Example 13.21: Calculating pH for Titration Solutions: Strong Acid+Strong Base
A titration is carried out for
25.00 mL of
0.100 M
\(\ce{HCl}\) (strong acid) with
0.100 M of a strong base
\(\ce{NaOH}\) the titration curve is shown in
Figure 21. Calculate the pH at these volumes of added base solution:
(a)
0.00 mL
(b)
12.50 mL
(c)
25.00 mL
(d)
37.50 mL
Solution
\(c_{\mathrm{\ce{HCl}}}\) \(= 0.100\ \mathrm{M}\)
\(c_{\mathrm{\ce{NaOH}}}\) \(= 0.100\ \mathrm{M}\)
\(V_{\mathrm{\ce{HCl}}}\) \(= 25.00\ \mathrm{mL}\)
\(V_{\mathrm{\ce{NaOH},a}}\) \(= 0.00\)
\(V_{\mathrm{\ce{NaOH},b}}\) \(= 12.50\ \mathrm{mL}\)
\(V_{\mathrm{\ce{NaOH},c}}\) \(= 25.00\ \mathrm{mL}\)
\(V_{\mathrm{\ce{NaOH},d}}\) \(= 37.50\ \mathrm{mL}\)
\(\mathrm{pH}_{\mathrm{a}}\) = ?
\(\mathrm{pH}_{\mathrm{b}}\) = ?
\(\mathrm{pH}_{\mathrm{c}}\) = ?
\(\mathrm{pH}_{\mathrm{d}}\) = ?
Part (a) Determining the pH before any
\(\ce{NaOH}\) is added is straightforward. We are looking at a strong acid with known concentration, so we know the concentration of
\(\ce{H3O+}\) and can calculate the pH
\(\mathrm{pH}_{\mathrm{a}}\) \(= -\mathrm{log}(\dfrac{c_{\mathrm{\ce{HCl}}}}{1\ \mathrm{M}})\)
\(\ \ \ =-\mathrm{log}(\dfrac{0.100\ \mathrm{M}}{1\ \mathrm{M}})\)
\(\ \ \ =-\mathrm{log}(0.1000)\)
\(\ \ \ =1.00\)
Plan for parts (b), (c), and (d)
The
\(\ce{HCl}\) solution provides
\(\ce{H3O+}\), and the added
\(\ce{NaOH}\) provides
\(\ce{OH-}\). The
\(\ce{H3O+}\) and
\(\ce{OH-}\) ions neutralize each other; the species that was in excess remains, and its concentration determines the pH. For each point in the titration, we have to determine the amount of
\(\ce{H3O+}\) and
\(\ce{OH-}\) present before neutralization to figure out what remains after neutralization. At first,
\(\ce{H3O+}\) will be in excess and the solution will be acidic. Eventually, sufficient
\(\ce{OH-}\) is added to neutralize all of the
\(\ce{H3O+}\) with
\(\ce{OH-}\) remaining, and the solution will be basic.
\(\ce{H3O+}\)\(\ce{ + }\)\(\ce{OH-}\)\(\ce{->}\)\(\ce{2H2O}\)\(\ce{ }\)
The amount of
\(\ce{H3O+}\) provided by the
\(\ce{HCl}\) solution is the same for all points in the titration (because we are not adding any more
\(\ce{HCl}\) solution during the titration):
\(n_{\mathrm{\ce{H3O+},initial}}\) \(= c_{\mathrm{\ce{HCl}}} \cdot V_{\mathrm{\ce{HCl}}}\)
\(\ \ \ =0.100\ \mathrm{M} \cdot 25.00\ \mathrm{mL}\)
\(\ \ \ =2.50\times 10^{-3}\ \mathrm{mol}\)
The amount of
\(\ce{OH-}\) provided by the
\(\ce{NaOH}\) solution and the total volume of the solution changes as the titration proceeds, so we have to calculate these separately for parts (b), (c) and (d)
Part (b)
Because strong bases dissociate completely, the amount of
\(\ce{OH-}\) ions provided by the
\(\ce{NaOH}\) solution is equal to the amount of
\(\ce{NaOH}\) added.
