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

4.77 is the pH of the given buffer .

Step-by-step explanation:

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

`pH=pK_a+\log(([salt])/([acid]))`

`pH=-\log[K_a]+\log(([salt])/([acid]))`

`pH=-\log[K_a]+\log(([CH_3CH_2COONa])/([CH_3CH_2COOH]))`

We are given:

`K_a` = Dissociation constant of propanoic acid =
`1.3* 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* 10^(-5)]+\log(([0.254 M])/([0.329]))`

pH = 4.77

4.77 is the pH of the given buffer .