Calculate the pH of 0.01 (M) CH 3 COOH at 25Â°C.( Given dissociation

1.	Calculate the pH of 0.01 (M) CH 3 COOH at 25Â°C.( Given dissociation constant of CH 3 COOH(ii)1.7510-5 ).
Answer» pKa = -log(10)[Ka] Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration. pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-] as [H+] = [CH3COO-] pKa = log(10)[CH3COOH] - 2log(10)[H+] For acetic acid pKa = 4.76 and -log(10)[H+] = pH 4.76 = log(10)(0.1) + 2pH => 2*pH = 4.76 + 1 => pH = 5.76/2 => pH = 2.88 The required pH of 0.1 M solution of acetic acid is 2.88

Calculate the pH of 0.01 (M) CH 3 COOH at 25Â°C.( Given dissociation constant of CH 3 COOH(ii)1.7510-5 ).

Answer»

pKa = -log(10)[Ka]

Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.

pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]

as [H+] = [CH3COO-]

pKa = log(10)[CH3COOH] - 2*log(10)[H+]

For acetic acid pKa = 4.76 and -log(10)[H+] = pH

4.76 = log(10)(0.1) + 2*pH

=> 2*pH = 4.76 + 1

=> pH = 5.76/2

=> pH = 2.88

The required pH of 0.1 M solution of acetic acid is 2.88