Weak acids/bases titrated with strong acids/bases
Twelve Examples

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Go to 15 weak acid/base titration problems

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Examples 1, 2, 3, & 4 are the titration of a weak acid with a strong base.
Examples 5, 6, 7, & 8 are the titration of a weak base with a strong acid.

Example 9 is the titration of the salt of a weak base (making the salt an acid) with a strong base.
Example 10 is the titration of the salt of a weak acid (making the salt a bzse) with a strong acid.

All ten of the above examples are multi-part problems.

Examples 11 and 12 are single-part problems that have interesting twists concerning how volumes are determined.


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.

  (x) (x)
1.77 x 10¯5 = –––––––
  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 / 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 the (now) buffered solution:

  7.80 x 10¯4
pH = 4.752 + log –––––––––
  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:

  1
pH = 4.752 + log –––
  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:

  (x) (x)
5.65 x 10¯10 = ––––––
  0.0581

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

pOH = 5.242

pH = 8.758


Example #2: 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 #3: 20.00 mL of a 0.150 M HCN solution is titrated with 0.100 M KOH solution. The Ka of HCN is 6.17 x 10¯10.

(a) What is the initial pH of the HCN solution?
(b) What is the volume of KOH needed to reach the equivalence point?
(c) What is the pH at the half-equivalence point?
(d) What is the pH of the solution after 20.00 mL of KOH is added?
(e) What is the pH at the equivalence point?
(f) What is the pH of the solution after 40.00 mL of KOH is added?

Solution to (a):

To calculate the pH of a weak acid, we will use a Ka calculation.

  [H+] [CN¯]
Ka = –––––––––
  [HCN]

  (x) (x)
6.17 x 10¯10 = ––––––
  0.150

x = 9.62 x 10¯6 M

pH = −log 9.62 x 10¯6 = 5.017

Solution to (b):

moles HCN ---> (0.150 mmol/mL) (20.00 mL) = 3.00 mmol

HCN and KOH react in a 1:1 molar ratio. Therefore, 3.00 mmol of KOH required.

volume KOH ---> 3.00 mmol / 0.100 mmol/mL = 30.0 mL

Since there is a 1:1 molar ratio, I could have used M1V1 = M2V2. I decided not to.

Solution to (c):

At the half-equivalence point, the amount of HCN and the amount of CN¯ are equal.

This means that the log portion of the Henderson-Hasselbalch Equation is zero.

Which means this:

pH = pKa

pH = −log 6.17 x 10¯10 = 9.210

Solution to (d):

A buffer has been formed. The Henderson-Hasselbalch Equation will be used.

mmol of HCN initially present ---> 3.00 mmol
mmoles KOH added ---> (0.100 mmol/mL) (20.00 mL) = 2.00 mmol

mmol of HCN remaining ---> 3.00 mmol − 2.00 mmol = 1.00 mmol
mmol of CN¯ formed ---> 2.00 mmol

pH = pKa + log [base / acid]

pH = 9.210 + log [2.00 / 1.00]

pH = 9.511

Solution to (e):

At the equivalence point, only CN¯ is present and it hydrolyzes:

CN¯ + H2O ⇌ HCN + OH¯

molarity of CN¯ ---> 3.00 mmol / 50.0 mL = 0.0600 M

The Kb of CN¯ = 1.00 x 10¯14 / 6.17 x 10¯10 = 1.62 x 10¯5

  [HCN] [OH¯]
Kb = –––––––––
  [CN¯]

  (x) (x)
1.62 x 10¯5 = ––––––
  0.0600

x = 0.000986 M

pOH = −log 0.000986 = 3.006

pH = 10.994

Solution to (f):

It took 30.00 mL of KOH to get to the equivalence point. That means 10.00 mL of 0.100 M KOH remains unreacted.

new molarity:

M1V1 = M2V2

(10.00 mL) (0.100 mol/L) = (60.00 mL) (x)

x = 0.016667 M

pOH = −log 0.016667 = 1.778

pH = 14 - 1.778 = 12.222

The weak base CN¯ is present in solution. Its contribution to the hydroxide concentration is negligible.


