Titration Calculations
There are a number of methods to use when determining the pH of a solution in a titration. Click on each step to see more details.
Step 1: Determine acid/base reaction type
Strong acid/strong base
- \(\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{H_2O}(l)\)
Weak acid/strong base
- \(\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{A^-}(aq)\)
Strong acid/weak base
- \(\mathrm{H^+}(aq) + \mathrm{B}(aq) \rightleftharpoons \mathrm{HB^+}(aq)\)
Step 2: Determine molar changes (use IRF table)
React acid and base (in moles)
Smaller amount is “wiped out”
Step 3: Determine final pH
If “strong” (H+ or OH–) remains
Determine Vfinal
Calculate final [H+] 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 Vfinal
Calculate concentration of “weak” species; [weak]
Use ICE table to find pH
Strong acid/strong base titration
Titrate 25.0 mL of 0.100 M HCl with 0.100 M NaOH. (at 25 °C)
Initial pH
What is the initial pH?
Since we have not added any base, there is no reaction yet. Simply use the pH equation.
Since we are dealing with a strong acid, it will dissociate completely into H+
(strong acids have a very large Ka).
\[\mathrm{HCl}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{Cl^-}(aq)\]
The concentration of HCl is equal to the concentration of H3O+. Note: I am using H+ below for convenience but it is equivalent to H3O+ in this context.
\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H^+}] \\ &= -\log (0.100) \\ &= 1.00 \end{align*}\]
Before equivalence point
What is the pH after 15.0 mL of NaOH is added?
Step 1: Determine acid/base reaction type
- This is a strong acid/strong base titration
- \(\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{H_2O}(l)\)
Step 2: Determine molar changes (use IRF table)
To determine molar changes, convert concentration (M) to moles
\[\begin{align*} n_{\mathrm{H}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~H}^{+} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0150~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00150~\mathrm{mol~OH}^{-} \end{align*}\]
React acid and base (in moles).
| Initial moles | 0.00250 | 0.00150 | |||
| Reaction | H+ | + | OH– | \(\longrightarrow\) | H2O |
| Final moles | 0.00100 | 0 |
Step 3: Determine final pH
A strong species (H+) remains. Find the total volume of solution and convert moles back into molarity (M). Get pH.
\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 15.0~\mathrm{mL} \\ &= 40.0~\mathrm{mL} \\ &= 0.04~\mathrm{L} \\[2ex] [\mathrm{H^+}] &= \dfrac{0.00100~\mathrm{mol}}{\mathrm{0.040~L}}\\[1.5ex] &= 0.025~M \\[2ex] \mathrm{pH} &= -\log [\mathrm{H^+}] \\ &= -\log (0.025) \\ &= 1.60 \end{align*}\]
At equivalence point
What is the pH after 25.0 mL of NaOH is added?
Step 1: Determine acid/base reaction type
- This is a strong acid/strong base titration
- \(\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{H_2O}(l)\)
Step 2: Determine molar changes (use IRF table)
To determine molar changes, convert concentration (M) to moles
\[\begin{align*} n_{\mathrm{H}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~H}^{+} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~OH}^{-} \end{align*}\]
React acid and base (in moles).
| Initial moles | 0.00250 | 0.00250 | |||
| Reaction | H+ | + | OH– | \(\longrightarrow\) | H2O |
| Final moles | 0 | 0 |
Step 3: Determine final pH
Since no strong species (H+ 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 given as (at 25 °C)
\[2\mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq)\quad K_{\mathrm{w}} = 1.00\times 10^{-14}\]
\[\begin{align*} [\mathrm{H_3O^+}] &= [\mathrm{OH^-}] = 1\times 10^{-7} \qquad \mathrm{(at~25~^{\circ}C)}\\[2ex] \mathrm{pH} &= [\mathrm{H_3O^+}] \\ &= -\log (1\times 10^{-7}) \\ &= 7 \end{align*}\]
Beyond the equivalence point
What is the pH after 30.0 mL of NaOH is added?
Step 1: Determine acid/base reaction type
- This is a strong acid/strong base titration
- \(\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \longrightarrow \mathrm{H_2O}(l)\)
Step 2: Determine molar changes (use IRF table)
To determine molar changes, convert concentration (M) to moles
\[\begin{align*} n_{\mathrm{H}^{+}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~H}^{+} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0300~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00300~\mathrm{mol~OH}^{-} \end{align*}\]
React acid and base (in moles).
| Initial moles | 0.00250 | 0.00300 | |||
| Reaction | H+ | + | OH– | \(\longrightarrow\) | H2O |
| 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.
