### Weak acids/bases titrated with strong acids/bases

Examples 1 & 3 are the titration of a weak acid with a strong base.
Examples 2 & 4 are the titration of a weak base with a strong acid.

Example 5 is the titration of the salt of a weak base (which is a weak acid) with a strong base.
Example 6 is the titration of the salt of a weak acid (which is a weak base) with a strong acid.

All six examples are multi-part problems.

I represented Ka and Kb expressions using different display techniques. I also did this with the log portion of the Henderson-Hasselbalch equation.

Example #1: Consider the titration of a 24.0-mL sample of 0.105 M CH3COOH with 0.130 M NaOH. What is . . .

(a) the initial pH?
(b) the volume of added base required to reach the equivalence point?
(c) the pH at 6.00 mL of added base?
(d) the pH at one-half of the equivalence point?
(e) the pH at the equivalence point?

Solution to part (a):

1) Insert values into the Ka expression for acetic acid. The Ka for acetic acid is 1.77 x 10-5.

1.77 x 10-5 = $\frac{\mathrm{\left(x\right) \left(x\right)}}{\mathrm{0.105}}$

x = 1.3633 x 10-3 M

pH = 2.865

Solution to part (b):

1) Calculate moles of acid present:

(0.105 mol/L) (0.0240 L) = 2.52 x 10-3 moles

2) Determine moles of base required to react equivalence point:

CH3COOH + NaOH ---> CH3COONa + H2O

There is a 1:1 molar ratio between acetic acid and sodium hydroxide.

Therefore, 2.52 x 10-3 moles of base required

3) Calculate volume of base solution required:
2.52 x 10-3 mol divided by 0.130 mol/L = 0.0194 L = 19.4 mL

Solution to part (c):

1) Calculate moles of acid and base in solution before reaction:

CH3COOH: 2.52 x 10-3 mol
NaOH: (0.00600 L) (0.130 mol/L) = 7.80 x 10-4 mol

2) Determine amounts of acid and acetate ion after reaction:

CH3COOH: 2.52 x 10-3 mol - 7.80 x 10-4 mol = 1.74 x 10-3 mol
CH3COONa: 7.80 x 10-4 mol

3) Use Henderson-Hasselbalch equation to determine pH of (now) buffered solution:

pH = 4.752 + log $\frac{\mathrm{7.80 x 10-4}}{\mathrm{1.74 x 10-3}}$

pH = 4.752 + log 0.4483

pH = 4.404

Note that the new molarities were not calculated for the log term, rather the mole amounts were used directly. This is because both mole amounts exist in the same 30.0 mL solution. There would be identical volume amounts in the numerator and denominator of the log term, so they cancel out.

Solution to part (d):

At one-half the equivalence, exactly one-half the acid (in this case) has been used up. The half that was used up was made into the salt (sodium acetate in this case). The two amounts (acid and salt) are equal in amount. Just below, I will use a '1' to symbolize the amount. Keep in mind that it is not the actual amount.

1) Use the Henderson-Hasselbalch Equation:

pH = 4.752 + log $\frac{1}{1}$

pH = 4.752

You may use the actual number of moles if you wish. It simply does not matter since the log term will always zero out at the half-equivalence point.

Note: pH = pKa at the half-equivalence point. Remember that. You stand a very good chance of being asked a half-equivalence point question on your test.

Solution to part (e):

The solution is now completely composed of a salt of a weak acid. The pH of this solution will be basic.

1) Calculate molarity of sodium acetate:

2.52 x 10-3 mol / 0.0434 L = 0.0581 M

2) Calculate the Kb of sodium acetate:

Kw = KaKb

1.00 x 10-14 = (1.77 x 10-5 ) (x)

x = 5.65 x 10-10

3) Calculate pH of the solution:

5.65 x 10-10 = $\frac{\mathrm{\left(x\right) \left(x\right)}}{\mathrm{0.0581}}$

x = 5.73 x 10-6 M (this is the hydroxide ion concentration)

pOH = 5.242

pH = 8.758

Example #2: Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl. Calculate the pH for (a) through (f). (Kb for CH3NH2 = 4.4 x 10¯4)

(a) after 0.0 mL of HCl added.
(b) after 20.0 mL of HCl added.
(c) after 65.0 mL of HCl added.
(d) at the equivalence point.
(e) after 300.0 mL of HCl added.
(f) At what volume of HCl added does the pH = 10.643?

