5.11 Calculating pH During Titration

When performing a titration, the pH of the solution is monitored via a pH probe, indicator, or some other form. Plotting the pH of solution vs. the amount of titrant added gives the characteristic titration curves (covered above).

We can also calculate the pH of the solution at any point of a titration and reproduce the titration curve plot without ever performing an actual titration!

These types of titration problems involving monoprotic acids can be broken down into four distinct pieces:

1. Initial pH
2. pH before equivalence point
3. pH at the equivalence point
4. pH beyond the equivalence point

Strong Acid/Strong Base Curve

Weak Acid/Strong Base Curve

The steps are outlined below.

Step 1: Determine acid/base reaction type

• Strong acid/strong base

  • \(\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{H_2O}(l)\)
  • \(\mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{2H_2O}(l)\)

• Weak acid/strong base

  • \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)

• Strong acid/weak base

  • \(\mathrm{H^+}(aq) + \mathrm{B}(aq) \rightleftharpoons \mathrm{HB^+}(aq)\)
  • \(\mathrm{H_3O^+}(aq) + \mathrm{B}(aq) \rightleftharpoons \mathrm{HB^+}(aq) + \mathrm{H_2O}(l)\)

Step 2: Determine molar changes (IRF table)

• React acid and base (in moles) if applicable

• Smaller amount is “wiped out”

Step 3: Determine final pH

• If “strong” (H₃O⁺ or OH^–) remains

  • Determine V_final

  • Calculate final [H₃O⁺] or [OH^–]

  • \(\mathrm{pH} = -\log{[\mathrm{H_3O^+}]}\) or \(\mathrm{pOH = -\log[\mathrm{OH^-}]}\)

• If “weak” pair present (HA/A^– or B/HB⁺), use Henderson-Hasselbalch

  • \(\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}}\)

  • \(\mathrm{pOH} = \mathrm{p}K_{\mathrm{b}} + \log\dfrac{[\mathrm{HB^+}]}{\mathrm{[B]}}\)

• If only one “weak” species present (HA, A^–, B, or HB⁺)

  • Determine V_final

  • Calculate concentration of “weak” species

  • Use ICE table to find pH

5.11.1 Strong Acid/Strong Base

Titrate 25.0 mL of 0.100 M HCl with 0.100 M NaOH. (at 25 °C).

Initial pH

What is the initial pH of solution?

Solution

Step 1: Recognize that this is a strong acid/strong base reaction.

Step 2: No molar changes occur since no strong base has been added yet.

Step 3: Determine final pH. Only a “strong” species is present (H₃O⁺).

Given that the initial solution contains a strong acid, recognize that all the strong acid reacts with water to produce hydronium and conjugate base.

\[\mathrm{HCl}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{Cl^-}(aq)\] Therefore, [H₃O⁺] = 0.100 M.

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log (0.100) \\ &= 1.00 \end{align*}\]

pH before equivalence point

What is the pH after 15.0 mL of NaOH is added?

Solution

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{H_3O}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0150~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00150~\mathrm{mol} \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00150
Reaction	H3O+	+	OH–	→	2H2O
Final moles	0.00100		0

Step 3: Determine final pH. Only a “strong” species is present (H₃O⁺).

A strong species (H₃O⁺) remains. Find the total volume of solution and convert moles back into molarity (M). Determine pH.

\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 15.0~\mathrm{mL} \\ &= 40.0~\mathrm{mL} \\ &= 0.04~\mathrm{L} \\[2ex] [\mathrm{H_3O^+}] &= \dfrac{0.00100~\mathrm{mol}}{\mathrm{0.040~L}}\\[1.5ex] &= 0.025~M \\[2ex] \mathrm{pH} &= -\log [\mathrm{H_3O^+}] \\ &= -\log (0.025) \\ &= 1.60 \end{align*}\]

pH at the equivalence point

What is the pH at the equivalence point?

