# Titration of a Polyprotic Weak Acid with Sodium Hydroxide

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- 70385

## Out of Class Assignment #4, Problem 2

**Starting with 30 ml of 0.1 M citric acid, calculate the initial pH and the pH at each 5 ml increment of 0.1 M NaOH until you are 10 ml past the last equivalence point. Plot the data and determine whether 99.9% of the citric acid has been neutralized at the last equivalence point. Also calculate the concentration of all species in solution at the second equivalence point.**

It is worth examining this problem in some detail since we have not done anything exactly like it in class. Essentially it consists of the titration of a polyprotic acid using a strong base. Citric acid is a common buffer but is an interesting example because the first two pK_{a} values are fairly close to each other.

If we look in the table we find out that citric acid (H_{3}cit) is a triprotic acid. The following three equilibrium reactions define the system.

\[\begin{align}

&\ce{H3cit + H2O \leftrightarrow H2cit- + H3O+} &&\mathrm{K_{a1} = 7.45\times 10^{-4}}\\

&\ce{H2cit^2- + H2O \leftrightarrow Hcit^2- + H3O+} &&\mathrm{K_{a2} = 1.73\times 10^{-5}}\\

&\ce{Hcit- + H2O \leftrightarrow cit^3- + H3O+} &&\mathrm{K_{a3} = 4.02\times 10^{-7}}

\end{align}\]

Even though the first two K_{a} values are fairly close to each other, we can still use only the K_{a1} expression to solve for the initial pH.

\[\begin{align}

& &&\ce{H3cit}\hspace{25px} + \hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{H2cit-} \hspace{25px} + &&\ce{H3O+} &&\mathrm{K_{a1}}\\

&\ce{Initial} &&0.1 &&0 &&0 && \\

&\ce{Equilibrium} &&0.1 - \ce{x} &&\ce{x} &&\ce{x} && \\

&\ce{Approximation} &&0.1 &&\ce{x} &&\ce{x} &&

\end{align}\]

\[\mathrm{K_{a1}=\dfrac{[H_2cit^-][H_3O^+]}{[H_3cit]}=\dfrac{(x)(x)}{0.1}=7.45\times10^{-4}}\]

\[\mathrm{x = [H_3O^+] = 0.00863 \hspace{60px} pH = 2.06}\]

If we check the approximation, it actually turns out that the value is too high and that we should have used a quadratic if we wanted the exact answer. But the value of 2.06 will suffice for now.

The next step is to consider what happens when we start adding sodium hydroxide to the solution. This will convert H_{3}cit into the other forms, and we can start the process by assuming it will occur in a stepwise manner (in other words, H_{3}cit will be converted into H_{2}cit^{–} by the base until all the H_{3}cit is used up, then H_{2}cit^{–} will be converted into Hcit^{2–}, etc.). This would allow us to construct the chart shown in Table 3 of the moles of each species that would occur over the course of the titration.

**Table 3.** Moles of each species in the titration of citric acid (0.1 M, 30 ml) with NaOH (0.1M).

ml NaOH | H_{3}cit |
H_{2}cit^{–} |
Hcit^{2–} |
cit^{3–} |
pH |
---|---|---|---|---|---|

0 | 0.0030 mol | 0 | 0 | 0 | 2.06 |

5 | 0.0025 | 0.0005 | 0 | 0 | |

10 | 0.0020 | 0.0010 | 0 | 0 | |

15 | 0.0015 | 0.0015 | 0 | 0 | |

20 | 0.0010 | 0.0020 | 0 | 0 | |

25 | 0.0005 | 0.0025 | 0 | 0 | |

30 | 0 | 0.0030 | 0 | 0 | |

35 | 0 | 0.0025 | 0.0005 | 0 | |

40 | 0 | 0.0020 | 0.0010 | 0 | |

45 | 0 | 0.0015 | 0.0015 | 0 | |

50 | 0 | 0.0010 | 0.0020 | 0 | |

55 | 0 | 0.0005 | 0.0025 | 0 | |

60 | 0 | 0 | 0.0030 | 0 | |

65 | 0 | 0 | 0.0025 | 0.0005 | |

70 | 0 | 0 | 0.0020 | 0.0010 | |

75 | 0 | 0 | 0.0015 | 0.0015 | |

80 | 0 | 0 | 0.0010 | 0.0020 | |

85 | 0 | 0 | 0.0005 | 0.0025 | |

90 | 0 | 0 | 0 | 0.0030 | |

95 | 0 | 0 | 0 | 0.0030 | |

100 | 0 | 0 | 0 | 0.0030 |

If we examine the increments from 5 ml to 25 ml, we see that we have appreciable quantities of H_{3}cit and H_{2}cit^{–}, which are both members of a conjugate pair. This region is a buffer solution and the pH can be determined using K_{a1}.

