Question 17.2: Calculating Concentrations of Species in a Weak Acid Solutio...
Calculating Concentrations of Species in a Weak Acid Solution Using K_{a} (Approximation Method)
What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0.10 \mathrm{M} nicotinic acid, \mathrm{HC}_{6} \mathrm{H}_{4} \mathrm{NO}_{2}, at 25^{\circ} \mathrm{C} ? What is the \mathrm{pH} of the solution? What is the degree of ionization of nicotinic acid? The acid-ionization constant, K_{a}, was determined in the previous example to be 1.4 \times 10^{-5}.
PROBLEM STRATEGY
You follow the three steps for solving equilibrium problems that were introduced in Example 15.7. In the last step, you solve the equilibrium-constant equation for the equilibrium concentrations. The resulting equation is quadratic, but because the equilibrium concentration of a weak acid is usually negligibly different from its starting value, the equation simplifies so that it involves only the square of the unknown, which is easily solved by taking the square root. (You will need to check that this assumption is valid.)
STEP 1 At the start (before ionization), the concentration of nicotinic acid, HNic, is 0.10 \mathrm{M} and that of its conjugate base, \mathrm{Nic}^{-}, is 0 . The concentration of \mathrm{H}_{3} \mathrm{O}^{+}is essentially zero (\sim 0), assuming that the contribution from the self-ionization of water can be neglected. In 1 \mathrm{~L} of solution, the nicotinic acid ionizes to give x \mathrm{~mol}
\mathrm{H}_{3} \mathrm{O}^{+}and x mol Nic { }^{-}, leaving (0.10-x) mol of nicotinic acid. These data are summarized in the following table:
The equilibrium concentrations of \mathrm{HNic}, \mathrm{H}_{3} \mathrm{O}^{+}, and \mathrm{Nic}^{-}are (0.10-x), x, and x, respectively.
STEP 2 Now substitute these concentrations and the value of K_{a} into the equilibrium-constant equation for acid ionization:
\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{Nic}^{-}\right]}{[\mathrm{HNic}]}=K_{a}
You get
\frac{x^{2}}{(0.10-x)}=1.4 \times 10^{-5}
STEP 3 Now solve this equation for x. This is actually a quadratic equation, but it can be simplified so that the value of x is easily found. Because the acidionization constant is small, the value of x is small. Assume that x is much smaller than 0.10 , so that
0.10-x \simeq 0.10
You will need to check that this assumption is valid after you obtain a value for x. The equilibrium-constant equation becomes
\frac{x^{2}}{0.10} \simeq 1.4 \times 10^{-5}
or
x^{2} \simeq 1.4 \times 10^{-5} \times 0.10=1.4 \times 10^{-6}
Hence,
x \simeq 1.2 \times 10^{-3}=0.0012
At this point, you should check to make sure that the assumption 0.10-x \simeq 0.10 is valid. You substitute the value obtained for x into 0.10-x.
0.10-x=0.10-0.0012=0.10 \quad \text { (to two significant figures) }
The assumption is indeed valid.
Now you can substitute the value of x into the last line of the table written in Step 1 to find the concentrations of species. The concentrations of nicotinic acid, hydronium ion, and nicotinate ion are 0.10 M, 0.0012 M, and 0.0012 M, respectively.
The \mathrm{pH} of the solution is
\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log (0.0012)=\mathbf{2 . 9 2}
The degree of ionization equals the amount per liter of nicotinic acid that ionizes (x=0.0012) divided by the total amount per liter of nicotinic acid initially present (0.10). Thus, the degree of ionization is 0.0012 / 0.10=\mathbf{0 . 0 1 2}.
ANSWER CHECK
As a quick check on your work, substitute the concentrations that you calculated for the acid (nicotinic), acid ion (nicotinate), and hydronium ion into the equilibrium expression for K_a. You should get the value of K_a given in the problem.
K_{a}=\frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{Nic}^{-}]}{[\mathrm{H}\mathrm{Nic}]}=\frac{0.0012\times0.00012}{0.10}=1.4\times10^{-5}
The pH, of course, should be in the acid range, below 7. Check that this is the case.