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rusak2 [61]
3 years ago
13

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.
Chemistry
1 answer:
Shtirlitz [24]3 years ago
7 0

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 .

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