\(\textrm{5 ml:} \hspace{32px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right )=3.128+\log\left( \dfrac{0.0005\: mol}{0.0025\: mol}\right )=2.43}\)

\(\textrm{10 ml:}\hspace{25px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right )=3.128+\log\left( \dfrac{0.0010\: mol}{0.0020\: mol}\right )=2.83}\)

\(\textrm{15 ml:} \hspace{25px} \mathrm{pH = pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right ) = 3.128 + \log\dfrac{0.0015\: mol}{0.0015\: mol} = 3.128}\)

Note that the pH at this increment is pK_{a1}.

\(\textrm{20 ml:} \hspace{25px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right )=3.128+\log\left( \dfrac{0.0020\: mol}{0.0010\: mol}\right )= 3.43}\)

\(\textrm{25 ml:} \hspace{25px} \mathrm{pH = pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right ) = 3.128 + \log\left( \dfrac{0.0025\: mol}{0.0005\: mol}\right ) = 3.83}\)

30 ml: This is the first equivalence point, since we have converted all of the H_{3}cit to H_{2}cit^{–}. At this point we have “all” of the first intermediate (H_{2}cit^{–}) and can calculate the pH using the expression (pK_{1} + pK_{2})/2

\[\mathrm{pH = \dfrac{pK_{a1}+ pK_{a2}}{2}=\dfrac{3.128+4.761}{2}=3.9}\]

If we examine the region from 35 to 55 ml, we have appreciable quantities of H_{2}cit^{–} and Hcit^{2–}, a buffer solution based on K_{a2}.

\(\textrm{35 ml:} \hspace{25px} \mathrm{pH = pK_{a2}+ \log\dfrac{[Hcit^{2-}]}{[H_2cit^-]} = 4.761 + \log\left( \dfrac{0.0005\: mol}{0.0025\: mol}\right ) = 4.06}\)

\(\textrm{40 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0010\: mol}{0.0020\: mol}\right ) = 4.46}\)

\(\textrm{45 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0015\: mol}{0.0015\: mol}\right ) = 4.761}\)

Note that at this point, the pH is equal to pK_{a2}.

\(\textrm{50 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0020\: mol}{0.0010\: mol}\right ) = 5.06}\)

\(\textrm{55 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0025\: mol}{0.0005\: mol}\right ) = 5.46}\)

60 ml: This is the second equivalence point, since we have converted all of the H_{2}cit^{-} to Hcit^{2-}. At this point we have “all” of the second intermediate (Hcit^{2-}) and can calculate the pH using the expression (pK_{2} + pK_{3})/2

\[\mathrm{pH = \dfrac{pK_{a2}+ pK_{a3}}{2}=\dfrac{4.761 + 6.396}{2}= 5.58}\]

If we examine the region from 65 to 85 ml, we have appreciable quantities of Hcit^{2-} and cit^{3-}, a buffer solution based on K_{a3}.

\(\textrm{65 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0005\: mol}{0.0025\: mol}\right ) = 5.70}\)

\(\textrm{70 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0010\: mol}{0.0020\: mol}\right ) = 6.09}\)

\(\textrm{75 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0015\: mol}{0.0015\: mol}\right ) = 6.396}\)

Note that at this point, the pH is equal to pK_{a3}.

