Question

Analyte: Acetic Acid 0.1M (15 mL) & water (100 mL) Titrant: ammonia 0.1 M (15 mL) Weak Acid: Initial Volume [ml] of itan 50 mL of ammonia Final Volume [mll of Titrant: A volume [ml] Elapsed time [s] Volume titrant/second Amount titrant added [ml] pH (measured) Weak Base: 0 mL of ammonia 50 mL of ammonia 180 seconds 0.28 mL/sec pH (calculated) 0.0 7.5 12 15 20 3.2 8.9 9.4 9.5 9.6 2 Volume at which indicator changed color: None

Answer 1

(i) Volume of titrant added = 0 mL

pH = 1/2 {pKa - Log[CH₃COOH]}

= 1/2 {4.74 - Log(0.1*15/(15+100))}

= 1/2 (4.74 + 1.88)

Therefore, pH = 3.31

(ii) Volume of titrant added = 7.5 mL

pH = 1/2 {pKa - Log[CH₃COOH]}

= 1/2 {4.74 - Log(0.1*(15-7.5)/(15+7.5+100))}

= 1/2 (4.74 + 2.21)

Therefore, pH = 3.48

(iii) Volume of titrant added = 12 mL

pH = 1/2 {pKa - Log[CH₃COOH]}

= 1/2 {4.74 - Log(0.1*(15-12)/(15+12+100))}

= 1/2 (4.74 + 2.63)

Therefore, pH = 3.68

(iv) Volume of titrant added = 15 mL

pH = 7 + 1/2 (pKa - pKb)

Here, pKa of acetic acid = pKb of ammonia

Therefore, pH = 7

(v) Volume of titrant added = 20 mL

pH = 14 - 1/2 {pKb - Log[NH₃]}

= 14 - 1/2 {4.74 - Log(0.1*(20-15)/(15+20+100))}

= 14 - 1/2 (4.74 + 2.43)

= 14 - 3.59

Therefore, pH = 10.41