A new compound has been isolated from an unusual plant and has been shown to hav

Question

A new compound has been isolated from an unusual plant and has been shown to have properties that make this chemical act as a weak monoprotic base. A 0.0356 g pure sample of this compound is dissolved in 25.00 mL of dis- tilled water. The pH of the water after the compound fully dissolves is found to be 11.42. This sample is then titrated with 0.01000 M HCl and gives an end point at pH 6.60 and after the addition of 22.67 mL of titrant. After the addition of 9.45 mL and 16.03 mL titrant, the pH of the sample/titrant mixture is determined to be 11.02 and 10.49, respectively. Estimate the molar mass and pKb for this compound. If you don't get the same pKb from all your calculation which value you think is the most reliable. s, state

Explanation / Answer

Lets denote base as BOH.

To get the end point,

concentration of HCl used = 0.01000 M

Volume of HCl used = 22.67 ml = 0.02267 L

Moles of HCl used = concentration*volume = 0.01000*0.02267 = 2.267*10-4 mol

Moles of HCl = moles of BOH = 2.267*10-4 mol

Mass of BOH used = 0.0356 g

Molar mass of BOH = mass/moles = 0.0356/2.267*10-4 = 157.04 g/mol

Molar mass of base = 157.04 g/mol

Initially,

Moles of BOH = 2.267*10-4 mol

Volume = 25.00 ml = 0.025 L

Concentration = moles/volume = 2.267*10-4 mol/0.025 = 0.0091 M

pH = 11.42

pOH = 14-11.42 = 2.58

[OH-] = 10-2.58 = 0.0026 M

BOH -----> B+ + OH-

ICE table:

Kb = [B+][OH-]/[BOH]

= 0.0026*0.0026/0.0065 = 0.00104

pKb = -log Kb = -log (0.00104) = 2.98

pKb = 2.98

After addition of 9.45 ml,

Total volume = 9.45+25.00 = 34.45 ml = 0.03445 L

Moles of HCl added = 0.01000*0.00945 = 9.45*10-5 mol

moles of BOH = 2.267*10-4 mol

Moles of BOH reacted = 9.45*10-5 mol

concentration = 9.45*10-5 mol/0.03445 L = 0.0027 M

moles of BOH remaining = 2.267*10-4 mol - 9.45*10-5 mol = 1.32*10-4 mol

concentration of BOH = moles/volume = 1.32*10-4 mol/0.03445 L = 0.0038 M

pH = 11.02

pOH = 14-11.02 = 2.98

[OH-] = 10-2.98 = 0.00105 M

ICE table:

Kb = [B+][OH-]/[BOH]

= 0.00105*0.00375/0.00275 = 0.00143

pKb = -log Kb = -log (0.00143) = 2.84

pKb = 2.84

After addition of 16.03 ml,

Total volume = 16.03+25.00 = 41.03 ml = 0.04103 L

Moles of HCl added = 0.01000*0.01603 = 1.60*10-4 mol

moles of BOH = 2.267*10-4 mol

Moles of BOH reacted = 1.60*10-4 mol

concentration = 1.60*10-4 mol/0.04103 L = 0.0039 M

moles of BOH remaining = 2.267*10-4 mol - 1.60*10-4 mol = 0.667*10-4 mol

concentration of BOH = moles/volume = 0.667*10-4 mol/0.04103 L = 0.0016 M

pH = 10.49

pOH = 14-10.49 = 3.51

[OH-] = 10-3.51 = 0.00031 M

ICE table:

Kb = [B+][OH-]/[BOH]

= 0.00421*0.00031/0.00129 = 0.0010

pKb = -log Kb = -log (0.0010) = 3

pKb = 3

The value which is most reliable is the value we got initially, without addition of any HCl.

So, pKb = 2.98

[BOH] [B+] [OH-] Initial 0.0091 0 0 Change -0.0026 +0.0026 +0.0026 Equilibrium 0.0065 0.0026 0.0026

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