Question

What is the pH of a buffer that consists of 0.254 M CH3CH2COONa and 0.329 M CH3CH2COOH? Ka of propanoic acid, CH3CH2COOH is 1.3 x 10-5 Enter your answer with two decimal places.

Answer 1

Answer:

4.77 is the pH of the given buffer .

Explanation:

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

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

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

$pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})$

We are given:

$K_a$ = Dissociation constant of propanoic acid = $1.3\times 10^{-5}$

$[CH_3CH_2COONa]=0.254 M$

$[CH_3CH_2COOH]=0.329 M$

pH = ?

Putting values in above equation, we get:

$pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})$

pH = 4.77

4.77 is the pH of the given buffer .