KEY2CHEM

Designing a Buffer

A buffer is a solution that resists changing its pH when strong acid or base is added to the solution. A buffer consists of a weak conjugate acid-base pair (where both partners in the pair can react with either acid or base, as appropriate) and its pH can be calculated using the Henderson Hasselbalch equation: \(pH = pK_a + log \frac{base}{acid}\), where \(pK_a\) is the \(pK_a\) of the conjugate acid partner of the weak conjugate acid-base pair (remember \(pK_a = -log K_a\)) and the “base” is the base of the weak conjugate acid base pair, and “acid” is the conjugate acid partner in the weak conjugate acid base pair. The buffer capacity is a measure of a buffer’s resistance to changes in pH. The greater the buffer capacity, due to increased absolute concentration of the components as well as the concentrations of the acid and base being closer to one another, the greater the buffer’s capacity.


Example 1.

 

A buffer is composed of \(\require{mhchem}\ce{NH4+}\) and\( \require{mhchem}\ce{NH3}\) in equal concentrations. If the \(K_a\) of \(\require{mhchem}\ce{NH4+} \) is \(5.6 \times 10 ^{-10}\), what is the pH of the buffer?

 

A. 7.00

B. 4.76

C. 9.24

 

 

 

Solution

C. 9.24

 

When \([\require{mhchem}\ce{NH3}] = [\require{mhchem}\ce{NH4+}]\), the ratio of \(\frac{\text{base}}{\text{acid}}\) in the Henderson-Hasselbalch equation is 1 (even if the specific values are not known, the information that they are equal to each other is given), since the concentration values are equal.

\(pH = pK_a + log\frac{[\require{mhchem}\ce{NH3}] }{[\require{mhchem}\ce{NH4+}]}\)

\(pH = -log(K_a) + log(\frac{[\require{mhchem}\ce{NH3}] }{[\require{mhchem}\ce{NH4+}]})\)

\(pH = -log(K_a) + log(1)\)

\(pH = -log(K_a) + 0\)

\(pH = -log(5.6 \times 10 ^{-10}) = 9.24\)

 


Example 2.

 

A buffer containing \( 1.0\text{ M }\) acetic acid (\(\require{mhchem}\ce{CH3COOH}\,, \;pK_a = 4.76\)) and its conjugate base, acetate (\(\require{mhchem}\ce{CH3COO-}\)) has a pH of 5.15. What is the \([\require{mhchem}\ce{CH3COO-}]\) in the buffer?

 

A. \(1.0 \text{ M}\)

B. \(2.5 \text{ M}\)

C. \(0.4 \text{ M} \)

 

 

Solution

 

B. \(2.5 \text{ M} \)

 

\(pH = pK_a + log\frac{[\require{mhchem}\ce{CH3COO-}] }{[\require{mhchem}\ce{CH3COOH}]}\)

\(5.15 = 4.76 + log\frac{[\require{mhchem}\ce{CH3COO-}] }{1.0\text{ M}}\)

\(0.39 = log\frac{[\require{mhchem}\ce{CH3COO-}] }{1.0\text{ M}}\)

\(2.45 = \frac{[\require{mhchem}\ce{CH3COO-}] }{1.0\text{ M}}\)

\(2.45 \text{ M }= [\require{mhchem}\ce{CH3COO-}] \)


Example 3.

If \(pH < pK_a\) for a buffer containing weak acid \(\require{mhchem}\ce{HA}\) and its conjugate base \(\require{mhchem}\ce{A-}\), which statement is true?

 

A. \([\require{mhchem}\ce{HA}] = [\require{mhchem}\ce{A-}]\)

B. \([\require{mhchem}\ce{HA}] > [\require{mhchem}\ce{A-}]\)

C. \([\require{mhchem}\ce{HA}] < [\require{mhchem}\ce{A-}] \)

 

 

 

 

Solution

 

B. \( [\require{mhchem}\ce{HA}] > [\require{mhchem}\ce{A-}]\)

 

\(pH = pK_a + log\frac{[\require{mhchem}\ce{A-}] }{[\require{mhchem}\ce{HA}]}\)

If \(pH < pK_a\), the ratio of \(\frac{[\require{mhchem}\ce{A-}] }{[\require{mhchem}\ce{HA}]}\) \(< 1\), so \([\require{mhchem}\ce{HA}] > [\require{mhchem}\ce{A-}]\)