1. Neutralization Reactions ch. 4.8 and ch. 15

2. The neutralization reaction is a simple double replacement:
HCl + NaOH → NaCl H2O
H2CO KOH → K2CO H2O
2 HNO3 + Mg(OH)2 → Mg(NO3) H2O
What is the net ionic reaction in each case?
H+ + OH- → H2O

3. Neutralization reactions occur when an acid and a base react with each other forming water and a salt. (4.8)
Acid Base → Salt + Water
The resulting pH of the solution is not necessarily 7, it depends on the stoichiometry (limiting reactant) as well as on the type of salt formed in the reaction.

4. Stoichiometry
If all of the moles of acid and base react, (no excess acid or base) then the pH will be determined from the type and molarity of the salt that is formed in solution.
If there is excess acid or base, then the resulting pH will be determined from the type and amount of acid or base remaining as well as the salt.

5. Titrations (15.4)
A titration is a laboratory procedure that uses the neutralization reaction to determine the concentration of an unknown acid or base
A solution of known concentration (standard) is titrated against the unknown solution with careful volume measurements. (volumetric analysis)

6. The titration reaches equivalence when equal moles of acid and base have reacted (“neutralization” – doesn’t mean pH = 7).
The endpoint (point of stopping the titration) is usually signaled by a change in the color of an indicator that has been added to one of the solutions or determined by observing a rapid and major change in the pH of the solution.

7. Indicators (15.5)
Usually weak acids or weak bases that are isolated from plant pigments.
Their acid color is different than their base color.
HIn H2O ⇄ In H3O+
(acid color) (base color)
Addition of more acid or base causes shifts in its equilibrium resulting in color changes.

8. Indicator Equilibrium Constant
𝐾 𝑎 = [𝐻 3 𝑂 + ][𝐼 𝑛 − ] [𝐻𝐼 𝑛 − ]
The eye needs about a 10X difference in the concentrations of the two forms of the indicator (acid vs. base) to notice one over the other…this means that most indicators will have a “range” of 2 pH units:
HIn: In- of 10+:1 will show acid color
HIn: In- of 1:1 will be midrange blend
HIn: In- of 1:10+ will show base color

9. Indicator Range
The “range” refers to the pH range where the indicator “changes color” (which it doesn’t, you just see more of one form over the other)
The Ka is related to the pH where this will happen (or Kb for pOH)
𝐾 𝑎 [ 𝐻 3 𝑂 + ] = [𝐼 𝑛 − ] [𝐻𝐼𝑛] (when the constant is 10x the hydronium, the base color will be visible, when the hydronium is ten times the constant, the acid color will be visible…when the pKa = pH, the indicator is a blend of both forms – called midrange)

10. Common indicators (p.746) Litmus Bromthymol blue Phenolphthalein
Methyl red

11. Types of Problems
Titrations (goal is to find the concentration of an unknown, or the amount needed to cause complete neutralization between two knowns)
Finding the resulting pH (reactant data is known) Three scenarios can occur:
1) equal moles acid and base
2) excess acid
3) excess base

12. Sample Titration Problem
What is the concentration of HCl if it requires 100 mL of
the acid to reach equivalence with 50.0 mL of 0.10 M NaOH?
What is known? HCl: mL ?M
NaOH: mL M
Balance equation: HCl + NaOH → NaCl H2O
Stoichio: 0.050𝐿 𝑏𝑎𝑠𝑒 𝑚𝑜𝑙 𝑏𝑎𝑠𝑒 𝐿 𝑏𝑎𝑠𝑒 1 𝑚𝑜𝑙 𝑎𝑐𝑖𝑑 1 𝑚𝑜𝑙 𝑏𝑎𝑠𝑒 = 𝑚𝑜𝑙 𝑎𝑐𝑖𝑑
Dividing the mol acid by the volume of acid (in L) gives molarity:
mol acid/0.100 L = M HCl

13. Find final pH
Titration calculations are solved the same way regardless of the acid or base used.
Problems involving the products of the reaction, like the pH at the end, are easiest to understand if they are grouped into classes.
We begin with Strong Acid + Strong Base

14. Strong Acid + Strong Base Reactions
There are 3 possibilities:
No excess reactants
Excess acid
Excess base
In all SA + SB reactions, remember that the salt produced will be Neutral!