Example #4: 20.00 mL of 0.250 M propanoic acid, C3H7COOH, is titrated with 0.125 M NaOH. Calculate the pH for the following five situations after the indicated amount of NaOH solution has been added. The Ka of propanoic acid is 1.34 x 10¯5

(a) 0 mL
(b) 10.00 mL
(c) 20.00 mL
(d) 40.00 mL
(e) 50.00 mL

Discussion:

Look at the molarities to notice that the NaOH value is half that of the acid. Based on that, you can identify (c) as the half-equivalence point and (d) as the equivalence point. You can also identify (e) as being flooded with excess NaOH and (b) as an ordinary buffer.

Solution to (a):

1) This is a standard Ka calculation:

C3H7COOH + H2O ⇌ H3O+ + C3H7COO¯

  [H3O+] [C3H7COO¯]
Ka = –––––––––––––––––
  [C3H7COOH]

  (x) (x)
1.34 x 10¯5 = ––––––
  0.250

x = 0.00183 M

pH = −log 0.00183 = 2.738

Note: I used the rule for rounding off with a five. I had 2.7375489 on my calculator.

Solution to (b):

1) A buffer calculation will be performed. First, calculate the mmole of propanoic acid and NaOH:

propanoic acid ---> (0.250 mmol/mL) (20.00 mL) = 5.00 mmol
NaOH ---> (0.125 mmol/mL) (10.00 mL) = 1.25 mmol

2) Propanoic acid is used up, sodium propanoate is formed. Calculate the amounts of each:

propanoic acid ---> 5.00 mmol − 1.25 mmol = 3.75 mmol
propanoate ---> 1.25 mmol

3) Use the Henderson-Hasselbalch Equation:

  [base]
pH = pKa + log –––––
  [acid]

  1.25
pH = 4.8729 + log –––––
  3.75

pH = 4.873 + (−0.477) = 4.396

Solution to (c):

1) The half-equivalence point. Calculate the amounts of propanoic acid and NaOH:

propanoic acid ---> (0.250 mmol/mL) (20.00 mL) = 5.00 mmol
NaOH ---> (0.125 mmol/mL) (20.00 mL) = 2.50 mmol

2) Propanoic acid is used up, sodium propanoate is formed. Calculate the amounts of each:

propanoic acid ---> 5.00 mmol − 2.50 mmol = 2.50 mmol
propanoate ---> 2.50 mmol

3) Use the Henderson-Hasselbalch:

  2.50  
pH = 4.8729 + log ––––– <--- note how log portion goes to zero, resulting in pH = pKa
  2.50  

pH = 4.873

Solution to (d):

1) The equivalence point. Here's the amounts of propanoic acid and NaOH:

propanoic acid ---> (0.250 mmol/mL) (20.00 mL) = 5.00 mmol
NaOH ---> (0.125 mmol/mL) (40.00 mL) = 5.00 mmol

2) Propanoic acid is used up, sodium propanoate is formed. Calculate the amounts of each:

propanoic acid ---> 5.00 mmol − 5.00 mmol = 0 mmol
propanoate ---> 5.00 mmol

3) We will do a Kb calculation, but we first need the actual molarity of the propanoate (and the Kb):

5.00 mmol / 60.00 mL = 0.083333 M <--- keep a couple extra digits

KaKb = Kw

Kb = Kw / Kw = 1.00 x 10¯14 / 1.34 x 10¯5 = 7.4627 x 10¯10

4) From here, it's a normal Kb calculation:

C3H7COO¯ + H2O ⇌ C3H7COOH + OH¯

  [C3H7COOH] [OH¯]
Kb = –––––––––––––––––
  [C3H7COO¯]

  (x) (x)
1.34 x 10¯5 = ––––––––
  0.083333

x = 0.000007886 M

5) pOH, then pH:

pOH = −log 0.000007886 = 5.103

pH = 14 − 5.103 = 8.897

Solution to (e):

1) An excess of NaOH:

propanoic acid ---> (0.250 mmol/mL) (20.00 mL) = 5.00 mmol
NaOH ---> (0.125 mmol/mL) (50.00 mL) = 6.25 mmol

2) Determine the molarity of the excess:

1.25 mmol / 70.00 mL = 0.017857 M

The presence of sodium propanoate is ignored because, in the presence of a strong base, its hydroxide contribution is negligible.