Solution

Step 1: Determine acid/base reaction type

• This is a strong acid/strong base titration
• \(\mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{2H_2O}(l)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{H_3O}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \end{align*}\]

Note that a volume of added base was not given. However, we implement the definition of the equivalence point (n_acid = n_base) and recognize that there must be an equal number of moles of base as there is acid! Therefore,

\[\begin{align*} n_{\mathrm{OH^-}} &= 0.00250~\mathrm{mol} \\[2ex] \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00250
Reaction	H3O+	+	OH–	→	2H2O
Final moles	0		0

Step 3: Determine final pH

Since no strong species (H₃O⁺ or OH^–) remains, the solution is neutral. All acid and base was converted to water. The only hydronium and hydroxide ions in solution come from pure water via the autoionization of water (at 25 °C)

\[2\mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq)\quad K_{\mathrm{w}} = 1.00\times 10^{-14}\]

\[\begin{align*} [\mathrm{H_3O^+}] &= [\mathrm{OH^-}] = 1\times 10^{-7}\\[2ex] \mathrm{pH} &= [\mathrm{H_3O^+}] \\ &= -\log (1\times 10^{-7}) \\ &= 7 \end{align*}\]

pH beyond the equivalence point

What is the pH after 30.0 mL of NaOH is added?

Solution

Step 1: Determine acid/base reaction type

• This is a strong acid/strong base titration
• \(\mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{2H_2O}(l)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{H_3O}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0300~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00300~\mathrm{mol} \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00300
Reaction	H3O+	+	OH–	→	2H2O
Final moles	0		0.0005

Step 3: Determine final pH

A strong species (OH^–) remains. Find the total volume of solution and convert moles back into molarity (M). Get pH.

\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 30.0~\mathrm{mL} \\ &= 55.0~\mathrm{mL}\\ &= 0.055~\mathrm{L} \\[2ex] [\mathrm{OH^-}] &= \dfrac{0.0005~\mathrm{mol}}{\mathrm{0.055~L}} \\[1.5ex] &= 0.0091~M \\[2ex] \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\ &= -\log (0.0091) = 2.04 \\[2ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH}\\[1ex] &= 14 - 2.04 \\[1ex] &= 11.96 \end{align*}\]

5.11.2 Weak Acid/Strong Base

Titrate 25.0 mL of 0.100 M CH₃COOH with 0.100 M NaOH (at 25 °C).
K_a(CH₃OOH) = 1.80 × 10^–5

Initial pH

What is the initial pH of solution?

Solution

Step 1: Determine acid/base reaction type

• This is a weak acid/strong base titration

Step 2: No molar changes occur since no strong base has been added yet.

Step 3: Determine final pH

This is a typical equilibrium problem via an ICE table.

	CH3OOH	+	H2O	⇌	H3O+	+	CH3OO–
I	0.100				≈0		0
C	–x				+x		+x
E	0.100 - x				x		x

Solve the equilibrium expression.

\[\begin{align*} \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{\mathrm{[HA]}} &= K_{\mathrm{a}} \\[1.5ex] \dfrac{x^2}{0.100-x} &= 1.80\times 10^{-5}\\[1.5ex] \dfrac{x^2}{0.100} &= 1.80\times 10^{-5}\\[1.5ex] x &= 1.34\times 10^{-3}\\[2ex] \dfrac{1.34\times 10^{-3}}{0.100} \times 100\% &= 1.34\% \end{align*}\]

Get pH. \[\begin{align*} \mathrm{pH} &= -\log[\mathrm{H_3O^+}] \\ &= -\log (1.3\times 10^{-3}) \\ &= 2.87 \end{align*}\]

pH before equivalence point

What is the pH after 15.0 mL of NaOH is added?

Solution

Step 1: Determine acid/base reaction type

• This is a weak acid/strong base titration
• \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{HA}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0150~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00150~\mathrm{mol} \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00150				0
Reaction	HA	+	OH–	→	H2O	+	A–
Final moles	0.00100		0				0.00150

Step 3: Determine final pH

If “weak” pair present (HA/A^– or B/HB⁺), determine of this is a buffer. If it is, use Henderson-Hasselbalch to get the pH.

\[\begin{align*} \dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]} = \dfrac{0.00150}{0.00100} = 1.5 \end{align*}\]

or

\[\begin{align*} \dfrac{[\mathrm{HA}]}{[\mathrm{A^-}]} = \dfrac{0.00100}{0.00150} = 0.667 \end{align*}\]

Since the ratio of acid to conjugate base is between 0.1 and 10, we have a buffer and can use Henderson-Hasselbalch!

