Titration Calculations

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 K_a).

\[\mathrm{HCl}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{Cl^-}(aq)\]

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).

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).

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).

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.

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

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.

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).

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!

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