# Weak Acids and Bases

# Acid-ionization Constant (*K*_{a})

_{a}

*K*) of this reaction is shown below:

_{a}

# Percentage Ionization

^{â€“5})

We will logically dissect the events that lead to the dissociation. At first we do not have any H^{+} and CH_{3}COO^{â€“} ions. Let's say we have a unit volume (a liter) of acetic acid and *xM *(mol/L) of it was ionized. We can represent the equation as follows:

With the concentrations at equilibrium we can substitute the concentrations in the acid-ionization constant expression.

Even though this has to be solved using a quadratic equation, we can skip that elaborate process by making some chemically acceptable assumptions. Since the acid-ionization constant is very small, we can assume the same with the value of *x*. So the expression changes as follows. For the MCAT, you won't be given problems that require extensive calculations. The problems will test mostly concepts, and when calculations are involved the numbers will usually be manageable ones.

Solving for *x* you will get roughly 1.34 x 10^{â€“3} = 0.00134 *M*.

The question also asks for pH of the solution.

pH = â€“ log [H^{+}]

The hydrogen ion concentration is 0.00134 *M*. By substituting in the pH formula you should get the following result.

The pH of the solution is 2.87.

# Base-ionization Constant (*K*_{b})

_{b}

*K*) for the above reaction as follows:

_{b}