Q5P - Organic Chemistry (Mcmurry)

Question

Calculate the percentages of dissociated and undissociated forms present in the following solutions:

(a) ${}^{\mathbf{0}\mathbf{.}\mathbf{0010}\mathbf{ }\mathbf{M}}$ glycolic acid (${}^{{\mathbf{HOCH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}}$;${}^{{\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{83}}$ ) at ${}^{\mathbf{pH}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{50}}$

(b) ${}^{\mathbf{0}\mathbf{.}\mathbf{0020}\mathbf{ }\mathbf{M}}$ propanoic acid ( ${}^{{\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{87}}$ ) at ${}^{\mathbf{pH}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{30}}$

Step-by-Step Solution

Answer

a) The percent dissociation of glycolic acid in the given solution at $pH=4.50$ is $82\%$.

(b) The percent dissociation of propanoic acid in the given solution $pH=5.30$ is $73\%$.

Step 1: Calculate the ratio of the conjugate base to the weak acid.

The given pH of this glycolic acid solution is 4.50.

The $p{K}_a$ for a weak glycolic acid is $3.83$.

The given concentration of this glycolic acid solution is $0.0010M$. It can be expressed in the following equation.

$\left[{\mathrm{HOCH}}_2{\mathrm{CO}}_{2}H\right]\mathbf{+}\left[{\mathrm{HOCH}}_2{{\mathrm{CO}}_2}^{-}\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0010}\mathbf{ }\mathbf{M}$            …(1)

Use the Henderson-Hasselbalch expression to find the ratio of the conjugate base to the weak acid as given below:

$\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\frac{\left[{\mathrm{HOCH}}_2{{\mathrm{CO}}_2}^{-}\right]}{\left[{\mathrm{HOCH}}_2{\mathrm{CO}}_{2}H\right]}$                    …(2)

Now insert these values into Equation (2).

$\mathbf{4}\mathbf{.}\mathbf{45}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{83}\mathbf{+}\mathbf{log}\frac{\left[{\mathrm{HOCH}}_2{{\mathrm{CO}}_2}^{-}\right]}{\left[{\mathrm{HOCH}}_2{\mathrm{CO}}_{2}H\right]}$

$\mathbf{log}\frac{\left[{\mathrm{HOCH}}_2{{\mathrm{CO}}_2}^{-}\right]}{\left[{\mathrm{HOCH}}_2{\mathrm{CO}}_{2}H\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{67}$

Take logarithm out on both sides of the equation.

$\frac{\left[{\mathrm{HOCH}}_2{{\mathrm{CO}}_2}^{-}\right]}{\left[{\mathrm{HOCH}}_2{\mathrm{CO}}_{2}H\right]}\mathbf{=}{\mathbf{10}}^{\mathbf{0}\mathbf{.}\mathbf{67}}$

$\frac{\left[{\mathrm{HOCH}}_2{{\mathrm{CO}}_2}^{-}\right]}{\left[{\mathrm{HOCH}}_2{\mathrm{CO}}_{2}H\right]}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{68}$

$\left[{\mathrm{HOCH}}_2{{\mathrm{CO}}_2}^{-}\right]=\mathbf{4}\mathbf{.}\mathbf{68}\mathbf{ }\mathbf{\left[}{\mathbf{HOCH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}\mathbf{\right]}$                …(3)

Step 2: Find the percent dissociation of glycolic acid in this solution.

Substitute Equation (3) into Equation (1) to find $\left[HOC{H}_{2}C{O}_{2}H\right]$.

$\mathbf{\left[}{\mathbf{HOCH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}\mathbf{\right]}\mathbf{+}\left(4.68\left[{\mathrm{HOCH}}_2{\mathrm{CO}}_{2}H\right]\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{010}\mathbf{ }\mathbf{M}$

$\mathbf{\left[}{\mathbf{HOCH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00018}\mathbf{ }\mathbf{M}$                    …(4)

Substitute Equation (4) into Equation (1) to calculate $\left[HOC{H}_{2}C{O_2}^{-}\right]$.