15. SA + SB (no excess)
What is the pH when 100 mL of 0.10 M HCl reacts with 100 mL of 0.10 M NaOH?
1) Known: HCl mL M
NaOH 100 mL M
2) Reaction: HCl NaOH → NaCl H2O
3) Find mols: (0.1L)(0.10M) (0.1L)(0.10M)
= 0.01 mol = 0.01 mol
4) react: mol mol mol
5) result: = = = 0.01 mol
6) analyze: (see next slide)

16. SA + SB (no excess)
6) analyze: If all that is remaining is 0.01 mol NaCl in a total volume of 200 mL (100 mL acid and 100 mL base mixed together)…the molarity of the NaCl is 0.01 mol/0.2 L = 0.05 M NaCl The pH is dependent on the only thing remaining: 0.05 M NaCl…it is a neutral salt, pH = 7

17. SA + SB (excess acid)
What is the pH when 200 mL of 0.10 M HCl reacts with 100 mL of 0.10 M NaOH?
1) Known: HCl mL M
NaOH 100 mL M
2) Reaction: HCl NaOH → NaCl H2O
3) Find mols: (0.2L)(0.10M) (0.1L)(0.10M)
= 0.02 mol = 0.01 mol (limiting!)
4) react: mol mol mol
5) result: = 0.01 mol = = 0.01 mol
6) analyze: (see next slide)

18. SA + SB (excess acid) 6) analyze:
If remaining is 0.01 mol HCl and 0.01 mol NaCl in a total volume of 300 mL (200 mL acid and 100 mL base mixed together)…both will be 0.03 M. Consider which will affect the pH. The strong acid will, the neutral salt won’t.
The pH will depend on the 0.03 M HCl.
pH = -log(0.03) = 1.5

19. SA + SB (excess base)
What is the pH when 100 mL of 0.10 M HCl reacts with 100 mL of 0.50 M NaOH?
1) Known: HCl mL M
NaOH 100 mL M
2) Reaction: HCl NaOH → NaCl H2O
3) Find mols: (0.1L)(0.10M) (0.1L)(0.50M)
= 0.01 mol (limiting!) = 0.05 mol
4) react: mol mol mol
5) result: = = 0.04 mol = 0.01 mol
6) analyze: (see next slide)

20. SA + SB (excess base) 6) analyze:
If remaining is 0.04 mol NaOH and 0.01 mol NaCl in a total volume of 200 mL (100 mL acid and 100 mL base mixed together…[NaOH]=0.2 M and
[NaCl]=0.05M.
Consider which will affect the pH. The strong base will, the neutral salt won’t.
The pH will depend on the 0.2 M NaOH.
pOH = -log(0.2) = 0.7, pH = 13.3

21. Weak Acid + Strong Base Reactions
There are 3 possibilities:
No excess reactants (basic salt remains)
Excess acid (weak acid and basic salt remain)
Excess base (strong base and basic salt remain)
In all WA + SB reactions, remember that the salt produced will be Basic!

22. WA + SB (excess base)
What is the pH when 100 mL of 0.10 M HA reacts with 100 mL of 0.50 M NaOH?
1) Known: HA mL M
NaOH 100 mL M
2) Reaction: HA NaOH → NaA H2O
3) Find mols: (0.1L)(0.10M) (0.1L)(0.50M)
= 0.01 mol (limiting!) = 0.05 mol
4) react: mol mol mol
5) result: = = 0.04 mol = 0.01 mol
6) analyze: (see next slide)

23. WA + SB (excess base) 6) analyze:
If remaining is 0.04 mol NaOH and 0.01 mol NaA in a total volume of 200 mL (100 mL acid and 100 mL base mixed together…[NaOH]=0.2 M and
[NaA]=0.05M.
Consider which will affect the pH. The strong base will overrule the basic salt.
The pH will depend on the 0.2 M NaOH.
pOH = -log(0.2) = 0.7, pH = 13.3

24. WA + SB (no excess)
What is the pH when 100 mL of 0.10 M HA reacts with 100 mL of 0.10 M NaOH? (Ka HA = 2 x 10-5)
1) Known: HA mL M
NaOH 100 mL M
2) Reaction: HA NaOH → NaA H2O
3) Find mols: (0.1L)(0.10M) (0.1L)(0.10M)
= 0.01 mol = 0.01 mol
4) react: mol mol mol
5) result: = = = 0.01 mol
6) analyze: (see next slide)

25. WA + SB (no excess) 6) analyze:
If all that is remaining is 0.01 mol NaA in a total volume of 200 mL (100 mL acid and 100 mL base mixed together)…the molarity of the NaA is 0.05 M.
NaA is a basic salt…the A- acts as the conj. base of a weak acid (HA), and it sets up an equilibrium with water.

