Question

What is the ph of a 0.40 m solution of benzoic acid, c6h5cooh? (the ka value for benzoic acid is 6.3×10−5?

Answer 1

Hello!

The dissociation reaction of Benzoic Acid is the following:

C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺

The ka value is expressed as follows:

$Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}$

After the dissociation, we can clear for the concentration of H₃O⁺ (x) in the following way (Considering that x is small).

$Ka= \frac{x*x}{[C_6H_5COOH]_i-x} \\ \\ x= \sqrt{[C_6H_5COOH]_i*Ka}= \sqrt{0,40M*6,3*10^{-5} } \\ \\ x=0,00502M=[H_3O^{+} ]$

So, to finish, we apply the definition of pH

$pH=-log[H_3O^{+}] =-log (0,00502M)=2,30$

So the pH of this solution is 2,30

Have a nice day!

Answer 2

The pH of a 0.40 M solution of benzoic acid is $\boxed{\text{2.30}}$.

Further Explanation:

Dissociation constant is a term used to describe the strengths of various acids. It is denoted by $\text{k}_\text{a}$.

pH is a quantity that is used to predict whether solution is acidic, alkaline or neutral. It is described as negative logarithm of concentration of hydrogen ion. Following is the expression for pH of solution.

$\text{pH}=-\text{log[H}^+]$                                     ...... (1)

Dissociation reaction for benzoic acid is written below.

$\text{C}_6\text{H}_5\text{COOH}\rightleftharpoons\text{C}_6\text{H}_5\text{COO}^{-}+\text{H}^{+}$

The formula for dissociation constant of benzoic acid is as follows:

$\text{k}_\text{a}=\dfrac{[\text{C}_6\text{H}_5\text{COO}^{-}][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]}$                       ...... (2)

Rearrange equation (2) to calculate $[\text{H}^+]$.

$[\text{H}^+]=\text{k}_\text{a}\dfrac{[\text{C}_6\text{H}_5\text{COOH}]}{[\text{C}_6\text{H}_5\text{COO}^{-}]}$                       ...... (3)

Consider x to be the concentration of $\text{H}^+$ ions produced by dissociation of benzoic acid. So concentration of benzoate ion becomes (0.40 M-x). But here, concentration of hydrogen ion is extremely large so 0.40 M can be neglected while writing the concentration term for benzoate ion and therefore its concentration also becomes x.

Substitute x for $[\text{H}^+]$, 0.40 for $[\text{C}_6\text{H}_5\text{COOH}]$, x for $[\text{C}_6\text{H}_5\text{COO}^{-}]$ and $6.3\times10^{-5}$ for $\text{k}_\text{a}$ in equation (3).

$\begin{aligned}{\text{x}&=(6.3\times10^{-5})}\left(\dfrac{\text{0.40 M}}{\text{x}}\right)\\\text{x}^2&=2.52\times10^{-5}\text{ M}}\end{aligned}$

Solving for x,

$\text{x}=\text{0.00502 M}$

Therefore hydrogen ion concentration comes out to be 0.00502 M.

Substitute 0.00502 M for $\text{[H}^+]$ in equation (1).

$\begin{aligned}{\text{pH}&=-\text{log(0.00502 M})\\&=2.30}\end{aligned}$

The pH of benzoic acid solution comes out to be 2.30.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Acids, bases and salts

Keywords: pH, ka, benzoic acid, C6H5COOH, H+, C6H5COO-, acidic, alkaline, neutral, H2O, 2.30, 0.40 M.