\(\textrm{80 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0020\: mol}{0.0010\: mol}\right ) = 6.70}\)

\(\textrm{85 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0025\: mol}{0.0005\: mol}\right ) = 7.09}\)

90 ml: This is the third equivalence point, since we have converted all of the Hcit^{2–} to cit^{3–}. To a first approximation we only have cit^{3–} in solution. This is a polybasic base, but as we have done before, we only need to consider the first reaction in the series to calculate the pH. The relevant reaction, which is the K_{b} value of K_{a3}, is shown below.

cit^{3–} + H_{2}O = Hcit^{2–} + OH^{–} K_{b} of K_{a3} = 2.5×10^{-8}

We need to calculate the concentration of cit^{3–} that is present in solution, recognizing that the titrant caused a dilution of the initial concentration of citric acid (30 ml of initial solution and 90 ml of additional titrant).

\[\textrm{Molarity of citrate = (0.0030 mol/0.120 L) = 0.025 M}\]

\[\begin{align}

& &&\ce{cit^3-}\hspace{25px} + \hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{Hcit^2-} \hspace{25px} + &&\ce{OH-} \\

&\ce{Initial} &&0.025 &&0 &&0 \\

&\ce{Equilibrium} &&0.025 - \ce{x} &&\ce{x} &&\ce{x} \\

&\ce{Approximation} &&0.025 &&\ce{x} &&\ce{x}

\end{align}\]

\[\mathrm{K_{b3} =\dfrac{[Hcit^{2-}][OH^-]}{[cit^{3-}]}=\dfrac{(x)(x)}{0.025} = 2.5\times10^{-8}}\]

\[\mathrm{x = [OH^-] = 2.5\times 10^{-5}}\]

\[\mathrm{pOH = 4.6 \hspace{60px} pH = 9.4}\]

Checking the approximation shows that it was valid in this case. Note that the pH at this equivalence point is basic, which is not surprising since cit^{3–} is a base.

\[\dfrac{2.5×10^{-5}}{0.025}× 100 = 0.1\%\]

95 ml: In this case we have a mixture of a strong base (NaOH) with a weaker base (citrate). The extra amount of strong base (5 ml or 0.0005 moles) will determine the pH.

\[\mathrm{[OH^-] =\dfrac{0.0005\: mol}{0.125\: L} = 4.0\times10^{-3}}\]

\[\mathrm{pOH = 2.4 \hspace{60px} pH = 11.6}\]

100 ml: Once again, the pH is determined by the amount of extra strong base present in the solution (10 ml or 0.0010 moles).

\[\mathrm{[OH^-] =\dfrac{0.0010\: mol}{0.130\: L} = 7.7\times10^{-3}}\]

\[\mathrm{pOH = 2.1 \hspace{60px} pH = 11.9}\]

We can now compile an entire chart (Table 4) of the changes that occur over this titration:

**Table 4.** Calculated pH values for the titration of citric acid (0.1 M, 30 mL) with NaOH (0.1 M).

ml NaOH | H_{3}cit |
H_{2}cit^{–} |
Hcit^{2–} |
cit^{3–} |
pH |
---|---|---|---|---|---|

0 | 0.0030 m | 0 | 0 | 0 | 2.06 |

5 | 0.0025 | 0.0005 | 0 | 0 | 2.43 |

10 | 0.0020 | 0.0010 | 0 | 0 | 2.83 |

15 | 0.0015 | 0.0015 | 0 | 0 | 3.128 (pK_{a1}) |

20 | 0.0010 | 0.0020 | 0 | 0 | 3.43 |

25 | 0.0005 | 0.0025 | 0 | 0 | 3.83 |

30 | 0 | 0.0030 | 0 | 0 | 3.94 (pK_{a1}+pK_{a2})/2 |

35 | 0 | 0.0025 | 0.0005 | 0 | 4.06 |

40 | 0 | 0.0020 | 0.0010 | 0 | 4.46 |

45 | 0 | 0.0015 | 0.0015 | 0 | 4.761 (pK_{a2}) |

50 | 0 | 0.0010 | 0.0020 | 0 | 5.06 |

55 | 0 | 0.0005 | 0.0025 | 0 | 5.46 |

60 | 0 | 0 | 0.0030 | 0 | 5.58 (pK_{a2}+pK_{a3})/2 |

65 | 0 | 0 | 0.0025 | 0.0005 | 5.70 |

70 | 0 | 0 | 0.0020 | 0.0010 | 6.09 |

75 | 0 | 0 | 0.0015 | 0.0015 | 6.396 (pK_{a3}) |

80 | 0 | 0 | 0.0010 | 0.0020 | 6.70 |

85 | 0 | 0 | 0.0005 | 0.0025 | 7.09 |

90 | 0 | 0 | 0 | 0.0030 | 9.40 |

95 | 0 | 0 | 0 | 0.0030 | 11.60 |

100 | 0 | 0 | 0 | 0.0030 | 11.90 |

It is especially helpful to plot these values versus the ml of titrant as shown in Figure 3.