Solution to (a):

No strong acid has been added, so this solution is simply a solution of a weak base. We use the Kb expression to solve for the hydroxide ion concentration and thence to the pH.

1) The reaction of interest is this one:

CH3NH2 + H2O ⇌ CH3NH3+ + OH¯

2) The relevant Kb expression is:

 [CH3NH3+] [OH¯] Kb = –––––––––––––––– [CH3NH2]

3) Solve the Kb expression for the hydroxide concentration:

 (x) (x) 4.4 x 10¯4 = –––––– 0.200

x = 0.00938083 M <--- some guard digits

4) Solve for the pOH, then the pH

pOH = -log 0.00938083 = 2.02776

pH = 14 - 2.02776 = 11.97

Solution to (b):

Some HCl has been added and some (not all) of the base has reacted. After the reaction is complete, there is a mixture of CH3NH3+ and CH3NH2 in solution. All the HCl has been used up and we have a buffer. The Henderson-Hasselbalch Equation will be used to solve this problem.

1) The chemical reaction of interest is this:

CH3NH2 + H3O+ ---> CH3NH3+ + H2O

Note that it is not an equilibrium, it goes to completion and all the H+ (from HCl) is used up. Later on, in a later part of the problem, the HCl will be in excess.

2) How much HCl was added to the solution?

HCl ---> (0.100 mol/L) (0.0200 L) = 0.00200 mol

3) The HCl reacts with CH3NH2 to make CH3NH3+:

CH3NH2 at start ---> (0.200 mol/L) (0.1000 L = 0.0200 mol

CH3NH2 after reacting with HCl ---> 0.0200 mol minus 0.00200 mol = 0.018 mol

4) How much CH3NH3+ was made?

0.00200 mol

due to the 1:1 stoichiometry between HCl and CH3NH2

5) Enter the Henderson-Hasselbalch!

 [CH3NH2] pH = pKa + log –––––––––– [CH3NH3+]
 0.0180 pH = 10.643 + log –––––– 0.0020

pH = 10.643 + 0.954

pH = 11.60

Notice that I did not bother to use molarities in the H-H. This is because both substances are in the same volume (120.0 mL in this case) and so their influence simply cancels out. In other words, the ratio of moles is the same exact value as the ratio of molarities.

Also take note that I used the pKa of the methylammonium ion. Since it is the conjugate acid to the base (the methyl amine), the Ka and Kb are related to each other by Kw:

KaKb = Kw

Solution to (c):

This problem is just like (b), except that 65.0 mL of HCl is added. The solution technique is the same as in (b) with just a slight change in amounts of chemical substances.

(0.100 mol/L) (0.0650 L) = 0.00650 mol

2) CH3NH2 and CH3NH3+ that are in solution:

CH3NH2 ---> 0.0200 mol - 0.00650 mol = 0.0135 mol
CH3NH3+ ---> 0.00650 mol

3) The Henderson-Hasselbalch:

pH = 10.643 + log (0.0135 / 0.00650)

pH = 10.643 + log 2.076923

pH = 10.32

Note that from (a) to (b) and then from (b) to (c), that pH is becoming more acidic. It's moving from 11.97 to 11.60 to 10.32. Since we are adding an acid (HCl), the pH is doing what it's supposed to do.

Yay!

Solution to (d):

At the equivalence point, all the base (the methyl amine) has been converted to its salt, the methylammonium ion. Being the salt of a weak base, the methylammonium ion will hydrolyze, forming an acidic pH at the equivalence point.

1) This is the reaction to consider:

CH3NH3+ + H2O ⇌ H3O+ + CH3NH2

2) We can write a Ka expression . . .

 [H3O+] [CH3NH2] Ka = –––––––––––––––– [CH3NH3+]

3) . . . and solve it:

 (x) (x) 2.2727 x 10¯11 = –––––––– 0.06667

x = 0.0000012309 M <--- [H3O+]

4) the pH:

pH = - log 0.0000012309 = 5.91

Notice the concentration of 0.06667 M. That is the [CH3NH3+]. There were 0.0200 moles of it (from all the CH3NH2 being neutralized and it was in 300.0 mL total solution. The other 200.0 mL came from the HCl solution required to neutralize all the methyl amine.

