Example 15.7.1: Calculating pH for Titration Solutions: Strong Acid/Strong Base

A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH the titration curve is shown in Figure 15.7.1 (below). Calculate the pH at these volumes of added base solution:

1. 0.00 mL
2. 12.50 mL
3. 25.00 mL
4. 37.50 mL

[reveal-answer q=”996546″]Show Solution[/reveal-answer]
[hidden-answer a=”996546″]

Since HCl is a strong acid, we can assume that all of it dissociates. The initial concentration of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] is [latex]{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}_{0}=0.100M[/latex]. When the base solution is added, it also dissociates completely, providing OH⁻ ions. The [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] and OH⁻ ions neutralize each other, so only those of the two that were in excess remain, and their concentration determines the pH. Thus, the solution is initially acidic (pH < 7), but eventually all the hydronium ions present from the original acid are neutralized, and the solution becomes neutral. As more base is added, the solution turns basic.

The total initial amount of the hydronium ions is:

[latex]\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}={\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}_{0}\times \text{0.02500 L}=\text{0.002500 mol}[/latex]

Once X mL of the 0.100-M base solution is added, the number of moles of the OH⁻ ions introduced is:

[latex]\text{n}{\left({\text{OH}}^{-}\right)}_{0}=0.100M\times \text{X mL}\times \left(\frac{\text{1 L}}{\text{1000 mL}}\right)[/latex]

The total volume becomes: [latex]V=\left(\text{25.00 mL}+\text{X mL}\right)\left(\frac{\text{1 L}}{\text{1000 mL}}\right)[/latex]

The number of moles of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] becomes:

[latex]\text{n}\left({\text{H}}^{\text{+}}\right)=\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{-}\right)}_{0}=\text{0.002500 mol}-0.100M\times \text{X mL}\times \left(\frac{\text{1 L}}{\text{1000 mL}}\right)[/latex]

The concentration of [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] is:

[latex]\begin{array}{l}\\ \\ \left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\dfrac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}=\dfrac{\text{0.002500 mol}-0.100M\times \text{X mL}\times \left(\frac{\text{1 L}}{\text{1000 mL}}\right)}{\left(\text{25.00 mL}+\text{X mL}\right)\left(\frac{\text{1 L}}{\text{1000 mL}}\right)}\\ =\dfrac{\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)-0.100M\times \text{X mL}}{\text{25.00 mL}+\text{X mL}}\end{array}[/latex]
[latex]\text{pH}=-\text{log}\left(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\right)[/latex]

The preceding calculations work if [latex]\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{-}\right)}_{0}>0[/latex] and so n(H⁺) > 0. When [latex]\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}=\text{n}{\left({\text{OH}}^{-}\right)}_{0}[/latex], the [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] ions from the acid and the OH⁻ ions from the base mutually neutralize. At this point, the only hydronium ions left are those from the autoionization of water, and there are no OH⁻ particles to neutralize them. Therefore, in this case:

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\left[{\text{OH}}^{-}\right],\left[{\text{H}}_{3}{\text{O}}^{+}\right]={K}_{\text{w}}=1.0\times {10}^{\text{-14}};\left[{\text{H}}_{3}{\text{O}}^{+}\right]=1.0\times {10}^{\text{-7}}[/latex]
[latex]\text{pH}=-\text{log}\left(1.0\times {10}^{\text{-7}}\right)=7.00[/latex]

Finally, when [latex]\text{n}{\left({\text{OH}}^{-}\right)}_{0}>\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}[/latex], there are not enough [latex]{\text{H}}_{3}{\text{O}}^{\text{+}}[/latex] ions to neutralize all the OH⁻ ions, and instead of [latex]\text{n}\left({\text{H}}^{\text{+}}\right)=\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{-}\right)}_{0}[/latex], we calculate: [latex]\text{n}\left({\text{OH}}^{-}\right)=\text{n}{\left({\text{OH}}^{-}\right)}_{0}-\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}[/latex]

In this case:

[latex]\begin{array}{l}\\ \\ \left[{\text{OH}}^{-}\right]=\frac{\text{n}\left({\text{OH}}^{-}\right)}{V}=\frac{0.100M\times \text{X mL}\times \left(\frac{\text{1 L}}{\text{1000 mL}}\right)-\text{0.002500 mol}}{\left(\text{25.00 mL}+\text{X mL}\right)\left(\frac{\text{1 L}}{\text{1000 mL}}\right)}\\ =\frac{0.100M\times \text{X mL}-\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)}{\text{25.00 mL}+\text{X mL}}\end{array}[/latex]
[latex]\text{pH}=14-\text{pOH}=14+\text{log}\left(\left[{\text{OH}}^{-}\right]\right)[/latex]

Let us now consider the four specific cases presented in this problem:

1. X = 0 mL
   [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\frac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}=\frac{\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)}{\text{25.00 mL}}=0.1M[/latex]pH = −log(0.100) = 1.000
2. X = 12.50 mL
   [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\frac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}=\frac{\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)-0.100M\times \text{12.50 mL}}{\text{25.00 mL}+\text{12.50 mL}}=0.0333M[/latex]pH = −log(0.0333) = 1.477
3. X = 25.00 mL
   Since the volumes and concentrations of the acid and base solutions are the same: [latex]\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}=\text{n}{\left({\text{OH}}^{-}\right)}_{0}[/latex], and pH = 7.000, as described earlier.
4. X = 37.50 mL
   In this case: [latex]\text{n}{\left({\text{OH}}^{-}\right)}_{0}>\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}[/latex][latex]\left[{\text{OH}}^{-}\right]=\frac{\text{n}\left({\text{OH}}^{-}\right)}{V}=\frac{0.100M\times \text{35.70 mL}-\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)}{\text{25.00 mL}+\text{37.50 mL}}=0.0200M[/latex]pH = 14 − pOH = 14 + log([OH⁻]) = 14 + log(0.0200) = 12.30