3) Calculate pOH, then pH:

pOH = −log 0.017857 = 1.748

pH = 14 − 1.748 = 12.252


Example #5: 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 − 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.

1) HCl added:

(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 methylamine) 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) By the way, the Ka was obtained by using this equation:

KaKb = Kw

(Ka) (4.4 x 10¯4) = 1.00 x 10¯14

Ka = (1.00 x 10¯14) / (4.4 x 10¯4)

Ka = 2.2727 x 10¯11

5) 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 − 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 #6: 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 − 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 #7: A titration of 35.00 mL of 0.1750 M pyridine using a 0.275 M HCl solution is carried out. Pyridine has a pKb = 8.77

(a) What is the initial pH of the pyridine solution (prior to any acid added)
(b) What is the pH of the pyridine solution after the addition of 5.00 mL of HCl
(c) What is the volume of HCl added at the half-equivalence point
(d) What is the pH of the pyridine solution at the half-equivalence point
(e) What is the pH of the pyridine solution after the addition of 21.00 mL of HCl
(f) What is the pH of the pyridine solution at the equivalence point
(g) What is the pH of the pyridine solution after the addition of 26.00 mL of HCl

Solution to (a):

Pyridine ionizes as follows:

C5H5N + H2O ⇌ C5H5NH+ + OH¯

We need a Kb expression:

  [C5H5NH+] [OH¯]
Kb = –––––––––––––––
  [C5H5N]

  (x) (x)  
1.698 x 10¯9 = –––––– <--- the Kb value came from 10¯8.77
  0.1750  

x = 0.000017238 M

pOH = −log 0.000017238 = 4.76 (to two sig figs)

pH = 14 − 4.764 = 9.236

Solution to (b):

millimoles pyridine before reaction ---> (0.1750 mmol/mL) (35.00 mL = 6.125 mmol
millimoles HCl before reaction ---> (0.275 mmol/mL) (5.00 mL) = 1.375 mmol

Reaction takes place in a 1:1 molar ratio:

mmol pyridine after reaction ---> 6.125 − 1.375 = 4.750 mmol
mmol pyridinium ion produced ---> 1.375 mmol

Use the Henderson-Hasselbalch Equation:

pH = 5.23 + log (4.750 / 1.375) <--- note the use of the pKa

pH = 5.23 + 3.45 = 8.68

Solution to (c):

mmole pyridine before reaction: 6.125 mmol

mmole pyridine used up at half-equivalence point: 3.0625 mmol

volume HCl solution required ---> 3.0625 mmol / 0.275 mmol/mL = 11.14 mL (included an extra digit)

Solution to (d):

At the half-equivalence point, the amounts of pyrdine and pyridinium are equal. This zeros out the log portion of the H-H Equation. Therefore:

pH = pKa

pH = 5.23

Solution to (e):

millimoles pyridine before reaction ---> (0.1750 mmol/mL) (35.00 mL = 6.125 mmol
millimoles HCl before reaction ---> (0.275 mmol/mL) (21.00 mL) = 5.775 mmol

Reaction takes place in a 1:1 molar ratio:

mmol pyridine after reaction ---> 6.125 − 5.775 = 0.350 mmol
mmol pyridinium ion produced ---> 5.775 mmol

Use the Henderson-Hasselbalch Equation:

pH = 5.23 + log (0.350 / 5.775) <--- note the use of the pKa

pH = 5.23 + (−1.28) = 3.95

Solution to (f):

At the equivalence point, there is no pyridine and 6.125 millimoles of pyridinium ion. The pyridinium ion dissociates as follows:

C5H5NH+ + H2O ⇌ C5H5N + H3O+

Let us write the Ka expression:

  [C5H5N] [H3O+]
Ka = –––––––––––––––
  [C5H5NH+]

The total volume of solution ---> 35.00 mL + 22.27 mL = 57.27 mL

The molarity of the pyridinium ion ---> 6.125 mmol / 57.27 mL = 0.10695 M (note some extra digits)

Calculate the [H3O+]:

  (x) (x)  
5.888 x 10¯6 = –––––– <--- the Ka came from 10¯5.23
  0.10695  

x = 0.00079355 M

pH = −log 0.00079355 = 3.10

Solution to (g):

22.27 mL of HCl solution was required to neutralize all the pyridine.