Make sure you recognize when you have to have a molarity in the calculation (when doing Ka or Kb calculations) and when you can use moles or molarity (when doing a Henderson-Hasselbalch calculation).

Solution to (e):

The HCl is now in excess. We need to determine how much HCl remains as some of it reacts with the CH3NH2. We will completely ignore the presence of the CH3NH3+, a weak acid, because its influence on the pH is completely overwhelmed by the presence of the HCl, a strong acid.

1) CH3NH2 present:

(0.200 mol/L) (0.1000 L) = 0.0200 mol

2) HCl present:

(0.100 mol/L) (0.3000 L) = 0.0300 mol

3) HCl present after reaction with the methyl amine:

0.0300 - 0.0200 = 0.0100 mol

4) New molarity of the HCl:

0.0100 mol / 0.4000 L = 0.025 M

5) pH of the HCl solution:

pH = -log 0.025 = 1.60

Solution to (f):

1) This is the chemical equation of interest:

CH3NH2 + H3O+ ⇌ CH3NH3+ + H2O

Some HCl was added, such that we wind up with a buffer of pH 10.643. By the way, we know it is a buffer because the pH at the equivalence point is 5.91. We are well away from that 5.91 value.

2) That means we can write a Henderson-Hasselbalch Equation for the above reaction:

 [CH3NH2] pH = pKa + log –––––––––– [CH3NH3+]

3) The usual situation is that we are solving for the pH. However, in this situation, the pH is known:

 [CH3NH2] 10.643 = 10.643 + log –––––––––– [CH3NH3+]

4) Since 10.643 minus 10.643 is zero, we have:

 [CH3NH2] log –––––––––– = 0 [CH3NH3+]

5) Antilog:

 [CH3NH2] –––––––––– = 1 [CH3NH3+]

6) The key point now is that the ratio of base to acid is 1:1 at a pH of 10.643. That means that exactly half the base was neutralized by HCl. The will give us the volume of HCl required:

moles CH3NH2 originally present
(0.200 mol/L) (0.1000 L) = 0.0200 mol

exactly 0.0100 mol of CH3NH2 was neutralized. (Remember, half the CH3NH2 gets neutralized.)

this requires 0.0100 mol of HCl

what volume of 0.100 M HCl contains 0.0100 mol?

0.0100 mol / 0.100 mol/L = 0.100 L = 100. mL

Often, in problems of this nature, one of the question parts asks for the pH at the half-equivalence point. in this case, the pH of the half-equivalence point was given and the amount of acid required to get there was asked for.

Example #3: Calculate the pH, when the following solutions are added to 100. mL 0.10 M HClO solution:

(a) 0 mL 0.10 M NaOH
(b) 75.0 mL 0.10 M NaOH
(c) 100. mL 0.10 M NaOH

The Ka of HClO is 3.0 x 10¯8

Solution to (a):

This is a standard weak acid calculation.

Ka = ([H+] [ClO¯]) / [HClO]

3.0 x 10¯8 = [(x) (x)] / 0.10

x = 5.477 x 10¯5 M

pH = -log 5.477 x 10¯5 = 4.26

Solution to (b):

This is a buffer calculation requiring the use of the Henderson-Hasselbalch Equation.

moles HClO ---> (0.10 mol/L) (0.100 L) = 0.010 mol
moles NaOH ---> (0.10 mol/L) (0.0750 L) = 0.00750 mol

HClO and NaOH react in a 1:1 molar ratio. The HClO is in excess.

moles HClO remaining ---> 0.010 mol - 0.00750 mol = 0.0025 mol
moles NaClO produced ---> 0.00750 mol

This problem can be solved by using the moles directly, since they are in the same ratio as the molarities would be.

pH = pKa + log (salt / acid)

pH = 7.523 + log (0.00750 / 0.0025)

pH = 8.00

Solution to (c):

1) It can be seen by examination that equimolar amounts of HClO and NaOH are present. This means that only NaClO is present after reaction. The problem becomes to determine the pH of the salt of an acid knowing the Ka of the acid. The first thing to be determine is the Kb of the salt.

