### Problems using the Henderson-Hasselbalch equation: a base and its salt

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

Problem #1:

A buffer is prepared containing 1.00 molar ammonia and 1.00 molar ammonium chloride. What is its pH?

Problem #2:

A buffer is prepared containing 0.800 molar ammonia and 1.00 molar ammonium chloride. What is its pH?

Problem #3:

A buffer is prepared containing 1.00 molar ammonia and 0.800 molar ammonium chloride. What is its pH?

Solution to Problem #1:

Solving problems with bases has (potentially) one more step that when working with acids. This happens if you are given the Kb for the base.

You can't use the Kb for the base, you have to use the Ka for the conjugate acid (and thence to the pKa.)

In this case, we know the Kb for ammonia; it is 1.77 x 10¯5. (By the way, the Kb for ammonia is very close in value to the Ka for acetic acid. This is a coincidence only, do not read anything into it.)

We need to know the Ka for the ammonium ion (it is the conjugate acid). Form that we get the pKa. We do that as follows:

Kw = KaKb

1.00 x 10¯14 = (x) (1.77 x 10¯5)

x = 5.65 x 10¯10

pKa = - log Ka = - log 5.65 x 10¯10 = 9.248

By the way, 9.258 has three significant figures. Since it is a logarithm, the significant figures are in the decimal portion. The 9 simply fixes the position of the decimal.

Now, we are ready to insert into the HH equation:

pH = 9.248 + log (1.00 / 1.00)

Since the log of 1 is zero, we have pH = 9.248

Before solving #2 and #3, a warning. Make sure you know which is the acid and which is the base. The acid is the substance with the proton. It goes in the denominator of the HH equation. The base is the substance without the proton. It goes in the numerator of the HH equation.

Solution to Problem #2:

x = 9.248 + log (0.800 / 1.00)

x = 9.248 - 0.097 = 9.151

Solution to Problem #3:

x = 9.248 + log (1.00 / 0.800)

x = 9.248 + 0.097 = 9.345

Discussion about the three above problems.

1) The pH of this buffer IS NOT the pH of a solution of ammonia only. The pH of a 1.00 molar solution of ammonia is 11.624.

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

3) Notice that, in example #1, the pH is less than the solution of just the pure base (9.258 as compared to 11.624). This is due to LeChatelier's Principle. Consider the ionization equation for ammonia:

NH3 + H2O ⇌ NH4+ + OH¯

Increasing the concentration of the ammonium (NH4+) will push the equiibrium back to the left, decreasing the concentration of OH¯. This makes the solution more acidic, making the pH of the buffer smaller than the pure ammonia solution.

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

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