Here is a way to test the validity of the statement made on page 719 in conjunction with the three key ideas governing the ionization of polyprotic acids. Determine the \(\mathrm{pH}\) of \(0.100 \mathrm{M}\) succinic acid in two ways: first by assuming that \(\mathrm{H}_{3} \mathrm{O}^{+}\) is produced only in the first ionization step, and then by allowing for the possibility that some \(\mathrm{H}_{3} \mathrm{O}^{+}\) is also produced in the second ionization step. Compare the results, and discuss the significance of your finding. $$\begin{array}{c} \mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-} \\ &K_{\mathrm{a}_{1}}=6.2 \times 10^{-5} \\ \mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{2-} \\ &K_{\mathrm{a}_{2}}=2.3 \times 10^{-6} \end{array}$$

Step by step solution

01

Calculate the pH assuming the first ionization only

Firstly, we need to set up an equilibrium expression for the first ionization of the acid: \(K_{a1} = \frac {[H_{3} O^{+}] [HC_{4} H_{4} O_{4}^{-}]} {[H_{2} C_{4} H_{4} O_{4}]}\). From the problem statement, we know that \(K_{a1} = 6.2 \times 10^{-5}\) and the initial concentration of the acid was 0.100 M. We can write down the change in concentrations as -x for \(H_{2} C_{4} H_{4} O_{4}\), +x for \(H_{3} O^{+}\) and \(HC_{4} H_{4} O_{4}^{-}\). After substituting the values in equilibrium expression, we can solve it to find out the value of 'x' which will be equal to the \(H_{3} O^{+}\) concentration.

02

Calculate the pH

Once we have the concentration of \(H_{3} O^{+}\), we can easily calculate the pH of the solution using the formula: \(\mathrm{pH} = -\log([H_{3} O^{+}])\).

03

Calculation considering both ionizations

Now let's repeat the process considering both ionization steps. But this time, we set up our equilibrium expression for the second ionization step: \(K_{a2} = \frac {[H_{3} O^{+}] [C_{4} H_{4} O_{4}^{2-}]} {[H_{4} C_{4} H_{4} O_{4}^{-}]}\). We can again set up our change in concentrations like we did before and can find out the new concentration for \(H_{3} O^{+}\).

04

Recalculate the pH

After getting the new concentration of \(H_{3} O^{+}\) considering both the ionizations, we can again calculate the pH of the solution using the formula: \(\mathrm{pH} = -\log([H_{3} O^{+}])\).

05

Compare the results

Finally, we need to compare the values of pH we got considering only the first ionization and both the ionizations respectively. We need to discuss the significance of these findings.

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