[/hidden-answer]

Check Your Learning

Example 15.7.2: Titration of a Weak Acid with a Strong Base

The titration curve shown in Figure 15.7.4 is for the titration of 25.00 mL of 0.100 M CH₃CO₂H with 0.100 M NaOH. The reaction can be represented as [latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H}+{\text{OH}}^{-}\longrightarrow {\text{CH}}_{3}{\text{CO}}_{2}{}^{-}+{\text{H}}_{2}\text{O}[/latex].

1. What is the initial pH before any amount of the NaOH solution has been added? K_a = 1.8 [latex]\times[/latex] 10⁻⁵ for CH₃CO₂H.
2. Find the pH after 25.00 mL of the NaOH solution have been added.
3. Find the pH after 12.50 mL of the NaOH solution has been added.
4. Find the pH after 37.50 mL of the NaOH solution has been added.

[reveal-answer q=”378310″]Show Solution[/reveal-answer]
[hidden-answer a=”378310″]

1. Assuming that the dissociated amount is small compared to 0.100 M, we find that:
   [latex]{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\approx \frac{{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}^{\text{2}}}{{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}_{0}}[/latex], and [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\sqrt{{K}_{a}\times \left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=\sqrt{1.8\times {10}^{-5}\times 0.100}=1.3\times {10}^{-3}[/latex][latex]\text{pH}=-\text{log}\left(1.3\times {10}^{-3}\right)=2.87[/latex]
2. After 25.00 mL of NaOH are added, the number of moles of NaOH and CH₃CO₂H are equal because the amounts of the solutions and their concentrations are the same. All of the CH₃CO₂H has been converted to [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex]. The concentration of the [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex] ion is:[latex]\frac{\text{0.00250 mol}}{\text{0.0500 L}}=\text{0.0500 M}{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex]The equilibrium that must be focused on now is the basicity equilibrium for [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}:[/latex][latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]so we must determine K_b for the base by using the ion product constant for water:[latex]{K}_{\text{b}}=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}[/latex][latex]{K}_{\text{a}}=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]},\text{so}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}=\frac{\left[{\text{H}}^{\text{+}}\right]}{{K}_{\text{a}}}[/latex].Since K_w = [H⁺][OH⁻]:[latex]{K}_{\text{b}}=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]}{{K}_{\text{a}}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=5.6\times {10}^{-10}[/latex]Let us denote the concentration of each of the products of this reaction, CH₃CO₂H and OH⁻, as x. Using the assumption that x is small compared to 0.0500 M, [latex]{K}_{\text{b}}=\frac{{x}^{\text{2}}}{0.0500M}[/latex], and then:[latex]x=\left[{\text{OH}}^{-}\right]=5.3\times {10}^{-6}[/latex]
   [latex]\text{pOH}=-\text{log}\left(5.3\times {10}^{-6}\right)=5.28[/latex]
   [latex]\text{pH}=14.00 - 5.28=8.72[/latex]Note that the pH at the equivalence point of this titration is significantly greater than 7.
3. In (1), 25.00 mL of the NaOH solution was added, and so practically all the CH₃CO₂H was converted into [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex]. In this case, only 12.50 mL of the base solution has been introduced, and so only half of all the CH₃CO₂H is converted into [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex]. The total initial number of moles of CH₃CO₂H is 0.02500L [latex]\times[/latex] 0.100 M = 0.00250 mol, and so after adding the NaOH, the numbers of moles of CH₃CO₂H and [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex] are both approximately equal to [latex]\frac{\text{0.00250 mol}}{2}=\text{0.00125 mol}[/latex], and their concentrations are the same.Since the amount of the added base is smaller than the original amount of the acid, the equivalence point has not been reached, the solution remains a buffer, and we can use the Henderson-Hasselbalch equation:[latex]\text{pH}=p{K}_{\text{a}}+\text{log}\frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]}=-\text{log}\left({K}_{\text{a}}\right)+\text{log}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=-\text{log}\left(1.8\times {10}^{-5}\right)+\text{log}\left(1\right)[/latex](as the concentrations of [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex] and CH₃CO₂H are the same)Thus:[latex]\text{pH}=-\text{log}\left(1.8\times {10}^{-5}\right)=4.74[/latex](the pH = the pK_a at the halfway point in a titration of a weak acid)
4. After 37.50 mL of NaOH is added, the amount of NaOH is 0.03750 L [latex]\times[/latex] 0.100 M = 0.003750 mol NaOH. Since this is past the equivalence point, the excess hydroxide ions will make the solution basic, and we can again use stoichiometric calculations to determine the pH: [latex]\left[{\text{OH}}^{-}\right]=\frac{\left(\text{0.003750 mol}-\text{0.00250 mol}\right)}{\text{0.06250 L}}=2.00\times {10}^{-2}M[/latex] So: [latex]\text{pOH}=-\text{log}\left(2.00\times {10}^{-2}\right)=\text{1.70, and pH}=14.00 - 1.70=12.30[/latex] Note that this result is the same as for the strong acid-strong base titration example provided, since the amount of the strong base added moves the solution past the equivalence point.

[/hidden-answer]

Check Your Learning