\(n_{\mathrm{\ce{OH-}\ initial\ b}}\) \(= c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH},b}}\)
\(\ \ \ =0.100\ \mathrm{M} \cdot 12.50\ \mathrm{mL}\)
\(\ \ \ =1.25\times 10^{-3}\ \mathrm{mol}\)
In the neutralization reaction,
\(\ce{H3O+}\) and
\(\ce{OH-}\) react in a ratio of 1:1, so the limiting reactant is simply the one present at lower amount. However, we will also show this by calculating how much
\(\ce{H3O+}\) and
\(\ce{OH-}\) remain after neutralization
\(n_{\mathrm{\ce{->},b}}\) \(= \mathrm{min}(\dfrac{n_{\mathrm{\ce{H3O+},initial}}}{1}, \dfrac{n_{\mathrm{\ce{OH-}\ initial\ b}}}{1})\)
\(\ \ \ =\mathrm{min}(\dfrac{2.50\times 10^{-3}\ \mathrm{mol}}{1}, \dfrac{1.25\times 10^{-3}\ \mathrm{mol}}{1})\)
\(\ \ \ =\mathrm{min}(2.500\times 10^{-3}\ \mathrm{mol}, 1.250\times 10^{-3}\ \mathrm{mol})\)
\(\ \ \ =1.25\times 10^{-3}\ \mathrm{mol}\)
\(n_{\mathrm{\ce{H3O+}\ remaining\ b}}\) \(= n_{\mathrm{\ce{H3O+},initial}} - n_{\mathrm{\ce{->},b}}\)
\(\ \ \ =2.50\times 10^{-3}\ \mathrm{mol} - 1.25\times 10^{-3}\ \mathrm{mol}\)
\(\ \ \ =1.25\times 10^{-3}\ \mathrm{mol}\)
\(n_{\mathrm{\ce{OH-}\ remaining\ b}}\) \(= n_{\mathrm{\ce{OH-}\ initial\ b}} - n_{\mathrm{\ce{->},b}}\)
\(\ \ \ =1.25\times 10^{-3}\ \mathrm{mol} - 1.25\times 10^{-3}\ \mathrm{mol}\)
\(\ \ \ =0.00000\)
So
\(\ce{OH-}\) is used up and some
\(\ce{H3O+}\) remains. For calculating the pH, we need a concentration, not an amount, so we will calculate the volume of the mixture at point (b), then calculate the concentration and then the pH:
\(V_{\mathrm{total\ b}}\) \(= V_{\mathrm{\ce{HCl}}} + V_{\mathrm{\ce{NaOH},b}}\)
\(\ \ \ =25.00\ \mathrm{mL} + 12.50\ \mathrm{mL}\)
\(\ \ \ =37.50\ \mathrm{mL}\)
\(c_{\mathrm{\ce{H3O+}\ remaining\ b}}\) \(= \dfrac{n_{\mathrm{\ce{H3O+}\ remaining\ b}}}{V_{\mathrm{total\ b}}}\)
\(\ \ \ =\dfrac{1.25\times 10^{-3}\ \mathrm{mol}}{37.50\ \mathrm{mL}}\)
\(\ \ \ =0.0333\ \mathrm{M}\)
\(\mathrm{pH}_{\mathrm{b}}\) \(= -\mathrm{log}(\dfrac{c_{\mathrm{\ce{H3O+}\ remaining\ b}}}{1\ \mathrm{M}})\)
\(\ \ \ =-\mathrm{log}(\dfrac{0.0333\ \mathrm{M}}{1\ \mathrm{M}})\)
\(\ \ \ =-\mathrm{log}(0.03333)\)
\(\ \ \ =1.48\)
Part (c)
\(n_{\mathrm{\ce{OH-}\ initial\ c}}\) \(= c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH},c}}\)
\(\ \ \ =0.100\ \mathrm{M} \cdot 25.00\ \mathrm{mL}\)
\(\ \ \ =2.50\times 10^{-3}\ \mathrm{mol}\)
\(n_{\mathrm{\ce{->},c}}\) \(= \mathrm{min}(\dfrac{n_{\mathrm{\ce{H3O+},initial}}}{1}, \dfrac{n_{\mathrm{\ce{OH-}\ initial\ c}}}{1})\)
\(\ \ \ =\mathrm{min}(\dfrac{2.50\times 10^{-3}\ \mathrm{mol}}{1}, \dfrac{2.50\times 10^{-3}\ \mathrm{mol}}{1})\)
\(\ \ \ =\mathrm{min}(2.500\times 10^{-3}\ \mathrm{mol}, 2.500\times 10^{-3}\ \mathrm{mol})\)
\(\ \ \ =2.50\times 10^{-3}\ \mathrm{mol}\)
\(n_{\mathrm{\ce{H3O+}\ remaining\ c}}\) \(= n_{\mathrm{\ce{H3O+},initial}} - n_{\mathrm{\ce{->},c}}\)
\(\ \ \ =2.50\times 10^{-3}\ \mathrm{mol} - 2.50\times 10^{-3}\ \mathrm{mol}\)
\(\ \ \ =0.00000\)
\(n_{\mathrm{\ce{OH-}\ remaining\ c}}\) \(= n_{\mathrm{\ce{OH-}\ initial\ c}} - n_{\mathrm{\ce{->},c}}\)
\(\ \ \ =2.50\times 10^{-3}\ \mathrm{mol} - 2.50\times 10^{-3}\ \mathrm{mol}\)
\(\ \ \ =0.00000\)
At point (c), the initial amounts of
\(\ce{H3O+}\) and
\(\ce{OH-}\) are equal. After neutralization, it is as if we had not added any strong acid or base to the solution. Because of the autodissociation of water, however, the
\(\ce{H3O+}\) concentration is not zero.
