Question

Consider the titration of 25.00 mL of 0.115M NH3 (ka= 5.70e-10 ; kb= 1.75e-5) with 0.0920 M HCl.

a) calculate the pH of the solution for the following volume of HCl: 6.25 mL

b) calculate the pH of the solution for the following volume of HCl: 31.25 mL

c) calculate the pH of the solution for the following volume of HCl: 50.0 mL

d) Which indicator from the table minimizes the titration error and why?

blue Indicator thymol blue bromophenol blue methyl orange | bromocresol green methyl red bromocresol purple bromothymol blue phenol red cresol red thymol blue phenolphthalein pka 1.7 4.1 3.7 4.7 5 Acid Color red yellow red yellow red yellow yellow yellow yellow yellow colorless Base Color pH Range yellow 1.2-2.8 3.0-4.6 yellow 3.1-4.4 blue | 3.8-5.4 yellow 4.2-6.3 purple 5.2-6.8 blue 6.0-7.6 blue 6.8-8.4 7.2-8.8 8.0-9.6 6.1 yellow red red 7.1 7.8 8.2 8.9 9.6 red 8.3-10.0

Answer 1

pKb of NH3 = -Log(1.75*10^-5) = 4.76

pKa of NH3 = -Log(Ka) = -Log(5.7*10^-10) = 9.24

Henderson-Hasselbulch equation: pH = pKa + Log([NH3]/[NH4Cl])

Millimoles of NH3 = 25 mL * 0.115 mmol/mL = 2.875 mmol

a) Millimoles of HCl = 6.25 mL * 0.092 mmol/mL = 0.575 mmol

i.e. pH = 9.24 + Log{(2.875-0.575)/0.575} = 9.85

b) Millimoles of HCl = 31.25 mL * 0.092 mmol/mL = 2.875 mmol

i.e. [NH4Cl] = 2.875 mmol/(25+31.25) mL = 0.0511 M

Here, pH = 7 - 1/2 (pKb + Log[NH4Cl])

i.e. pH = 7 - 1/2 {4.76 + Log(0.0511)} = 5.27

c) Millimoles of HCl = 50 mL * 0.092 mmol/mL = 4.6 mmol

Now, [HCl] = [H⁺] = (4.6-2.875) mmol/(25+50) mL = 0.023 M

Therefore, pH = -Log[H⁺] = -Log(0.023) = 1.64

d) Methyl orange minimizes the titration error because it is the basic indicator (MeOH) and the titration involved strong acid (HCl).