Problems using the Henderson-Hasselbalch equation: an acid and its salt

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As a reminder, here is the Henderson-Hasselbalch Equation:

Problem #1:

A buffer is prepared containing 1.00 molar acetic acid and 1.00 molar sodium acetate. What is its pH?

Problem #2:

A buffer is prepared containing 0.800 molar acetic acid and 1.00 molar sodium acetate. What is its pH?

Problem #3:

A buffer is prepared containing 1.00 molar acetic acid and 0.800 molar sodium acetate. What is its pH?

Solution to Problem #1:

We must know the pKa of acetic acid. Most times, the problem will provide the pKa. If the problem provides the Ka, you must convert it to the pKa.

Be aware, your teacher may force a situation where you must look up the Ka. Even in this era of fairly easy Internet access, try one of the appendices of your textbook (if you have one). Tables of Ka values are a standard feature of most textbooks.

The Ka of acetic acid is 1.77 x 10¯5

pKa = - log Ka = - log 1.77 x 10¯5 = 4.752

Next, we simply insert the appropriate values into the HH equation:

x = 4.752 + log (1.00 / 1.00)

Since the log of 1 is zero, we have pH = 4.752

Solution to Problem #2:

x = 4.752 + log (1.00 / 0.800)

x = 4.752 + 0.097 = 4.849

Solution to Problem #3:

x = 4.752 + log (0.800 / 1.00)

x = 4.752 - 0.097 = 4.655

Discussion about the three above problems.

1) The pH of this buffer IS NOT the pH of a solution of acetic acid only. The pH of a 1.00 molar solution of acetic acid is 2.376.

2) You can consider the above buffer examples to be mixtures of an acid and its conjugate base.

3) Notice that, in example #1, the pH is less than the solution of just the pure acid (4.752 as compared to 2.376). This is due to LeChatelier's Principle. Consider the dissociation equation for acetic acid:

HAc <===> H+ + Ac¯

Increasing the concentration of the acetate (Ac¯) will push the equiibrium back to the left, decreasing the concentration of H+. This makes the solution less acidic, making the pH of the buffer larger than the pure acid solution.

4) Notice that with the larger amount of base compared to acid (Problem #2) the pH is more basic (it is larger) than the 1:1 ratio pH in Problem #1.

5) Notice that with the larger amount of acid compared to base (Problem #3) the pH is more acidic (it is smaller) than the 1:1 ratio pH in Problem #1.


Three more buffer calculations with an acid and its salt

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