Nicole, a botanist, discovered a monoprotic acid in an extract from a tropical p
ID: 844204 • Letter: N
Question
Nicole, a botanist, discovered a monoprotic acid in an extract from a tropical plant and performed the following experiments: She titrates 100.0 mL of the plant extract with 0.100 M NaOH. The titration midpoint is reached at 25.6 mL of NaOH. The pH at this point is 3.59. By evaporating the solvent from the extract, she determines that there is only 0.124 g of acid per 10.0 mL of plant extract. What is the Ka value for the acid? What is the pH of the plant extract? What is the pH after the following additions of 0.100 M NaOH: 30.0 mL, 51.2 mL, 55.0 mL ? What is the molecular weight of the plant extract?
Explanation / Answer
v1 = 100ml
N2 = 0.1
V2 = 25.6
0.124g acid in 10ml extract So in 100ml = 0.124/10*100 g acid = 1.24g acid
1.24g acid + 0.1M 25.5 ml NaOH giving pH 3.59
((1.24g/M)/1000*100)acid +( 0.1/1000*25) base giving pH = -log (H) =3.59
X-0.0025=0.0275
X=0.029
((1.24g/M)/1000*100)acid giving 0.029 H+
Ka = [conc of H]2/ [CONC OF ACID]-[H] = (0.029)2 / (0.124-0.029) = 0.009
pH of plant extract = -log (0.029) = 1.53
0.1M NaOH 30 ml = 0.1/1000*30
pH = -log (H+) = -log (0.029-(0.1/1000*30)) = 3.65
0.1M NaOH 51.2ml = 0.1/1000*51.2
pH = -log (0.029-(0.1/1000*51.2)) = 3.73
0.1M NaOH 55ml = 0.1/1000*55
pH = -log (0.029-(0.1/1000*55)) = 3.75
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.