$\mathbf{0}\mathbf{.}\mathbf{00018}\mathbf{ }\mathbf{M}\mathbf{+}\mathbf{\left[}{\mathbf{HOCH}}_{\mathbf{2}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0010}\mathbf{ }\mathbf{M}$

$$\begin{gathered} \mathbf{\left[}{\mathbf{HOCH}}_{\mathbf{2}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0010}\mathbf{ }\mathbf{M}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{00018}\mathbf{ }\mathbf{M} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00082}\mathbf{M} \end{gathered}$$

Use the below expression to find the percent dissociation of glycolic acid in this solution.

$$\begin{gathered} \mathbf{Percent}\mathbf{ }\mathbf{dissociation}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{HOCH}}_{\mathbf{2}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{HOCH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}\mathbf{\right]}\mathbf{+}\mathbf{\left[}{\mathbf{HOCH}}_{\mathbf{2}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{00082}\mathbf{M}}{\mathbf{0}\mathbf{.}\mathbf{0010}\mathbf{M}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{82}\mathbf{\%} \end{gathered}$$

Hence, the percent dissociation of glycolic acid in the given solution at $pH = 4.50$ is $82\%$.

Step 3: Find the ratio of propanoate to propanoic acid.

The given pH of this propanoic acid solution is 5.30.

The pK_a for a weak propanoic acid is 4.87.

The given concentration of this propanoic acid solution is $0.0020M$. It can be expressed in the following equation.

$\left[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CO}}_{2}H\right]\mathbf{+}\left[{\mathrm{CH}}_3{\mathrm{CH}}_2{{\mathrm{CO}}_2}^{-}\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0020}\mathbf{ }\mathbf{M}$            …(5)

Use the Henderson-Hasselbalch expression to find the ratio of propanoate to propanoic acid as given below:

$\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\frac{[{\mathrm{CH}}_3{\mathrm{CH}}_2{{\mathrm{CO}}_2}^{-}]}{[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CO}}_{2}H]}$                    …(6)

Substitute these values into Equation (6).

$\mathbf{5}\mathbf{.}\mathbf{30}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{87}\mathbf{+}\mathbf{log}\frac{[{\mathrm{CH}}_3{\mathrm{CH}}_2{{\mathrm{CO}}_2}^{-}]}{[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CO}}_{2}H]}$

$\mathbf{log}\frac{[{\mathrm{CH}}_3{\mathrm{CH}}_2{{\mathrm{CO}}_2}^{-}]}{[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CO}}_{2}H]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{43}$

Take logarithm out on both sides of the equation.

$\frac{[{\mathrm{CH}}_3{\mathrm{CH}}_2{{\mathrm{CO}}_2}^{-}]}{[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CO}}_{2}H]}\mathbf{=}{\mathbf{10}}^{\mathbf{0}\mathbf{.}\mathbf{43}}$

$\frac{[{\mathrm{CH}}_3{\mathrm{CH}}_2{{\mathrm{CO}}_2}^{-}]}{[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CO}}_{2}H]}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{69}$

$\mathbf{[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{]}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{69}\mathbf{[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}\mathbf{]}$                …(7)

Step 4: Calculate the percent dissociation of propanoic acid in this solution.

Substitute Equation (7) into Equation (5) to obtain $\left[C{H}_{3}C{H}_{2}C{O}_{2}H\right]$.

$\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}\mathbf{\right]}\mathbf{+}\left(2.69\left[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CO}}_{2}H\right]\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0020}\mathbf{ }\mathbf{M}$

$\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00054}\mathbf{ }\mathbf{M}$                    …(8)

Substitute Equation (8) into Equation (5) to determine $\left[C{H}_{3}C{H}_{2}C{O_2}^{-}\right]$

$\mathbf{0}\mathbf{.}\mathbf{00054}\mathbf{ }\mathbf{M}\mathbf{+}\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0020}\mathbf{ }\mathbf{M}$

$$\begin{gathered} \mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0020}\mathbf{ }\mathbf{M}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{00054}\mathbf{ }\mathbf{M} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00146}\mathbf{ }\mathbf{M} \end{gathered}$$

Use the following formula to calculate the percent dissociation of propanoic acid in this solution.

$$\begin{gathered} \mathbf{Percent}\mathbf{ }\mathbf{dissociation}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}\mathbf{H}\mathbf{\right]}\mathbf{+}\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{00146}\mathbf{M}}{\mathbf{0}\mathbf{.}\mathbf{0020}\mathbf{M}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{73}\mathbf{\%} \end{gathered}$$

Therefore, the percent dissociation of propanoic acid in the given solution at $pH = 5.30$ is $73\%$.