Acid-Base Buffers: Calculating the pH of a Buffered Solution

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• 0:05 Buffers
• 1:54 pH of a Buffered Solution
• 4:17 pH of a Buffered…
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Lesson Transcript
Instructor: Amy Meyers

Amy holds a Master of Science. She has taught science at the high school and college levels.

Learn what a buffer is, how it works, and why we benefit from having our blood buffered. Learn how to calculate the pH of a buffered solution before an acid or base is added and how the pH changes after an acid or base is added.

Buffers

Your blood has a pH of 7.4. Keeping your blood pH constant is very important to your health. If your blood pH becomes too low, under 7.35, you get acidosis. If your blood pH gets too high, greater than 7.45, you get alkalosis. Symptoms of acidosis include seizures, heart palpitations, shortness of breath, loss of consciousness, and coma. As you can see, you probably don't want acidosis. Your body controls your blood pH by using buffer solutions.

A buffer is a solution that contains equal amounts of a weak acid and its conjugate base or a weak base and its conjugate acid. A buffer solution works because it keeps the pH of a solution from changing very much. The HX is the acid. The X- is the conjugate base. A buffer equation made of the two is: HX + H2O <--> H3O+ + X-

If you add a base to this buffered solution, the base reacts with the H3O+ and removes it from the solution. If you remember back to Le Chatelier's principle, it says that the equilibrium will shift right to adjust and make more of the H3O+. This keeps the pH from changing too much, thus 'buffering' it. If you had added an acid, it would have joined with the X- to make HX. The reaction would have shifted left.

The equilibrium constant expression is Ka = [H3O+][X-] / [HX]

pH of a Buffer Solution

To calculate the pH of a buffer solution, the equation is pH = -log(Ka). You have seen these equations before, so to calculate the pH of a buffered solution when an acid or base is added, you use the steps you've learned before. The big deal about these reactions is that the [H+] and [OH-] concentrations are no longer equal.

Example:

Look at the equation CH3COOH <--> CH3COO- + H+

Ka = [CH3COO-][H+] / [CH3COOH]

Suppose you have a solution of 0.25 M acetic acid, CH3COOH, and 0.6 M sodium acetate, C2H3NaO2, which is its conjugate base. The acetic acid is in equilibrium, but the [CH3COO-] does not equal [H+].

What you can assume, though, is that the weak acid, which in this case is the acetic acid, will barely dissociate. In this case, the [CH3COO-] is the same as the initial concentration of sodium acetate, or 0.6 M. The Ka for acetic acid is 1.76 x 10^-5. We can solve for [H+].

Ka = [CH3COO-][H+] / [CH3COOH]
1.76 x 10^-5 = (0.6 M)[H+] / (0.25 M)
[H+] = 7.33 x 10^-6
pH = -log[H+]
pH = -log(7.33 x 10^-6)
pH = 5.13

pH of a Buffer Solution When Acid Is Added

Example:

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