\(K_{\mathrm{w}}\) \(= 1.00\times 10^{-14}\)
\(\mathrm{pH}_{\mathrm{c}}\) \(= -\mathrm{log}(\sqrt{K_{\mathrm{w}}})\)
\(\ \ \ =-\mathrm{log}(\sqrt{1.00\times 10^{-14}})\)
\(\ \ \ =-\mathrm{log}(1.0000\times 10^{-7})\)
\(\ \ \ =7.000\)
Part (d)
\(n_{\mathrm{\ce{OH-}\ initial\ d}}\) \(= c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH},d}}\)
\(\ \ \ =0.100\ \mathrm{M} \cdot 37.50\ \mathrm{mL}\)
\(\ \ \ =3.75\times 10^{-3}\ \mathrm{mol}\)
\(n_{\mathrm{\ce{->},d}}\) \(= \mathrm{min}(\dfrac{n_{\mathrm{\ce{H3O+},initial}}}{1}, \dfrac{n_{\mathrm{\ce{OH-}\ initial\ d}}}{1})\)
\(\ \ \ =\mathrm{min}(\dfrac{2.50\times 10^{-3}\ \mathrm{mol}}{1}, \dfrac{3.75\times 10^{-3}\ \mathrm{mol}}{1})\)
\(\ \ \ =\mathrm{min}(2.500\times 10^{-3}\ \mathrm{mol}, 3.750\times 10^{-3}\ \mathrm{mol})\)
\(\ \ \ =2.50\times 10^{-3}\ \mathrm{mol}\)
\(n_{\mathrm{\ce{H3O+}\ remaining\ d}}\) \(= n_{\mathrm{\ce{H3O+},initial}} - n_{\mathrm{\ce{->},d}}\)
\(\ \ \ =2.50\times 10^{-3}\ \mathrm{mol} - 2.50\times 10^{-3}\ \mathrm{mol}\)
\(\ \ \ =0.00000\)
\(n_{\mathrm{\ce{OH-}\ remaining\ d}}\) \(= n_{\mathrm{\ce{OH-}\ initial\ d}} - n_{\mathrm{\ce{->},d}}\)
\(\ \ \ =3.75\times 10^{-3}\ \mathrm{mol} - 2.50\times 10^{-3}\ \mathrm{mol}\)
\(\ \ \ =1.25\times 10^{-3}\ \mathrm{mol}\)
At point (d),
\(\ce{OH-}\) is in excess over
\(\ce{H3O+}\), i.e. after neutralization, some
\(\ce{OH-}\) will remain and determine the pH.
\(V_{\mathrm{total\ d}}\) \(= V_{\mathrm{\ce{HCl}}} + V_{\mathrm{\ce{NaOH},d}}\)
\(\ \ \ =25.00\ \mathrm{mL} + 37.50\ \mathrm{mL}\)
\(\ \ \ =62.50\ \mathrm{mL}\)
\(c_{\mathrm{\ce{OH-}\ remaining\ d}}\) \(= \dfrac{n_{\mathrm{\ce{OH-}\ remaining\ d}}}{V_{\mathrm{total\ d}}}\)
\(\ \ \ =\dfrac{1.25\times 10^{-3}\ \mathrm{mol}}{62.50\ \mathrm{mL}}\)
\(\ \ \ =0.0200\ \mathrm{M}\)
Now we can first calculate the
\(\ce{pOH}\), and from that the pH
\(\mathrm{pOH}_{\mathrm{d}}\) \(= -\mathrm{log}(\dfrac{c_{\mathrm{\ce{OH-}\ remaining\ d}}}{1\ \mathrm{M}})\)
\(\ \ \ =-\mathrm{log}(\dfrac{0.0200\ \mathrm{M}}{1\ \mathrm{M}})\)
\(\ \ \ =-\mathrm{log}(0.02000)\)
\(\ \ \ =1.70\)
\(\mathrm{pH}_{\mathrm{d}}\) \(= 14.00 - \mathrm{pOH}_{\mathrm{d}}\)
\(\ \ \ =14.00 - 1.70\)
\(\ \ \ =12.30\)
To summarize, the initial pH is
1.00, which then goes up to
1.48,
7.00 and
12.30 in the course of the titration.