\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 30.0~\mathrm{mL} \\ &= 55.0~\mathrm{mL}\\ &= 0.055~\mathrm{L} \\[2ex] [\mathrm{OH^-}] &= \dfrac{0.0005~\mathrm{mol}}{\mathrm{0.055~L}} \\[1.5ex] &= 0.0091~M \\[2ex] \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\ &= -\log (0.0091) = 2.04 \\[2ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH}\\[1ex] &= 14 - 2.04 \\[1ex] &= 11.96 \end{align*}\]
Weak acid/strong base titration
Titrate 25.0 mL of 0.100 M CH3COOH with 0.100 M NaOH (at 25 °C). Ka(CH3COOH) = 1.80 ×10–5
Initial pH
What is the initial pH?
Since we have not added any base, there is no reaction yet. We have a weak acid so we must resort to an ICE table.
| CH3COOH | + | H2O | \(\rightleftharpoons\) | H3O+ | + | CH3COO– | |
|---|---|---|---|---|---|---|---|
| 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. \[\begin{align*} \mathrm{pH} &= -\log[\mathrm{H_3O^+}] \\ &= -\log (1.3\times 10^{-3}) \\ &= 2.87 \end{align*}\]
Before equivalence point
What is the pH after 15.0 mL of NaOH is added?
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 (use IRF table)
To determine molar changes, convert concentration (M) to moles
\[\begin{align*} n_{\mathrm{HA}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~HA} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0150~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00150~\mathrm{mol~OH}^{-} \end{align*}\]
React acid and base (in moles).
| Initial moles | 0.00250 | 0.00150 | 0 | ||||
| Reaction | HA | + | OH– | \(\longrightarrow\) | 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
\[V_{\mathrm{tot}} = 25.0~\mathrm{mL} + 15.0~\mathrm{mL} = 40.0~\mathrm{mL}\]
\[\begin{align*} [\mathrm{H_3O^+}] &= \dfrac{0.00100~\mathrm{mol~H_3O^+}}{0.040~\mathrm{L}}\\[1.5ex] &= 0.0250~M\\[2ex] [\mathrm{OH^-}] &= \dfrac{0.00150~\mathrm{mol~A^-}}{0.040~\mathrm{L}}\\[1.5ex] &= 0.0375~M \end{align*}\]
\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}} \\ &= -\log(1.80\times 10^{-5}) + \log\dfrac{0.0375}{0.0250} \\ &= 4.92 \end{align*}\]
At equivalence point
What is the pH after 25.0 mL of NaOH is added?
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 (use IRF table)
To determine molar changes, convert concentration (M) to moles
\[\begin{align*} n_{\mathrm{HA}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~HA} \\[2ex] n_{\mathrm{OH}^{-}} &= (0.0250~\mathrm{L})\left(0.100~\mathrm{mol~L}^{-1}\right)\\ &= 0.00250~\mathrm{mol~OH}^{-} \end{align*}\]
React acid and base (in moles).
| Initial moles | 0.00250 | 0.00250 | 0 | ||||
| Reaction | HA | + | OH– | \(\longrightarrow\) | 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).
\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 25.0~\mathrm{mL} \\ &= 50.0~\mathrm{mL} \\ &= 0.0500~\mathrm{L} \\[2ex] [\mathrm{A^-}] &= \dfrac{0.00250~\mathrm{mol}}{0.0500~\mathrm{L}}\\[1.5ex] &= 0.0500~M \end{align*}\]
Set up an ICE table for A–, a conjugate base, reacting with water.
| A– | + | H2O | \(\rightleftharpoons\) | 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, Ka, to the base-ionization constant, Kb, 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.
\[\begin{align*} \dfrac{[\mathrm{OH^-}][\mathrm{HA}]}{[\mathrm{A^-}]} &= K_{\mathrm{b}}\\[1.5ex] \dfrac{x^2}{0.0500-x} &= 5.6\times 10^{-10}\\[1.5ex] \dfrac{x^2}{0.0500} &= 5.6\times 10^{-10} \\[1.5ex] x &= 5.29\times 10^{-6} = [\mathrm{OH^-}] = [\mathrm{HA}]\\[2ex] \dfrac{5.29\times 10^{-6}}{0.0500} \times 100\% &= 0.01\% \end{align*}\]
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?
