Question

Which base in Table 14.3 is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.

Short answer

\(C{H_3}N{H_2} \) would be the best choice for our buffer solution.

Step-by-step solution

Which acid is appropriate for preparation of buffer solution

A buffer solution of any pH value can, in theory, be made using any acid- base pair and modifying the ratio of acid to base concentration, according to the Henderson-Hasselbach equation. However, not all of these buffers would have the same capacity. A capacity of a buffer is definedas the amount of acid or base that can be added to the buffer solution before a significant change in pH occurs. This capacity depends on the concentrations of the acid and the base component of the buffer. Therefore, we must choose the components of our buffer to fit the desired pH of the solution. A general rule is, choose the closer the pKa of the acid component to the desired pH, the better a choice the buffer is. Once we know this, we can conclude that we need to determine the pK_a values for the acidic component of each of the buffers in the table and see which value is the closest to 10.65. However, we only have K_bvalues in the table. We have to first calculate the Ka values from the K_b values. We do this using the equation:

\({K_a} = \frac{{{K_w}}}{{{K_b}}}\)

Where K_w is the ionic water product, a constant that equals 10^-14.

\({K_a}\left( {{{\left( {C{H_3}} \right)}_2}NH_2^ + } \right) = \frac{{{K_w}}}{{{K_b}\left( {{{\left( {C{H_3}} \right)}_2}NH} \right)}}\)

\(\begin{align}{K_a}\left( {{{\left( {C{H_3}} \right)}_2}NH_2^ + } \right) &= \frac{{1{0^ - }14}}{{5.9 \times 1{0^{ - 4}}}}\\{K_a}\left( {{{\left( {C{H_3}} \right)}_2}NH_2^ + } \right) &= 1.7 \times 1{0^{ - 11}}\\{K_a}\left( {C{H_3}NH_3^ + } \right) &= \frac{{{K_w}}}{{{K_b}\left( {C{H_3}N{H_2}} \right)}}\end{align}\)

\(\begin{align}{K_a}\left( {C{H_3}NH_3^ + } \right) &= \frac{{1{0^{ - 14}}}}{{4.4 \times 1{0^{ - 4}}}}\\{K_a}\left( {C{H_3}NH_3^ + } \right) &= 2.3 \times 1{0^{ - 11}}\end{align}\)

\(\begin{align}{K_a}\left( {{{\left( {C{H_3}} \right)}_3}N{H^ + }} \right) &= \frac{{{K_w}}}{{{K_b}\left( {{{\left( {C{H_3}} \right)}_3}\;N} \right)}}\\{K_a}\left( {{{\left( {C{H_3}} \right)}_3}N{H^ + }} \right) &= \frac{{1{0^{ - 14}}}}{{6.3 \times 1{0^{ - 5}}}}\\{K_a}\left( {{{\left( {C{H_3}} \right)}_3}N{H^ + }} \right) &= 1.6 \times 1{0^{ - 10}}\\{K_a}\left( {NH_4^ + } \right) &= \frac{{{K_w}}}{{{K_b}\left( {N{H_3}} \right)}}\\{K_a}\left( {NH_4^ + } \right) = \frac{{1{0^{ - 14}}}}{{1.8 \times 1{0^{ - 5}}}}\\{K_a}\left( {NH_4^ + } \right) &= 5.6 \times 1{0^{ - 10}}\\{K_a}\left( {{C_6}{H_5}NH_3^ + } \right) &= \frac{{{K_w}}}{{{K_b}\left( {{C_6}{H_5}N{H_2}} \right)}}\\{K_a}\left( {{C_6}{H_5}NH_3^ + } \right) &= \frac{{1{0^{ - 14}}}}{{4.3 \times 1{0^{ - 10}}}}\\{K_a}\left( {{C_6}{H_5}NH_3^ + } \right) &= 2.3 \times 1{0^{ - 5}}\end{align}\)

Now that we know the K_avalues, we can calculate the pK_a value for each acid using the formula:

\(\begin{align}p{K_a} &= - log{K_a}\\p{K_a}\left( {{{\left( {C{H_3}} \right)}_2}NH_2^ + } \right) &= - log\left( {1.7 \times 1{0^{ - 11}}} \right)\\p{K_a}\left( {{{\left( {C{H_3}} \right)}_2}NH_2^ + } \right) &= 10.77\\p{K_a}\left( {C{H_3}NH_3^ + } \right) &= - log\left( {2.3 \times 1{0^{ - 11}}} \right)\\p{K_a}\left( {C{H_3}NH_3^ + } \right) &= 10.64\end{align}\)

\(\begin{align}p{K_a}\left( {{{\left( {C{H_3}} \right)}_3}N{H^ + }} \right) &= - log\left( {1.6 \times 1{0^{ - 10}}} \right)\\p{K_a}\left( {{{\left( {C{H_3}} \right)}_3}N{H^ + }} \right) &= 9.80\\p{K_a}\left( {NH_4^ + } \right) &= - log\left( {5.6 \times 1{0^{ - 10}}} \right)\\p{K_a}\left( {NH_4^ + } \right) &= 9.25\\p{K_a}\left( {{C_6}{H_5}NH_3^ + } \right) &= - log\left( {2.3 \times 1{0^{ - 5}}} \right)\\p{K_a}\left( {{C_6}{H_5}NH_3^ + } \right) &= 4.64\end{align}\)

We can see from these pK_a value that the pK_a value for \(C{H_3}N{H_2}\) is the closest to our desired pH value of 10.65 and would, therefore, be the best choice for our buffer solution.

Final answer

\(C{H_3}N{H_2} \) would be the best choice for our buffer solution.