1. Imagine that you start with 17.4 mL of a 0.26 M solution of CH 3 COOH (acetic

Question

1. Imagine that you start with 17.4 mL of a 0.26 M solution of CH3COOH (acetic acid) and you titrate it to the equivalence point with a 0.4M solution of barium hydroxide. What is the pH at the equivalence point of this titration?

2. When you titrate 11.6 mL of a 0.46 M solution of CH3CH2CH2NH2 (propylamine), it required 16.0 mL of hydrochloric acid solution. Assuming that you have reached the equivalence point, what is the pH of the resulting solution?

Weak Acid Ka Weak Base Kb CH3COOH (Acetic Acid) 1.8 X 10-5 NH3 (Ammonia) 1.76 X 10-5 C6H5COOH (Benzoic Acid) 6.5 X 10-5 C6H5NH2 (Aniline) 3.9 X 10-10 CH3CH2CH2COOH (Butanoic Acid) 1.5 X 10-5 (CH3CH2)2NH (Diethyl amine) 6.9 X 10-4 HCOOH (Formic Acid) 1.8 X 10-4 C5H5N (Pyridine) 1.7 X 10-9 HBrO (Hypobromous Acid) 2.8 X 10-9 CH3CH2NH2 (Ethyl amine) 5.6 X 10-4 HNO2 (Nitrous Acid) 4.6 X 10-4 (CH3)3N (Trimethyl amine) 6.4 X 10-5 HClO (Hypochlorous Acid) 2.9 X 10-8 (CH3)2NH (Dimethyl amine) 5.4 X 10-4 CH3CH2COOH (Propanoic Acid) 1.3 X 10-5 CH3CH2CH2NH2 (Propyl amine 3.5 X 10-4 HCN (Hydrocyanic Acid) 4.9 X 10-10

Explanation / Answer

1. at equivalence point

No of moles of CH3COOH = nO OF MOLES OF Ba(OH)2

No of moles of CH3COOH = 17.4/1000*0.26 = 0.00454 mol

No of moles of Ba(OH)2 = 0.00454 mol

volume of Ba(OH)2 = 0.00454/0.4 = 11.35 ml

concentration of Ba(OH)2 = 11.35/(11.35+17.4)*0.4 = 0.16 M

pH = 7+1/2(pka+logC)

   = 7+1/2(4.74+log0.16)

   = 8.97

2.

concentration of salt = 11.6/(11.6+16)*0.46 = 0.193

pH = 7-1/2(pkb+log C)

= 7-1/2(4.7+log0.193)

= 5

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