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 (use 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– | \(\longrightarrow\) | 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.
\[\begin{align*} V_{\mathrm{final}} &= 25.0~\mathrm{mL} + 30.0~\mathrm{mL}\\ &= 55.0~\mathrm{mL}\\ &= 0.055~\mathrm{L}\\[2ex] [\mathrm{OH^-}] &= \dfrac{0.0005~\mathrm{mol}}{\mathrm{0.055~L}}\\[1.5ex] &= 0.0091~M \\[2ex] \mathrm{pOH} &= -\log [\mathrm{OH^-}] \\ &= -\log (0.0091) = 2.04 \\[2ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH}\\ &= 14 - 2.04 \\ &= 11.96 \end{align*}\]
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.
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?
| V of OH– (mL) | nOH– (mol) | H+ remain (mol) | Vtotal (mL) | [H+] (mol L–1) | pH | |
|---|---|---|---|---|---|---|
| 0 | 0 | 0.00250 | 25.0 | 0.100 | 1.00 | |
| 5.0 | 0.00050 | 0.00200 | 30.0 | 0.0667 | 1.18 | |
| 10.0 | 0.00100 | 0.00150 | 35.0 | 0.0429 | 1.36 | |
| 15.0 | 0.00150 | 0.00100 | 40.0 | 0.0250 | 1.60 | |
| 20.0 | 0.00200 | 0.00050 | 45.0 | 0.011 | 1.96 | |
| 25.0 | 0.00250 | 0 | 50.0 | 1.0×10–7 | 7.00 |
| V of OH– (mL) | nOH– (mol) | Excess OH– (mol) | Vtotal (mL) | [OH–] (mol L–1) | pOH | pH |
|---|---|---|---|---|---|---|
| 30.0 | 0.00300 | 0.00050 | 55.0 | 0.0091 | 2.04 | 11.96 |
| 35.0 | 0.00350 | 0.00100 | 60.0 | 0.0167 | 1.778 | 12.22 |
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? Ka(CH3COOH) = 1.80 ×10–5
| V of OH– (mL) | nOH– (mol) | CH3COOH remain (mol) | CH3COO– made (mol) | Vtotal (mL) | pH |
|---|---|---|---|---|---|
| 0 | 0 | 0.00250 | 0 | 25.0 | 2.87 |
| 5.0 | 0.00050 | 0.00200 | 0.00050 | 30.0 | 4.14 |
| 10.0 | 0.00100 | 0.00150 | 0.00100 | 35.0 | 4.56 |
| 15.0 | 0.00150 | 0.00100 | 0.00150 | 40.0 | 4.92 |
| 20.0 | 0.00200 | 0.00050 | 0.00200 | 45.0 | 5.34 |
| 25.0 | 0.00250 | 0 | 0.00250 | 50.0 | 8.72 |
| V of OH– (mL) | nOH– (mol) | Excess OH– (mol) | Vtotal (mL) | [OH–] (mol L–1) | pOH | pH |
|---|---|---|---|---|---|---|
| 30.0 | 0.00300 | 0.00050 | 55.0 | 0.0091 | 2.04 | 11.96 |
| 35.0 | 0.00350 | 0.00100 | 60.0 | 0.0167 | 1.778 | 12.22 |
Timed Assessments
The following TAS videos present problems for solving pH in the following order:
- Initial pH
- pH before equivalence point
- pH at equivalence point
- pH after equivalence point
pOH is also included.
Strong Acid/Strong Base Titration I
Answers
- pH = 0.70; pOH = 13.30
- pH = 0.92; pOH = 13.08
- pH = 7.00; pOH = 7.00
- pH = 12.46; pOH = 1.54
Strong Acid/Strong Base Titration II
Answers
- pH = 0.60; pOH = 13.40
- pH = 0.72; pOH = 13.28
- pH = 7.00; pOH = 7.00
- pH = 12.55; pOH = 1.45
Weak Acid/Strong Base Titration I
Answers
- pH = 2.00; pOH = 12.00
- pH = 3.37; pOH = 10.63
- pH = 8.02; pOH = 5.98
- pH = 12.43; pOH = 1.57
Weak Acid/Strong Base Titration II
Answers
- pH = 1.80; pOH = 12.20
- pH = 3.39; pOH = 10.61
- pH = 8.34; pOH = 5.66
- pH = 12.70; pOH = 1.30