KaKb = Kw

(3.0 x 10¯8) (Kb) = 1.0 x 10¯14

Kb = 3.333 x 10¯7

2) We need to know the molarity of the NaClO:

0.010 mol / 0.200 L = 0.050 M

3) Use the Kb expression:

Kb = ([HClO] [OH¯]) / [ClO¯]

3.333 x 10¯7 = [(x) (x)] / 0.050

x = 0.0001291 M

4) Since this is a hydroxide concentration, we determine the pOH and use that and pKw to determine the pH:

pOH = - log 0.0001291 = 3.89

pH = 14.00 - 3.89 = 10.11

Example #4: 500. mL of a solution containing 1.50 M NH3(aq) is mixed with 500. mL of a solution containing 0.500 M of HCl(aq). (Kb for NH3 = 1.77 x 10¯5). Dtermine the pH of:

(a) the final solution.
(b) the solution in (a) after 40.0 mL of 0.200 M NaOH has been added.
(c) the solution in (a) after 24.0 mL of 0.500 M HCl has been added.

Solution to (a):

1) The chemical reaction of interest is this (written in net-ionic form):

NH3 + H+ ---> NH4+

2) It should be fairly obvious that the H+ is the limiting reagent. However, let us check:

NH3 ---> (1.50 mol/L) (0.500 L) = 0.750 mol
H+ ---> (0.500 mol/L) (0.500 L) = 0.250 mol

Since NH3 and H+ react in a 1:1 molar ratio, it can be seen that H+ is the limiting reagent.

We can also intuit that H+ is the limiting reagent from the problem type. If the H+ is in excess, we wind up with a solution of a strong acid. That's not the usual place to start in a buffer calculation. Moving on . . .

3) Determine moles of NH3 remaining and moles of NH4+ formed:

NH3 ---> 0.750 minus 0.250 = 0.500 mol
NH4+ ---> 0.250 mol

4) Use the Kb of ammonia to get the Ka of ammonium ion:

KaKb = Kw

(Ka) (1.77 x 10¯5) = 1.00 x 10¯14

Ka = 5.65 x 10¯10 <--- use this to get the pKa

5) Unleash the Henderson-Hasselbalch:

pH = pKa + log [base / acid]

pH = 9.248 + log [0.500 / 0.250)

pH = 9.248 + 0.301 = 9.549

Solution to (b):

1) Let us determine how many moles of NaOH are added:

(0.200 mol/L) (0.0400 L) = 0.00800 mol

2) The hydroxide will react with NH4+ to form more NH3. The NH4+ amount will go down.

NH3 ---> 0.500 + 0.00800 = 0.508 mol
NH4+ ---> 0.250 - 0.00800 = 0.242 mol

3) Unleash the Henderson-Hasselbalch:

pH = pKa + log [base / acid]

pH = 9.248 + log [0.508 / 0.242)

pH = 9.248 + 0.322 = 9.570

Solution to (c):

1) Let us determine how many moles of HCl are added:

(0.500 mol/L) (0.0240 L) = 0.0120 mol

2) The hydrogen ion will react with NH3 to form more NH4+. The NH3 amount will go down.

NH3 ---> 0.500 - 0.0120 = 0.488 mol
NH4+ ---> 0.250 + 0.0120 = 0.262 mol

3) Unleash the Henderson-Hasselbalch:

pH = pKa + log [base / acid]

pH = 9.248 + log [0.488 / 0.262)

pH = 9.248 + 0.270 = 9.518

Example #5: Calculate the pH at the points indicated below if 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. (Ka for aniline hydrochloride is 2.4 x 10-5)

C6H5NH3+(aq) + OH¯(aq) ---> C6H5NH2(aq) + H2O(ℓ)
(a) before the titration begins
(b) at the equivalence point
(c) at the midpoint of the titration
(d) after 20. mL of NaOH has been added
(e) after 30. mL of NaOH has been added

Solution to part (a):

Aniline hydrochloride (formula is C6H5NH3+Cl¯) is a salt of the weak base aniline. The salt will form an acidic solution.

