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A weak polyprotic acid is an acid that is usually considered as weak acid in its monoprotic form (only one H^{+} in the molecule), but instead has more than one H^{+} in the molecule, therefore making it a polyprotic acid.
By definition, an acid donates a proton (H^{+}), so a polyprotic acid donates protons. However, a polyprotic acid differs from a monoprotic acid because it has more than one acidic H^{+}, so it has the ability to donate multiple protons. As a weak polyprotic acid, it does not completely dissociate. Here are some examples of weak polyprotic acids:
H_{3}PO_{4} - triprotic acid
As an acid, a polyprotic acid has a very small acid dissociation constant (K_{a}), which measures the strength of the acid. K_{a} corresponds to the reaction of a weak acid with water and can be used to determine the pH of a solution. In the figure below, water serves as the base because it accepts a proton, H^{+}, from the phosphoric acid to become a hydronium ion. Phosphoric acid becomes a conjugate base because it loses a proton.
The acid dissociation constant can be attained by the following equation:
\[ K_a=\dfrac{Concentration of Products}{Concentration of Reactants} \]
or even by acid dissociation constant at a logarithmic scale, also known as pK_{a}:
\[ pK_a=-log_{10}K_a \]
The equation can be manipulated into
\[ K_a=10^{-pK_a}\]
pKa also be used to determine the pH of a solution given the concentrations of the conjugate base and undissociated acid. The equation is as follows:
\[ pH=pK_a+log\dfrac{[A^-]}{[HA]} \]
\[ {A^-}=conjugate base \]
\[ HA=undissociated acid \]
There are as many acid ionization constants as there are acidic protons. Therefore, there are three acid ionization constants for phosphoric acid.
Example |
For example, phosphoric acid...
\(H_{3}PO_{4}+H_{2}O \rightleftharpoons H_{3}O^++H_{2}PO_{4}^-\) \({K_{a1}}=\dfrac{[H_{3}O^+][H_{2}PO_{4}^-]}{[H_{3}PO_{4}]}=5.0*10^{-3}\)
\(H_{2}PO_{4}^-+H_{2}O \rightleftharpoons H_{3}O^+ + HPO_{4}^ {2-}\) \({K_{a2}}=\dfrac{[H_{3}O^+][HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}=6.3*10^{-7}\)
\(HPO_{4}^{2-}+H_{2}O \rightleftharpoons H_{3}O^+ + PO_{4}^ {3-}\) \({K_{a3}}=\dfrac{[H_{3}O^+][PO_{4}^{3-}]}{[HPO_{4}^{2-}]}=3.5*10^{-13}\)
Note that the acid dissociation constant of the first proton, indicated by \(K_{a1}\), is the largest of all the successive acid dissociation constants. Acid dissociation constants, along with information from a titration, give the information needed to determine the pH of the solution.
The purpose of titration is to find the concentration of an unknown solution by adding a known volume of a solution with a known concentration to the unknown concentration of a solution. After finding the concentration of this unknown solution, one can find the pH of the solution, given information about the acid dissociation constant(s). Figure 2 below shows the typical lab titration setup prior to adding any titrant to the analyte.
When an acid is titrated, there is an equivalence, or stoichiometric, point, which is when the moles of the strong base added equal of the moles of weak acid present. However, when a weak polyprotic acid is titrated, there are multiple equivalence points because the equivalence point will occur when an H^{+} is dissociated. Therefore, the number of equivalence points depends on the number of H^{+} atoms that can be removed from the molecule. Note that when the weak polyprotic acid dissociates, the proton (H^{+}) combines with H_{2}O to form H_{3}O^{+}.
The midpoint, also indicated in the figure, is when the number of moles of strong base added equals half of the moles of the weak acid that are present. For this reason, the midpoint is half of the equivalence point. Notice that there are as many midpoints as there are equivalence points. At the midpoint, pH equals the value of pKa because there is 50:50 mixture of the weak acid and the strong base. To quantify this, the Henderson-Hasselbalch Approximation can be used:
\[ pH=pK_a+log\dfrac{[A^-]}{[HA]} \]
Since the solution is a 50/50 mixture, then the concentrations of both A- and HA are equal. Therefore, \( \frac{[A^-]}{[HA]}=1 \). Plugging it back into the original equation, you get \( pH=pK_a+log(1) \). Since \( log(1)=0 \), the equation becomes \( pH=pK_a \).
Example |
This next example shows what occurs when titrating the weak polyprotic acid H_{3}A with a strong base, like LiOH and NaOH. Since there are 3 acidic protons in this example, there is expected to be three equivalence points. (Note: This is disregarding the base used in the titration which would change your products depending upon the base used)
\[ H_{3}A+H_{2}O \rightleftharpoons H_{3}O^++ H_{2}A^- \]
As illustrated above in Figure 3.1, adding 10 mL of the titrant to the weak polyprotic acid is need to reach the first equivalence point.
\[ H_{2}A^-+H_{2}O \rightleftharpoons H_{3}O^++HA^{2-} \]
Figure 3.2 above illustrates that adding another 10 mL (total of 20 mL) to the weak polyprotic acid solution will allow for another H^{+} to dissociate. Another equivalence points also means yet another midpoint.
\[ HA^{2-}+H_{2}O \rightleftharpoons H_{3}O^++A^{3-} \]
In Figure 3.3 above, the titration is finally complete because there are three equivalence points, with the third being attained by adding yet another 10 mL (total of 30 ml) of the titrant.
The following example below, we can conclude that the graph of a weak polyprotic acid will show not one (as the graph of a weak acid with a strong base titration graph would look), but multiple equivalence points.
4. \(mol H_{2}SO_{3}=Molarity*Volume=0.10 M*0.07 L=0.007\)
_{ }\(mol KOH=Molarity*Volume=0.10 M*.05 L=0.005\)
After this titration, 0.002 mol H_{2}SO_{3} remain and 0.005 mol HSO_{3}^{-} form.
\(H_{2}SO_{3}+H_{2}O \rightleftharpoons H_{3}O^++HSO_{3}^{-}\)
\(K_{a1}=\dfrac{[H_{3}O^+][HSO_{3}^{-}]}{[H_{2}SO_{3}]}=5.9*10^{-3}\)
\([H_{3}O^+]=2.36*10^{-3}\)
\(pH=-log[H_{3}0^+]=-log(2.36*10^{-3})=2.63\)
5. \(mol H_{2}CO_{3}=Molarity*Volume=0.10 M*.03L=0.003\)
\(mol LiOH=Molarity*Volume=0.10 M*.05L=0.005"} }}
After this titration, 0.002 mol HCO_{3}^{-} remain and 0.001 mol CO_{3}^{2-} form.
\(HCO_{3}^{-1}+H_{2}O \rightleftharpoons H_{3}O^++CO_{3}^{2-}\)
\(K_{a2}=\dfrac{[H_{3}O^+][CO_{3}^{2-}]}{[HCO_{3}^{-1}]}=5.9*10^{-7}\)
\([H_{3}O^+]=1.18*10^{-6}\)
\(pH=-log[H_{3}0^+]=-log(1.18*10^{-6})=5.92\)
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