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}} \\[1ex] &= -\log(1.80\times 10^{-5}) + \log\dfrac{0.00150}{0.00100} \\[1ex] &= 4.92 \end{align*}\]

Why did you use moles??

I used mole quantities here in the Henderson-Hasselbalch equation because converting to concentration would net the same answer. Consider the two examples:

Example 1: Mole quantities

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}} \\[1.5ex] &= -\log(1.80\times 10^{-5}) + \log\dfrac{0.00150}{0.0010} \\[1.5ex] &= 4.92 \end{align*}\]

Example 2: Molar concentration quantities

\[V_{\mathrm{tot}} = 25.0~\mathrm{mL} + 15.0~\mathrm{mL} = 40.0~\mathrm{mL}\]

\[\begin{align*} [\mathrm{H_3O^+}] &= \dfrac{0.00100~\mathrm{mol~H_3O^+}}{0.040~\mathrm{L}}\\[1.5ex] &= 0.0250~M\\[2ex] [\mathrm{OH^-}] &= \dfrac{0.00150~\mathrm{mol~A^-}}{0.040~\mathrm{L}}\\[1.5ex] &= 0.0375~M \end{align*}\]

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}} \\ &= -\log(1.80\times 10^{-5}) + \log\dfrac{0.0375}{0.0250} \\ &= 4.92 \end{align*}\]

pH at equivalence point

What is the pH at the equivalence point?

Solution

Step 1: Determine acid/base reaction type

• This is a weak acid/strong base titration
• \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{HA}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol} \\[2ex] \end{align*}\]

Note that a volume of added base was not given. However, we implement the definition of the equivalence point (n_acid = n_base) and recognize that there must be an equal number of moles of base as there is acid! Therefore,

\[\begin{align*} n_{\mathrm{OH^-}} &= 0.00250~\mathrm{mol} \\[2ex] \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00250				0
Reaction	HA	+	OH–	→	H2O	+	A–
Final moles	0		0				0.00250

Step 3: Determine final pH

Only one “weak” species (A^–) is present. Convert moles back into molarity (M).

First, determine the volume of base added.

\[\begin{align*} V_{\mathrm{NaOH}} &= \dfrac{0.00250~\mathrm{mol}}{0.100~\mathrm{mol~L^{-1}}} \\[1.5ex] &= 0.025~\mathrm{L} \\[1.5ex] &= 25.0~\mathrm{mL} \end{align*}\]

Next, find the total volume of solution and determine the concentration of A^–.

\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 25.0~\mathrm{mL} \\ &= 50.0~\mathrm{mL} \\ &= 0.0500~\mathrm{L} \\[2ex] [\mathrm{A^-}] &= \dfrac{0.00250~\mathrm{mol}}{0.0500~\mathrm{L}}\\[1.5ex] &= 0.0500~M \end{align*}\]

Set up an ICE table for A^–, a conjugate base, reacting with water.

	A–	+	H2O	⇌	OH–	+	HA
I	0.0500				≈0		0
C	–x				+x		+x
E	0.0500 - x				x		x

NOTE: Because we are analyzing a base-ionization reaction (since we have a base reacting with water), you must convert the acid-dissociation constant, K_a, to the base-ionization constant, K_b, of the corresponding conjugate base!

\[\begin{align*} K_{\mathrm{b}} &= \dfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}} \\[1.5ex] &= \dfrac{1.00\times 10^{-14}}{1.80\times 10^{-5}} \\[1.5ex] &= 5.6\times 10^{-10} \end{align*}\]

Set up the equilibrium expression and solve.

