [OH-] with the formula from the reference table.
Kw = [H3O+] [OH-] = 1.0 x 10^-14 at 25C
Is this solution acidic or basic?
[H3O+] [OH-] = 1.0 x 10^-14
(1.0 x 10^-3)[OH-] = 1.0 x 10^-14
[OH-] = (1.0 x 10^-14)/(1.0 x 10^-3)
[OH] = 1.0 x 10^-11 M
Acidic:
1.0 x 10-3 > 1.0 x 10-11
[H3O+] > [OH-]

9. And…
If [OH-] increases,
Then [H3O+] must decrease in order for the ion product constant (still just kw the equilibrium constant) to remain
1.0 x 10^-14.

10. In a basic solution [H3O+] < [OH-]
This change would make a basic solution.
A basic solution contains a base that creates additional OH- ions causing the [OH-] to increase and the [H3O-] to decrease.
In a basic solution [H3O+] < [OH-]

11. Walk Through
If [OH-] = 1.0 x 10^-12, solve the ion product constant expression for [H3O+]. Use the Kw formula from the reference table.
Is this solution acidic, basic or neutral?
[H3O+] [OH-] = 1.0 x 10^-14
[H3O+] (1.0 x 10^-12) = 1.0 x 10^-14
[H3O+] = (1.0 x 10^-14)/(1.0 x 10^-12)
[H3O+] = 1.0 x 10-2 M
Basic:
1.0 x 10-2 < 1.0 x 10-12
[H3O-] > [OH-]

12. Notice:
Changing [H3O+] in an aqueous solution produces an inverse change in [OH-] and vise versa.

13. Summarizing Kw: A neutral solution contains
[H3O+] = [OH-] = 1.0 x 10^-7 M (at 25C)
An acidic solution contains
[H3O+] > [OH-]
A basic solution contains
[OH-] > [H3O+]

14. Practice Problem!
Calculate [OH-] at 25C for each solution and determine if the solution is acidic, basic, or neutral.
[H3O+] = 7.5 x 10-5 M
[H3O+] = 1.5 x 10-9 M
[H3O+] = 1.0 x 10-7 M

15. Answers: (7.5 x 10-5 )[OH-] = (1.0 x 10-14)
a) [OH-] = 1.3 x M, acidic solution
b) [OH-] = 6.7 x 10-6 M, basic solution
c) [OH-] = 1.0 x 10-7 M, neutral solution
(7.5 x 10-5 )[OH-] = (1.0 x 10-14)
[OH-] = (1.0 x )/(7.5 x 10-5 )
(1.5 x 10-9 )[OH-] = (1.0 x 10-14)
[OH-] = (1.0 x )/(1.5 x 10-9 )
(1.0 x 10-7)[OH-] = (1.0 x 10-14)
[OH-] = (1.0 x )/(1.0 x 10-7)

16. The pH scale: A Way to Quantify Acidity and Basicity
pH is defined as the negative of the log of the hydronium ion concentration.
Formula: pH = -log[H3O+]
Note: write pH to two decimals places because only the numbers to the right of the decimal are significant in a logarithmic
A solution with [H3O+] = 1.0 x 10-7 M (neutral at 25C) has a pH of:
pH = -log[H3O+]
= -log(1.0 x 10-7 )
= 7

17. In General… pH < 7 Acidic Solution pH > 7 Basic Solution
pH = Neutral Solution

18. Because the pH scale is a logarithmic scale, an increase of 1 on the pH scale corresponds to a factor of 10 decrease in [H3O+].
ACIDIC
BASIC

19. Practice Problems!
Calculate the pH of each solution at 25C and indicate whether the solution is acidic or basic.
[H3O+] = 1.8 x 10-4 M
[OH-] = 1.3 x 10-2 M
2) Calculate the [H3O+] for a solution with a pH of 4.80

20. Answers 1a) pH = -log[H3O+] = -log(1.8 x 10-4 ) = -(-3.74)
= 3.74 (acidic)
pH = -log[H3O+]
= -log(7.7 x 10-13)
= -(-12.11)
= 12.11(basic)
1b) First find [H3O+] using Kw.
[H3O+] [OH-] = 1.0 x 10-14
[H3O+](1.3 x 10-2 M) = 1.0 x 10-14
[H3O+] = (1.0 x 10-14)/(1.3 x 10-2 M)
[H3O+] = 7.7 x M

21. 2) pH = -log[H3O+]
4.8 = -log[H3O+]
-4.8 = log10[H3O+]
= [H3O+]
[H3O+] = 1.6 x 10-5 M
Divide by -1
Take the base of the log (base = 10) and raise it to the power of Then set that equal to [H3O+] and plug into your calculator.

22. pOH & Other p Scales
The pOH scale is comparable to the to the pH scale but is defined with respect to [OH-] instead of [H3O+]. It can be found when the [OH-] is known.
pOH = -log[OH-]
On the pOH scale… pOH < 7 basic
pOH > 7 acidic
pOH = 7 neutral
Ex: A solution with an [OH-] of 1.0 x 10-3 M has a pOH of 3 (basic).