**Figure 3.** pH versus ml titrant for the titration of citric acid (0.1 M, 30 mL) with NaOH (0.1 M).

There are several things worth noting in this plot. One is the way that the first two equivalence points blend together and there are no clear breaks in the plot. The only equivalence point in this titration that is readily observable is the third. The other is to note that citric acid has a significant buffer region that stretches from a pH of about 2.5 to 5.5. Citric acid is commonly used as a buffer for this pH region.

It is also worth examining what would be observed for a similar plot of a different triprotic acid. The data in Table 5 is for an identical titration of phosphoric acid.

**Table 5.** pH values for the titration of phosphoric acid (0.1 M, 30 ml) with NaOH (0.1 M).

ml NaOH | H_{3}PO_{4} |
\(\ce{H2PO4-}\) | \(\ce{HPO4^2-}\) | \(\ce{PO4^3-}\) | pH |
---|---|---|---|---|---|

0 | 0.0030 mol | 0 | 0 | 0 | - |

5 | 0.0025 | 0.0005 | 0 | 0 | 1.45 |

10 | 0.0020 | 0.0010 | 0 | 0 | 1.85 |

15 | 0.0015 | 0.0015 | 0 | 0 | 2.148 (pK_{a1}) |

20 | 0.0010 | 0.0020 | 0 | 0 | 2.45 |

25 | 0.0005 | 0.0025 | 0 | 0 | 2.85 |

30 | 0 | 0.0030 | 0 | 0 | 4.673 (pK_{a1}+pK_{a2})/2 |

35 | 0 | 0.0025 | 0.0005 | 0 | 6.50 |

40 | 0 | 0.0020 | 0.0010 | 0 | 6.90 |

45 | 0 | 0.0015 | 0.0015 | 0 | 7.198 (pK_{a2}) |

50 | 0 | 0.0010 | 0.0020 | 0 | 7.50 |

55 | 0 | 0.0005 | 0.0025 | 0 | 7.90 |

60 | 0 | 0 | 0.0030 | 0 | 9.789 (pK_{a2}+pK_{a3})/2 |

65 | 0 | 0 | 0.0025 | 0.0005 | 11.68 |

70 | 0 | 0 | 0.0020 | 0.0010 | 12.08 |

75 | 0 | 0 | 0.0015 | 0.0015 | 12.38 (pK_{a3}) |

80 | 0 | 0 | 0.0010 | 0.0020 | 12.68 |

85 | 0 | 0 | 0.0005 | 0.0025 | - |

90 | 0 | 0 | 0 | 0.0030 | - |

95 | 0 | 0 | 0 | 0.0030 | 11.60 |

100 | 0 | 0 | 0 | 0.0030 | 11.90 |

It is worth realizing that a few data points have been omitted since there is a problem at the beginning and again at about 75 ml of titrant. In the early part, the acid is strong enough that a fairly significant proportion dissociates. At the latter part of the titration, the base is so strong that we really do not convert all of the \(\ce{HPO4^2-}\) to \(\ce{PO4^3-}\) as implied. Even with this problem, we can examine a generalized plot for the titration of phosphoric acid with sodium hydroxide (Figure 4).

**Figure 4.** pH versus ml titrant for the titration of phosphoric acid (0.1 M, 30 mL) with NaOH (0.1 M).

Note here that the first two equivalence points are obvious, whereas the third equivalence point will not be distinguishable because of the very high pK_{a3} value. The concentration of citric or phosphoric acid can be determined through a titration with sodium hydroxide, provided you realize which equivalence points can be successfully monitored during the titration.

Now we can examine the last two parts of the homework problem. The first is whether 99.9% of the species is in the form cit^{3–} at the third equivalence point. Going back to the calculation at 90 ml of titrant, we determined that [cit^{3–}] was 0.025 M and [Hcit^{2–}] was 2.5×10^{-5} M.

\[\dfrac{2.5 \times 10^{-5}}{0.025} \times 100 = 0.1\%\]

If 0.1% is in the form Hcit^{2–}, then 99.9% is in the form cit^{3–} and it just makes it.

