Question

Consider a solution that contains both${\mathit{C}}_{\mathbf{5}}{\mathit{H}}_{\mathbf{5}}\mathit{N}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }{\mathit{C}}_{\mathbf{5}}{\mathit{H}}_{\mathbf{5}}\mathit{N}\mathit{H}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{.}$ Calculate the ratio$\left[C_5{H}_{5}N\right]\mathbf{/}\left[C_5{H}_{5}N{H}^{+}\right]$if the solution has the following pH values

a)$\mathit{p}\mathit{H}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{50}$

b)$\mathit{p}\mathit{H}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{00}$

c)$\mathit{p}\mathit{H}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{23}$

d)$\mathit{p}\mathit{H}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{50}$

Short answer

1. The ratio of $\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}$for $\text{pH}=4.50$is$0.186$.
2. The ratio of$\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}$ for $\text{pH}=5.00$is$0.589$.
3. The ratio of $\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}$for $\text{pH}=5.23$is$1$.
4. The ratio of$\frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}$for $\text{pH}=5.50$is$1.86$.

Step-by-step solution

Definition of pH

The pH is a quantitative measureof the acidity or basicity of aqueous or liquid solutions.

Calculations

a)

$$\begin{gathered} {\text{K}}_{\text{a}}=5.9\times {10}^{-6} \\ {\text{pK}}_{\text{a}}=-\log (5.9\times {10}^{-6}) \\ {\text{pK}}_{\text{a}}=5.23 \\ \text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]} \end{gathered}$$

$$\begin{gathered} 4.5=5.23+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]} \\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}=\text{antilog}(-0.73) \\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}=0.186 \end{gathered}$$

b)

$$\begin{gathered} {\text{K}}_{\text{a}}=5.9\times {10}^{-6} \\ {\text{pK}}_{\text{a}}=-\log (5.9\times {10}^{-6}) \\ {\text{pK}}_{\text{a}}=5.23 \\ \text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]} \end{gathered}$$

$$\begin{gathered} 5.00=5.23+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]} \\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}=\text{antilog}(-0.23) \\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}=0.589 \end{gathered}$$

c)

$$\begin{gathered} {\text{K}}_{\text{a}}=5.9\times {10}^{-6} \\ {\text{pK}}_{\text{a}}=-\log (5.9\times {10}^{-6}) \\ {\text{pK}}_{\text{a}}=5.23 \\ \text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]} \end{gathered}$$

$$\begin{gathered} 5.23=5.23+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]} \\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}=\text{antilog}\left(0\right) \\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}=1 \end{gathered}$$

d)

$$\begin{gathered} {\text{K}}_{\text{a}}=5.9\times {10}^{-6} \\ {\text{pK}}_{\text{a}}=-\log (5.9\times {10}^{-6}) \\ {\text{pK}}_{\text{a}}=5.23 \\ \text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]} \end{gathered}$$

$$\begin{gathered} 5.50=5.23+\log \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]} \\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}=\text{antilog}(0.27) \\ \frac{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\right]}{\left[{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right]}=1.86 \end{gathered}$$