24. We can derive a relationship between pH and pOH at 25C from the expression for Kw:
[H3O+] [OH-] = 1.0 x 10-14
Take the log of both sides:
log([H3O+] [OH-]) = log(1.0 x 10-14)
log[H3O+] + log[OH-] =
-log[H3O+] - log[OH-] = 14.00
pH + pOH = 14
Note: the sum of pH and pOH is always 14 at 25C. Therefore, a solution with a pH of 3 has a pOH of 11.

25. The pKa Scale pKa = -logKa
The pKa of a weak acid is another way to quantify its strength.
The smaller the pKa , the stronger the acid.

26. 15.6 - Finding the [H3O+] and pH of Strong and Weak Acid Solutions

27. A solution containing a strong or weak acid has two potential sources of [H3O+]:
The ionization of the acid itself
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
The autoionization of water
H2O(l) + H2O(l) H3O+(aq) + OH-
The autoionization of water contributes a negligibly small amount of H3O+ compared to the ionization of the strong or weak acid. In a strong or weak acid solution, the additional H3O+ from the acid causes the autoionization equilibrium to shift left. Therefore, the autoionization of water produces even less H3O+ than in pure water and so it can be neglected. We just focus on amount of H3O+ produced by the acid.

28. Strong Acids Because… Strong acids completely ionize in solution
& we can ignore the ions contributed by the autoionization of water
The concentration of H3O+ in a strong acid solution is equal to the concentration of the strong acid.
Ex: 0.10 M HCl [H3O+] = 0.10 M pH = -log[H3O+] = -log(0.10) = 1.00

30. Weak Acids
The concentration of H3O+ is not equal to the concentration of a weak acid.
Calculating the [H3O+] formed by the ionization of a weak acid requires solving an equilibrium problem (Think Chapter 14 ICE Tables).
Since we can still ignore the contribution of H3O+ from the autoionization of water, we only have to determine the concentration of H3O+ formed by:
HA(aq) + H2O(l) H3O+(aq) + A-(aq)

32. “The pH of an acid solution depends on both the strength of the acid and the concentration of the acid”(College Board)

33. Consider a 0. 10 M solution of the generic weak acid HA
Consider a 0.10 M solution of the generic weak acid HA. We ignore the autoionization of water and just determine the concentration of H3O+ formed by the following equilibrium:
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[HA]
[H3O+]
[A-]
Initial
0.10
~0.00
0.00
Change -x +x
Equilibrium x x

34. Find the value of x by using the equilibrium expression to set up an equation with x as the only variable.
Ka = ([H3O+][A-]) / [HA]
Ka= x2 / x
(The value of the acid ionization constant will be given for you to substitute in for Ka.)

35. Practice Problem!
Find the [H3O+] of a M HCN solution. Ka= 4.9 x
[HCN]
[H3O+]
[CN-]
Initial
0.100
Change -x +x
Equilibrium x x

36. Answer Ka = ([H3O+][CN-]) / [HCN] = x2 / (0.100 - x)
[H3O+] = 7.0 x 10-6
“x is small” approximation allows you to cross out x and ignore it
(7.0 x 10-6 / 0.1) x 100 = 7.0 x 10 -3%
Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in approximation. The ratio should be less than 0.05 or 5%.
(if x is small is not valid you solve the equation normally with the quadratic equation)

37. Practice Problem!
Find the pH of a 2.00 M HNO2 solution. Ka = 4.6 x 10-4.
Hint
Think about what value you are missing for the calculation of pH
[HNO2]
[H3O+]
[NO2-]
Initial
2.00
Change -x +x
Equilibrium x x

38. Answer Ka = ([H3O+][NO2-]) / [HNO2] = x2 / (2.00 - x)
[H3O+] = 3.0 x 10-2
“x is small” approximation allows you to cross out x and ignore it
( 3.0 x 10-2 / 2.00) x 100 = 1.5%
Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in approximation. The ratio should be less than 0.05 or 5%.

39. Answer continued
pH = -log[H3O+]
= -log[3.0 x 10-2]
= =1.52

40. Practice Problem! (Last one)
A weak acid (HA) solution has a pH of Find the Ka value for the acid.
[HA]
[H3O+]
[A-] I
0.100 C
-3.5 x 10-5
+3.5 x 10-5 E
x 10-5
3.5 x 10-5
pH = -log[H3O+]
4.45 = -log[H3O+]
[H3O+] = 3.5 x 10-5 M

41. Ka = ([H3O+][A-]) / [HA]
= ((3.5 x 10-5)(3.5 x 10-5))/ 0.100
= 1.3 x 10-8

42. Formulas Recap (On Reference Table)
Kw = [H3O+] [OH-] = 1.0 x at 25C
pH = -log[H3O+]
pOH = -log[OH-]
pH + pOH = 14

43. Thanks for Listening! (Good Luck on the test!)