The other part of the problem was to calculate the concentration of all species in solution at the second equivalence point. The first thing we ought to do is compile a list of what all the species are so we know what we have to calculate. In doing this, we can ignore any spectator ions such as sodium. That means there are six species whose concentration we need to calculate.

H_{3}cit H_{2}cit^{–} Hcit^{2–} cit^{3–} H_{3}O^{+} OH^{–}

Since we already calculated the pH of this solution (5.58), we can readily calculate the concentration of H_{3}O^{+} and OH^{–}.

[H_{3}O^{+}] = 2.63×10^{-6}

[OH^{–}] = 3.80×10^{-9}

We also said that to a first approximation it was “all” Hcit^{2–} (0.0030 mol). With 30 ml of initial solution and 60 ml of titrant, we have a total volume of 90 ml.

[Hcit^{2–}] = 0.0030 mol/0.090 L = 0.033 M

Since we now have [H_{3}O^{+}] and [Hcit^{2–}], we can use the appropriate K_{a} expressions to calculate the three other citrate species.

Use the K_{a3} expression to calculate [cit^{3–}]:

\[\mathrm{K_{a3} =\dfrac{[cit^{3-}][H_3O^+]}{[Hcit^{2-}]}=\dfrac{[cit^{3-}](2.63\times10^{-6})}{0.033} = 4.02\times10^{-7}}\]

\[\mathrm{[cit^{3-}] = 0.005\: M}\]

Use the K_{a2} expression to calculate [H_{2}cit^{–}]:

\[\mathrm{K_{a2} =\dfrac{[Hcit^{2-}][H_3O^+]}{[H_2cit^{2-}]}=\dfrac{(0.033)(2.63\times10^{-6})}{[Hcit^{2-}]} = 1.733\times10^{-5}}\]

\[\mathrm{[H_2cit^–] = 0.005\: M}\]

Using the value of H_{2}cit^{–} that was just calculated, we can substitute this into the K_{a1} expression and calculate the concentration of H_{3}cit.

\[\mathrm{K_{a1} =\dfrac{[H_2cit^{2-}][H_3O^+]}{[H_3cit]} =\dfrac{(0.005)(2.63\times10^{-6})}{[H_3cit]} = 7.45\times10^{-4}}\]

\[\mathrm{[H_3cit] = 1.77 \times 10^{-5}}\]

One last set of things to examine are the calculated values for [H_{2}cit^{–}], [Hcit^{2–}] and [cit^{3–}].

[Hcit^{2–}] = 0.033 M

[H_{2}cit^{–}] = 0.005 M

[cit^{3–}] = 0.005 M

What we need to appreciate is that there is a problem with these numbers. We started this calculation by claiming that “all” of the material was in the form of Hcit^{2–}. What these calculations show is that there are appreciable amounts of H_{2}cit^{–} and cit^{3–} in solution. If these two values were accurate, it would mean that the concentration of Hcit^{2–} could only be 0.023 M.

Why does this happen? It has to do with how close the pK_{a} values are for citric acid. The approximation that we could examine this as a stepwise manner, where we proceeded from one reaction to the other and that intermediates were overwhelmingly the predominant form at the equivalence points, breaks down in this case because of how close the pK_{a} values are. The interesting part of this is that the pH of the solution would be 5.58, and that we will get exactly equivalent concentrations of H_{2}cit^{–} and cit^{3–}, although they will not be exactly 0.005 M. In reality, I do not think we would ever try to calculate the exact amount of each of these species at a pH like this, although later in the course we are going to come back to the citric acid situation and see a way to calculate the exact concentration of species present provided we know the pH.

One thing to keep in mind is that often times we do not use equilibrium calculations to arrive at exact values of substances. For one thing, concentrations are an approximation of activities and this may not always be a good one. For another, we often use equilibrium calculations to provide ballpark values to let us know whether a particular process we may be considering is even feasible. In this case, for example, these values show that we could never use the second equivalence point in a citric acid titration for measurement purposes. This does not mean that citric acid cannot be used as a buffer, because it frequently is. However, if we prepare a citric acid buffer (or any buffer), we do not rely on calculated amounts to ensure that the pH is where we want it, we use a pH meter to monitor the buffer and use small amounts of a strong acid or strong base to adjust the pH to the value we want.