Than means this ---> 26.00 − 22.27 = 3.73 mL of HCl remains unreacted

(3.73 mL) (0.275 mol/L) = (61.00 mL) (x)

x = 0.016816 M <--- the new HCl concentration

Since HCl is a strong acid, it ionizes 100% and this is the new pH:

pH = −log 0.016816 = 1.77


Example #8: A 0.750 M solution of aniline is prepared and then titrated with 0.450 M nitric acid. Consider the following questions.

(a) What is the pH of the initial solution? What are the conjugate acid and base in the initial solution?
(b) If you have 75.0 mL of the initial aniline solution, what volume of nitric acid must be added to reach the endpoint and what will the pH be at the end point?
(c) Determine the volume of nitric acid you must add to reach the midpoint of the buffer region and the pH at the midpoint of the buffer region.
(d) If you add 5.00 mL of nitric acid when at the midpoint of the buffer region, what is the new pH of the solution?
(e) What mass of anilinium chloride must you add to the original 75.0 mL of aniline to prepare a buffer whose pH is 5.00
?

Discussion:

Aniline is a weak base, which means a Kb value is required for some of the above calculations. Since it is not provided in the problem, we look it up and find several values, of which 4.0 x 10¯10 seems to be mentioned the most often. Let's go with it.

Solution to (a):

1) This is a standard Kb calculation. Let us write the chemical equation and the Kb expression:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH¯

  [C6H5NH3+] [OH¯]
Kb = –––––––––––––––
  [C6H5NH2]

2) Insert values and solve:

  (x) (x)
4.0 x 10¯10 = –––––––
  0.750

x = 1.732 x 10¯5 M

3) Continue on to the pH:

pOH = −log 1.732 x 10¯5 = 4.76

pH = 14 − 4.76 = 9.24

4) The conjugate acid is the anilinium ion (C6H5NH3+) and the base is aniline (C6H5NH2). Note they they differ by one proton and that the acid has the proton.

Solution to (b):

1) Write the chemical equation for neutralization:

HNO3 + C6H5NH2 ---> C6H5NH3+ + NO3¯

The key point is the 1:1 molar ratio between the two reactants.

2) Determine millimoles, then volume (in mL) of acid required:

moles aniline ---> (0.750 mmol/L) (75.0 mL) = 56.25 mmol

56.25 mmol of nitric acid required

volume nitric acid ---> 56.25 mmol / 0.450 mmol/mL = 125 mL

3) We will need the concentration of the anilinium ion for the pH calculation:

56.25 mmol / 200. mL = 0.28125 M (I'll keep all digits and round off at the pH

4) The anilinium is the salt of a weak base. It will be an acid in solution. Write the hydrolysis equation and the Ka expression:

C6H5NH3+ + H2O ⇌ C6H5NH2 + H3O+

  [C6H5NH2] [H3O+]
Ka = –––––––––––––––
  [C6H5NH3+]

5) We need the Ka of the anilinium ion:

KaKb = Kw

(Ka) (4.0 x 10¯10) = 1.00 x 10¯14

Ka = 2.5 x 10¯5

6) Use the Ka expression to get the [H3O+], then get the pH:

  (x) (x)
2.5 x 10¯5 = ––––––––
  0.28125

x = 0.00265165 M

pH = −log 0.00265165 = 2.58

Solution to (c):

1) The mid-pont of the buffer region is also known as the half-equivalence point.

28.125 mmol of HNO3 is required to reach the mid-point of the buffer region. This comes from 56.25 / 2.

volume needed ---> 28.125 mmol / 0.450 mmol / mL = 62.5 mL

You could also have done this: 125 mL / 2 = 62.5 mL

2) The pH at the half-equivalence point:

pH = pKa

pH = −log 2.5 x 10¯5 = 4.60

Solution to (d):

1) millimoles of nitric acid added:

(0.450 mmol/mL) (5.00 mL) = 2.25 mmol

2) Millimoles of aniline and anilinium that remain after reaction:

aniline ---> 28.125 − 2.25 = 25.875 mmol
anilinium ---> 28.125 + 2.25 = 30.375 mmol

3) This is a buffer solution. Deploy the Henderson-Hasselbalch!!