C6H5NH3+(aq) + H2O(ℓ) ⇌ C6H5NH2(aq) + H3O+(aq)

2.4 x 10-5 = $\frac{\mathrm{\left(x\right) \left(x\right)}}{\mathrm{0.100}}$

x = 0.00155 M

pH = 2.810

Solution to part (b):

1) Calculate moles of aniline hydrochloride initially present:

(0.100 mol/L) (0.0500 L) = 0.00500 mol

2) Calculate the volume of NaOH solution required to reach equivalence point:

(0.185 mol/L) (x) = 0.00500 mol

x = 0.0270 L = 27.0 mL

3) Calculate new molarity of aniline:

0.00500 mol / 0.0770 L = 0.064935 M

4) Calculate Kb for aniline:

KaKb = Kw

(2.4 x 10-5) (x) = 1.00 x 10-14

x = 4.17 x 10-10

5) Calculate pH of solution:

4.17 x 10-10 = [(x) (x)] / 0.064935

x = 5.20364 x 10-6 M

pOH = -log 5.20364 x 10-6 = 5.284

pH = 8.716

Solution to part (c):

A better term for midpoint of the titration is half-equivalence point. At the half-equivalence point, exactly half the weak acid (in this case) has been titrated. The half that has been titrated has been converted into a base (in our case, named aniline). This solution is a buffer, so we use the Henderson-Hasselbalch Equation:

pH = pKa + log $\frac{\mathrm{\left[base\right]}}{\mathrm{\left[acid\right]}}$

However, the base/acid ratio at the half-equivalence always equals one, therefore:

pH = pKa

pH = - log 2.4 x 10-5

pH = 4.62

Solution to part (d):

1) Calculate moles of each substance before reacting:

aniline hydrochloride: (0.100 mol/L) (0.0500 L) = 0.00500 mol
NaOH: (0.185 mol/L) (0.0200 L) = 0.00370 mol

2) Calculate moles of each substance after reacting:

aniline hydrochloride: 0.00500 mol - 0.00370 mol = 0.00130 mol
aniline: 0.00370 mol

3) Use the Henderson-Hasselbalch Equation:

pH = pKa + log $\frac{\mathrm{\left[base\right]}}{\mathrm{\left[acid\right]}}$

pH = 4.620 + log $\frac{\mathrm{0.00370}}{\mathrm{0.00130}}$

pH = 5.07 (to two sig figs)

Solution to part (e):

27.0 mL of the NaOH solution goes to converting aniline hydrochloride into aniline, a weak base. 3.0 mL of 0.185 M NaOH are left over. Anytime you have a mixture of a strong base and a weak base, ignore the weak and concentrate on the strong.

1) Find new molarity of NaOH:

M1V1 = M2V2

(0.185 mol/L) (3.0 mL) = (x) (80.0 mL)

x = 0.0069375 M

2) Find pOH, then pH:

pOH = - log 0.0069375 = 2.159

pH = 14 - pOH = 11.841

Example #6: Consider the titration of 20.0 mL of 0.243 M of KX with 0.106 M HCl. The pKa of HX is 10.39. Give all pH values to 0.01 pH units.

(a) What is the pH of the original solution before addition of any acid?
(b) How many mL of acid are required to reach the equivalence point?
(c) What is the pH at the equivalence point?
(d) What is the pH of the solution after the addition of 27.9 mL of acid?
(e) What is the pH of the solution after the addition of 55.0 mL of acid?

Solution to (a):

1) KX is the salt of a weak acid. It hydrolyzes as follows:

X¯ + H2O ⇌ HX + OH¯

2) X¯ is a base, so we write the Kb expression:

Kb = $\frac{\mathrm{\left[HX\right] \left[OH¯\right]}}{\mathrm{\left[X¯\right]}}$

The Kb is determined from the pKa:

Since pKa = 10.39, the pKb = 3.61

Kb = 10-pKb = 10-3.61 = 2.4547 x 10¯4

3) Solve for the hydroxide concentration:

2.4547 x 10¯4 = $\frac{\mathrm{\left(x\right) \left(x\right)}}{\mathrm{0.243}}$

x = 0.010512 M <--- kept a couple guard digits

4) The pOH, then the pH:

pOH = -log 0.010512 = 1.98

pH = 12.02

Solution to (b):

1) KX and HCl react in a 1:1 molar ratio:

KX + HCl ---> HX + KCl

2) Moles of KX that got titrated:

(0.243 mol/L) (0.0200 L) = 0.00486 mol

3) How many mL of 0.106 M HCl are required to deliver 0.00486 mol?