26. WA + SB (no excess)
Equilibrium of A-: R) A- + H2O ⇄ HA + OH- I) 0.05 M 0 0 C) -x +x +x E) 0.05 – x x x Need a Kb = Kw/Ka = 5 x = x2/0.05 – x x = [OH-] = 5 x 10-6 … pOH = 5.3…pH = 8.7

27. WA + SB (excess acid)
What is the pH when 200 mL of 0.10 M HA reacts with 100 mL of 0.10 M NaOH?
1) Known: HA mL M
NaOH 100 mL M
2) Reaction: HA NaOH → NaA H2O
3) Find mols: (0.2L)(0.10M) (0.1L)(0.10M)
= 0.02 mol = 0.01 mol (limiting!)
4) react: mol mol mol
5) result: = 0.01 mol = = 0.01 mol
6) analyze: (see next slide)

28. WA + SB (excess acid) 6) analyze:
If remaining is 0.01 mol HA and 0.01 mol NaA in a total volume of 300 mL (200 mL acid and 100 mL base mixed together)…both will be 0.03 M.
Both will affect the pH…they are a conj. acid/base pair and will set up an equilibrium in the water.

29. WA + SB (excess acid)
Equilibrium of HA with A-: R) HA + H2O ⇄ A- + H3O+ I) 0.03 M 0.03M 0 C) -x +x +x E) 0.03 – x x x x Ka = 2 x 10-5 = (x)( x)/(0.03 – x) x = [H3O+] = 2 x 10-5 … pH = 4.7

30. Buffer Solutions
Any solution that contains appreciable amounts of a weak acid and its conjugate base, or of a weak base and its conjugate acid. More on this later…but that is what you made in the previous slide…

31. Strong Acid + Weak Base Reactions
There are 3 possibilities:
No excess reactants (acidic salt remains)
Excess acid (strong acid and acidic salt remain)
Excess base (weak base and basic salt remain)
In all SA + WB reactions, remember that the salt produced will be Acidic!

32. SA + WB (excess acid)
What is the pH when 200 mL of 0.10 M HCl reacts with 100 mL of 0.10 M NH3 ?
1) Known: HCl mL M
NH mL M
2) Reaction: HCl NH3 → NH4Cl H2O
3) Find mols: (0.2L)(0.10M) (0.1L)(0.10M)
= 0.02 mol = 0.01 mol (limiting!)
4) react: mol mol mol
5) result: = 0.01 mol = = 0.01 mol
6) analyze: (see next slide)

33. SA + WB (excess acid) 6) analyze:
If remaining is 0.01 mol HCl and 0.01 mol NH4Cl in a total volume of 300 mL (200 mL acid and 100 mL base mixed together)…both will be 0.03 M.
Both are acids (HCl and NH4+) but the strong acid will overrule the acidic salt.
The pH will depend on the 0.03 M HCl.
pH = -log(0.03) = 1.5

34. SA + WB (no excess)
What is the pH when 100 mL of 0.10 M HCl reacts with 100 mL of 0.10 M NH3 ? (Kb NH3 = 2 x 10-5)
1) Known: HCl mL M
NH mL M
2) Reaction: HCl NH3 → NH4Cl H2O
3) Find mols: (0.1L)(0.10M) (0.1L)(0.10M)
= 0.01 mol = 0.01 mol
4) react: mol mol mol
5) result: = = = 0.01 mol
6) analyze: (see next slide)

35. SA + WB (no excess) 6) analyze:
If all that is remaining is 0.01 mol NH4Cl in a total volume of 200 mL (100 mL acid and 100 mL base mixed together)…the molarity of the NH4Cl is 0.05 M.
NH4Cl is an acidic salt…the NH4+ acts as the conj. acid of a weak base (NH3 ), and it sets up an equilibrium with water.

36. SA + WB (no excess)
Equilibrium of NH4+ : R) NH4+ + H2O ⇄ NH3 + H3O+ I) 0.05 M 0 0 C) -x +x +x E) 0.05 – x x x Need a Ka = Kw/Kb = 5 x = x2/0.05 – x x = [H3O+] = 5 x 10-6 … pH = 5.3

37. SA + WB (excess base)
What is the pH when 100 mL of 0.10 M HCl reacts with 100 mL of 0.50 M NH3 ?
1) Known: HCl mL M
NH mL M
2) Reaction: HCl NH3 → NH4Cl H2O
3) Find mols: (0.1L)(0.10M) (0.1L)(0.50M)
= 0.01 mol (limiting!) = 0.05 mol
4) react: mol mol mol
5) result: = = 0.04 mol = 0.01 mol
6) analyze: (see next slide)

38. SA + WB (excess base) 6) analyze:
If remaining is 0.04 mol NH3 and 0.01 mol NH4Cl in a total volume of 200 mL (100 mL acid and 100 mL base mixed together…[NH3]=0.2 M and
[NH4Cl ]=0.05M.
Both will affect the pH…they are a conj. acid/base pair and will set up an equilibrium in the water.

39. SA + WB (excess acid)
Equilibrium of NH3 with NH4+ : R) NH3 + H2O ⇄ NH4+ + OH- I) 0.03 M 0.03M 0 C) -x +x +x E) 0.03 – x x x Kb = 2 x 10-5 = (x)( x)/(0.03 – x) x = [OH-] = 2 x 10-5 … pOH = 4.7…pH = 9.3