\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{HA}]}{[\mathrm{A^-}]} &= K_{\mathrm{b}}\\[1.5ex] \dfrac{x^2}{0.0500-x} &= 5.6\times 10^{-10}\\[1.5ex] \dfrac{x^2}{0.0500} &= 5.6\times 10^{-10} \\[1.5ex] x &= 5.29\times 10^{-6} (0.01\%) \end{align*}\]

Get pH.

\[\begin{align*} \mathrm{pOH} &= -\log[\mathrm{OH^-}] \\ &= -\log (5.29\times 10^{-6})\\ &= 5.28\\[2ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH}\\ &= 14 - 5.28\\ &= 8.72 \end{align*}\]

Beyond the equivalence point

What is the pH after 30.0 mL of NaOH is added?

Solution

Step 1: Determine acid/base reaction type

• This is a weak acid/strong base titration
• \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)

Step 2: Determine molar changes (IRF table)

To determine molar changes, convert concentration (M) to moles

\[\begin{align*} n_{\mathrm{HA}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~HA} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0300~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00300~\mathrm{mol~OH}^{-} \end{align*}\]

React acid and base (in moles).

Initial moles	0.00250		0.00300				0
Reaction	HA	+	OH–	→	H2O	+	A–
Final moles	0		0.0005				0.00250

Step 3: Determine final pH

A strong species (OH^–) remains. Find the total volume of solution and convert moles back into molarity (M). Get pH.

\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 30.0~\mathrm{mL}\\ &= 55.0~\mathrm{mL}\\ &= 0.055~\mathrm{L}\\[2ex] [\mathrm{OH^-}] &= \dfrac{0.0005~\mathrm{mol}}{\mathrm{0.055~L}}\\[1.5ex] &= 0.0091~M \\[2ex] \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\ &= -\log (0.0091) = 2.04 \\[2ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH}\\ &= 14 - 2.04 \\ &= 11.96 \end{align*}\]

5.11.3 Practice

Below are tables for two different titrations (strong acid/strong base and weak acid/strong base). Try to replicate the pH in the last column by using only the data in column 1.

5.11.3.1 Strong acid/strong base titration

Titrate 25.0 mL of 0.100 M HCl with 0.100 M NaOH. (at 25 °C) What is the pH after X mL (from column 1) of NaOH is added?

VOH– (mL)	nOH– (mol)	nH+ remain (mol)	Vtot (mL)	[H+] (M)		pOH	pH
0	0	0.0025	25	0.1		13.00	1.00
5	0.0005	0.002	30	0.0667		12.82	1.18
10	0.001	0.0015	35	0.0429		12.64	1.36
15	0.0015	0.001	40	0.025		12.40	1.60
20	0.002	0.0005	45	0.011		12.04	1.96
25	0.0025	0	50	1.0 ×10–7		7.00	7.00

VOH– (mL)	nOH– (mol)	Excess OH– (mol)	Vtot (mL)	[OH–] (M)		pOH	pH
30	0.003	0.0005	55	0.0091		2.04	11.96
35	0.0035	0.001	60	0.0167		1.78	12.22

5.11.3.2 Weak acid/strong base titration

Titrate 25.0 mL of 0.100 M CH₃COOH with 0.100 M NaOH (at 25 °C). What is the pH after X mL (from column 1) of NaOH is added? K_a(CH₃COOH) = 1.80 ×10^–5

VOH– (mL)	nOH– (mol)	HA remain (mol)	A– made (mol)	Vtot (mL)		pOH	pH
0	0	0.0025	0	25		11.13	2.87
5	0.0005	0.002	0.0005	30		9.86	4.14
10	0.001	0.0015	0.001	35		9.44	4.56
15	0.0015	0.001	0.0015	40		9.08	4.92
20	0.002	0.0005	0.002	45		8.66	5.34
25	0.0025	0	0.0025	50		5.28	8.72

VOH– (mL)	nOH– (mol)	Excess OH– (mol)	Vtot (mL)	[OH–] (M)		pOH	pH
30	0.003	5e-04	55	0.0091		2.04	11.96
35	0.0035	0.001	60	0.0167		1.778	12.22

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