0.00486 mol / 0.106 mol/L = 0.04585 L = 45.85 mL

I decided to ignore the solution path that uses millimoles.

Solution to (c):

N.B. The solution at the equivalence point is the solution of a weak acid (HX). The KCl plays no role in the pH calculation.

1) Calculate the molarity of HX in the solution at the equivalence point:

0.00486 mol / 0.06585 L = 0.0738 M

The 0.06585 L came from the 45.85 mL of HCl solution and the 20.0 mL of KX solution.

2) Write the Ka expression for HX and solve:

HX + H2O ⇌ H3O+ + X¯

Ka = $\frac{\mathrm{\left[H3O+\right] \left[X¯\right]}}{\mathrm{\left[HX\right]}}$

4.0738 x 10¯11 = $\frac{\mathrm{\left(x\right) \left(x\right)}}{\mathrm{0.0738}}$

N.B. the Ka came from 10¯pKa = 10¯10.39

x = 0.0000235 M

pH = -log 0.0000235 = 4.63

Solution to (d):

At this point the titration is only partially done (remember, it required 45.85 mL of HCl to reach the equivalence point). The 27.9 mL in this part of the problem will create a buffer and we will use the Henderson-Hasselbalch equation to get the pH.

1) Determine moles of HCl added:

(0.106 mol/L) (0.0279 L) = 0.0029574 mol

2) The HCl was added to 0.00486 mol of KX. How much HX and X¯ are in the solution?

moles HX = 0.0029574 mol (all the HCl reacted)
moles X¯ remaining ---> 0.00486 mol - 0.0029574 mol = 0.0019026 mol

3) Unleash the Henderson-Hasselbalch!

pH = pKa + log $\frac{\mathrm{\left[base\right]}}{\mathrm{\left[acid\right]}}$

pH = 10.39 + log $\frac{\mathrm{0.0019026}}{\mathrm{0.0029574}}$

N.B. I didn't have to calculate molarities because the ratio of the moles involved is equal to the ratio of molarities.

pH = 10.39 + (-0.19)

pH = 10.20

This pH fits. We are titrating a base (X¯) with an acid (HCl). The pH has moved from 12 to 10 on its way to 4.6.

Solution to (e):

The solution has been titrated past the endpoint and there is now excess strong acid in the solution. This overwhelms the presence of the weak acid (the HX) and the solution acts like a solution with only HCl present.

1) Calculate the volume of unreacted HCl:

55.0 mL - 45.85 mL = 9.15 mL

2) Calculate the new molarity of the 9.15 mL of HCl:

M1V1 = M2V2

(0.106 mol/L) (9.15 mL) = (x) (75.0 mL)

x = 0.012932 M

N.B. The 75 mL comes from 20 mL of KX being titrated with 55 mL of acid.

3) Calculate the pH:

pH = -log 0.012932 = 1.89

N.B. HCl is a strong acid, so it dissociates 100%

Bonus Example: A 0.552 g sample of ascorbic acid (Vitamin C) was dissolved in water to a total volume of 20.0 mL and titrated with 0.1103 M KOH. The equivalence point was reached at 28.42 mL and the pH of the solution at 10.0 mL of added base was 3.72. What is: (a) the molar mass of Vitamin C and (b) its Ka?

Solution to part a:

1) Determine moles of base used to reach equivalence point:

(0.1103 mol/L) (0.02842 L) = 0.003134726 mol

2) Assuming ascorbic acid is monoprotic (a necessary assumption to solve the problem!), calculate its molar mass:

0.552 g / 0.003134726 mol = 176 g/mol

Solution to part b:

1) Determine amounts of acid and salt at 10.0 mL of added base:

acid: 0.003134726 mol x (18.42/28.42) = 0.002031726 mol
base: 0.003134726 mol x (10.0/28.42) = 0.001103 mol

2) Use the Henderson-Hasselbalch Equation to determine the pKa:

3.72 = x + log $\frac{\mathrm{0.001103}}{\mathrm{0.002031726}}$

3.72 = x + (-0.265)

x = 3.98 (to two, not three, sig figs)

10-pKa yields the Ka