  [base]
pH = pKa + log –––––
  [acid]

  25.875
pH = 4.60 + log –––––
  30.375

pH = 4.60 + (−0.070)

pH = 4.53

Solution to (e):

1) Addition of anilinium to the aniline solution will create a buffer. A comment first, then on to the Henderson-Hasselbalch:

I will do this calculation using millimoles rather than molarity. This is because the millimoles of aniline and anilinium are in the same ratio as they would be using molarities.

2) The Henderson-Hasselbalch:

  56.25
5.00 = 4.60 + log –––––
  x

  56.25
0.40 = log –––––
  x

  56.25
2.512 = –––––
  x
x = 22.3925 mmol of C6H5NH3Cl

3) Calculate grams:

(0.0223925) (129.5892 g/mol) = 2.9 g (to two sig figs)

Example #9: 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)

  (x) (x)
2.4 x 10¯5 = –––––
  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:

  [base]
pH = pKa + log –––––
  [acid]

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:

  0.00370
pH = 4.620 + log –––––––
  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 #10: 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:

  [HX] [OH¯]
Kb = –––––––––––
  [X¯]

3) The Kb is determined from the pKa:

pKa = 10.39, therefore pKb = 3.61

Kb = 10¯pKb = 10¯3.61 = 2.4547 x 10¯4

4) Solve for the hydroxide concentration:

  (x) (x)
2.4547 x 10¯4 = –––––
  0.243

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

5) 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¯

  [H3O+] [X¯]
Ka = –––––––––––
  [HX]

  (x) (x)
4.0738 x 10¯11 = ––––––
  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!

  0.0019026
pH = 10.39 + log –––––––––
  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%


Example #11: Calculate the pH at the equivalence point for the titration of 0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of methylamine is 4.4 x 10¯4.

Solution:

1) At the equivalence point, all the methyamine is used up (as well as the added HCl) and the only major chemical species of interest in the solution is methyammonium ion (CH3NH3+). It hydrolyzes and this is the chemical equation describing that:

CH3NH3+ + H2O ⇌ CH3NH2 + H3O+

2) We will use a Ka calculation to determine the pH:

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

3) Notice that no volumes of reactant solutions are given. This would be a problem, but the molarities of the two reactants are equal. This means that the volumes of the two reactant solutions required to reach the equivalence point will be equal. This means the total volume doubles and that the concentration of the product (the methyammonium ion) will be 0.090 M.

4) We require the Ka of the methyammonium ion:

KaKb = Kw

Ka = Kw / Kb

Ka = (1.00 x 10¯14) / (4.4 x 10¯4)

Ka = 2.2727 x 10¯11

5) Solve the Ka expression:

  (x) (x)
2.2727 x 10¯11 = –––––––
  0.090

x = 1.43 x 10¯6 M <--- the [H3O+]

6) Determine the pH:

pH = −log 1.43 x 10¯6

pH = 5.844664

To two sig figs, 5.84


Example #12: A 0.0370 M solution of ammonia was titrated with HCl to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur? Kb of ammonia = 1.77 x 10¯5.

Solution:

1) The ammonia and HCl react in a 1:1 molar ratio:

HCl(aq) + NH3(aq) ---> NH4Cl(aq)

2) Assume 1.00 L of the original solution is titrated. This means two things:

(a) 0.0370 mol of ammonia was converted into 0.0370 mol of ammonium ion.

(b) 1.50 L is the final volume after the titration is finished.

3) Determine the molarity of the ammonium chloride solution:

0.0370 mol / 1.50 L = 0.024667 M

4) Determine the Ka of ammonium ion:

KaKb = Kw

Ka = Kw / Kb

Ka = 1.00 x 10¯14 / 1.77 x 10¯5

Ka = 5.64972 x 10¯10

5) Ammonium ion hydrolyzes as follows:

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

6) Write the Ka expression, put values in, and solve it:

  [H3O+] [NH3]
Ka = –––––––––––
  [NH4+]

  (x) (x)
5.64972 x 10¯10 = ––––––––
  0.024667

x = (5.64972 x 10¯10) (0.024667)

x = 0.00000373312 M

7) Calculate the pH:

pH = −log 0.00000373312 = 5.428

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:

  0.001103
3.72 = pKa + log –––––––––––
  0.002031726

3.72 = pKa + (−0.265)

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

10¯pKa yields the Ka


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