Simplify the expression by clearing the fraction.
5.5 [acid] = 0.100 mol/L - [acid]
Solve the expression for [acid].
5.5 [acid] + [acid] = 0.100 mol/L
6.5 [acid] = 0.100 mol/L
0.100 mol/L
[acid] = --------------
6.5
[acid] = 0.0154 mol/L
or equivalently, 1.54 X 10-2 mol/L
To determine the amount of salt required, substitute the calculated value of
the acid concentration in the expression relating the salt and acid concentration to the
desired mol/L concentration of the buffer.
[salt] = 0.100 mol/L - [acid]
[salt] = 0.100 mol/L - 0.0154 mol/L
[salt] = 0.0846 mol/L
or equivalently, 8.46 x 10-2 mol/L
d. Example 4. What amount of salt/base is required to prepare a 0.100 mol/L
ammonia buffer at a pH of 9.8? The pKb for ammonium hydroxide is 4.75.
Solution. After reading the problem carefully, note that you are to prepare a
basic buffer of a certain pH. Select an expression that will allow you to solve for the
unknown quantity.
[salt]
pOH = pKb + log ------
[base]
Convert pH to pOH.
14 = pH + pOH
pOH = 14 - pH
pOH = 14 - 9.8
pOH = 4.2
MD0837
9-19