1.

What is the difference in pH for 1/3 and 2/3 stages of neutralisation of 0.1 M CH3COOH with 0.1 M NaOH.

Answer»

CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) ….. Ka = 1.8 × 10⁻⁵ 

HENDERSON Hasselbalch equation : 
pH = pKa + log([CH₃COO⁻]/[CH₃COOH]) 

In the 1/3 STAGE of the NEUTRALIZATION : 
1/3 of CH₃COOH is changed to CH₃COO⁻. 
Hence, [CH₃COO⁻]/[CH₃COOH] = (1/3) / [1 - (1/3)] = 1/2 
pH = pKa + log([CH₃COO⁻]/[CH₃COOH]) 
pH = -log(1.8 × 10⁻⁵) + log(1/2) 
pH = 4.44 
 In the 2/3 stage of the neutralization : 
2/3 of CH₃COOH is changed to CH₃COO⁻. 
Hence, [CH₃COO⁻]/[CH₃COOH] = (2/3) / [1 - (2/3)] = 2 
pH = pKa + log([CH₃COO⁻]/[CH₃COOH]) 
pH = -log(1.8 × 10⁻⁵) + log(2) 
pH = 5.05   ..


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