Question

Calculate the ph of a buffer solution that contains 0.25 m benzoic acid (c6h5co2h) and 0.15 m sodium benzoate (c6h5coona). (ka = 6.5 × 10–5 for benzoic acid)

Answer 1

Answer: The pH of the buffer solution is 3.97

Explanation:

$pK_a$ is defined as the negative logarithm of $K_a$

Mathematically,

$pK_a=-\log(K_a)$

We are given:

$K_a=6.5\times 10^{-5}$

Putting values in above equation, we get:

$pK_a=-\log(6.5\times 10^{-5})\\\\pK_a=4.19$

The chemical equation for the reaction of benzoic acid (weak acid) and sodium hydroxide (strong base) follows:

$C_6H_5COOH+NaOH\rightarrow C_6H_5COONa+H_2O$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[C_6H_5COONa]}{[C_6H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.19

$[C_6H_5COONa]=0.15M$

$[C_6H_5COOH]=0.25M$

pH = ?

Putting values in above equation, we get:

$pH=4.19+\log(\frac{0.15}{0.25})\\\\pH=3.97$

Hence, the pH of the buffer solution is 3.97

Answer 2

Hello!

To solve this problem we'll use the Henderson-Hasselbach equation, but first we need the vale for the pKa of Benzoic acid, which is pKa= -log(Ka)=4,19

Now, we apply the equation as follows:

$pH=pKa + log ( \frac{[C_6H_5COONa]}{[C_6H_5COOH]} )=4,19+log( \frac{0,15M}{0,25M} )=3,97$

So, the pH of this solution of Sodium Benzoate and Benzoic Acid is 3,97

Have a nice day!