Question

For propanoic acid (HC₃H₅O₂, K_a= 1.3 ×10^-5), determine the concentration of all species present, the pH, and the percent dissociation of a 0.100 M solution.

Short answer

Propanoic acid dissociate into [H⁺] and [C₃H₅O₂^-] ions.

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]=\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{{\text{O}}_{\text{2}}}^{\text{-}}\right]=1.14\times {10}^{-3}\text{\hspace{0.33em}M}\\ \left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\right]=0.09886\text{\hspace{0.33em}M}\\ \text{pH}=2.943\end{array}$

$\begin{array}{l}\text{Percent\hspace{0.33em}Dissociation\hspace{0.33em}=\hspace{0.33em}}1.14\\ \end{array}$

Step-by-step solution

Calculating theAmount Dissociated 

Propanoic acidsolutiondissociates as,

$\begin{array}{l}{\text{HC}}_3{\text{H}}_{\text{5}}{{\text{O}}_{\text{2}}}_{\text{(aq)}}\rightleftarrows {{\text{\hspace{0.33em}H}}^{\text{+}}}_{\text{(aq)}}{\text{+C}}_3{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}\\ 0.10000\\ 0.10-\text{x\hspace{0.33em}\hspace{1em}\hspace{1em}\hspace{1em}x\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

$\begin{array}{l}{\text{K}}_a=\frac{{\text{x}}^2}{0.10-x}=1.3\times {10}^{-5}\\ {\text{x}}^2=1.3\times {10}^{-6}\Rightarrow \text{x}=1.14\times {10}^{-3}\text{\hspace{0.33em}M}\end{array}$

(K_a is small, so 0.1-x ~0.1)

Calculating pH and Concentration of all the Species

xis the amount dissociated or the amount of [H⁺] and [C₃H₅O₂^-] ions formed.

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]=\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{{\text{O}}_{\text{2}}}^{\text{-}}\right]=x=1.14\times {10}^{-3}\text{\hspace{0.33em}M}\\ \left[{\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\right]=0.1-x=0.1-0.00114=0.09886\text{\hspace{0.33em}M}\\ \text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]=-\log (1.14\times {10}^{-3})=2.943\end{array}$

Calculation of Percentage Dissociation

$\begin{array}{l}\text{Percent\hspace{0.33em}Dissociation\hspace{0.33em}=\hspace{0.33em}}\frac{\text{Amount\hspace{0.33em}Dissociated}}{\text{Total\hspace{0.33em}Amount}}\times 100=\frac{1.14\times {10}^{-3}}{0.1}\times 100=1.14\%\\ \end{array}$