Question

A solution is prepared by dissolving 0.56 g of benzoic acid (C₆H₅CO₂H, K_a= 6.4 × 10^-5) in enough water tomake 1.0 L of solution. Calculate [C₆H₅CO₂H], [C₆H₅CO₂^-], [H⁺], [OH^-], and the pH of this solution.

Short answer

Following data was obtained after calculation:

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]=\left[{\text{C}}_6{\text{H}}_{\text{5}}{{\text{CO}}_{\text{2}}}^{\text{-}}\right]=6.083\times {10}^{-4}\text{\hspace{0.33em}\hspace{0.33em}M}\\ \left[{\text{C}}_6{\text{H}}_{\text{5}}{\text{CO}}_{\text{2}}\text{H}\right]=3.992\times {10}^{-3}\text{\hspace{0.33em}M}\\ \text{pH}=3.215\end{array}$

$\left[O{\text{H}}^{-}\right]=1.64\times {10}^{-11}$

Step-by-step solution

Calculating the Concentration of Benzoic Acid

Benzoic acid mass taken = 0.56 g

$\begin{array}{l}\text{Moles\hspace{0.33em}of\hspace{0.33em}Benzoic\hspace{0.33em}Acid}=\frac{{\text{Mass}}^{}}{\text{Molar\hspace{0.33em}Mass}}=\frac{{\text{0.56}}^{}}{\text{122}}=4.6\times {10}^{-3}\\ \text{Molarity}=\frac{\text{Moles}}{\text{Volume(L)}}=\frac{4.6\times {10}^{-3}}{1}=4.6\times {10}^{-3}\text{\hspace{0.33em}M}\end{array}$

Calculating theAmount Dissociated 

Benzoic acidsolutiondissociates as,

$\begin{array}{l}{\text{C}}_6{\text{H}}_{\text{5}}{\text{CO}}_{\text{2}}{\text{H}}_{\text{(aq)}}\rightleftarrows {{\text{\hspace{0.33em}H}}^{\text{+}}}_{\text{(aq)}}{\text{+C}}_6{\text{H}}_{\text{5}}{\text{CO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}\\ 4.6\times {10}^{-3}00\\ 4.6\times {10}^{-3}-\text{x\hspace{0.33em}\hspace{1em}\hspace{1em}\hspace{1em}x\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

$\begin{array}{l}{\text{K}}_a=\frac{{\text{x}}^2}{4.6\times {10}^{-3}-x}=6.4\times {10}^{-5}\\ \\ {\text{x}}^2+6.4\times {10}^{-5}-2.9\times {10}^{-7}=0\Rightarrow \text{x}=6.083\times {10}^{-4}\text{\hspace{0.33em}M}\end{array}$

Calculating pH and Concentration of all the Species 

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

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]=\left[{\text{C}}_6{\text{H}}_{\text{5}}{{\text{CO}}_{\text{2}}}^{\text{-}}\right]=x=6.083\times {10}^{-4}\text{\hspace{0.33em}\hspace{0.33em}M}\\ \left[{\text{C}}_6{\text{H}}_{\text{5}}{\text{CO}}_{\text{2}}\text{H}\right]=0.0046-0.0006083=0.003992\text{\hspace{0.33em}M}\\ \text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]=-\log (6.083\times {10}^{-4})=3.215\end{array}$

$\begin{array}{l}{\text{K}}_{\text{w}}=\left[{\text{H}}^{\text{+}}\right]\left[O{\text{H}}^{-}\right]=1\times {10}^{-14}\\ \left[O{\text{H}}^{-}\right]=\frac{1\times {10}^{-14}}{\left[{\text{H}}^{+}\right]}=\frac{1\times {10}^{-14}}{6.083\times {10}^{-4}}=1.64\times {10}^{-11}\\ \end{array}$