Question

25.00 mL 0.100 M CH3NH2 is titrated with 0.100 M HCl. Calculate the pH of this the titration solution after the addition of a) 12.5 mL and b) 25.00 mL of the titrant has been added. Based on these pH values, select an appropriate indicator for the titration from table at the end of lecture notes for Chapter 16.

Answer 1

millimoles of CH3NH2 = 25 x 0.1 = 2.5

a) millimoles of HCl = 12.5 x 0.1 = 1.25

CH3NH2 + HCl ---------------------> CH3NH3+Cl-

2.5              1.25                                0        ---------------------> initial

1.25              0                                  1.25 ------------------------> equilibrium

here base CH3NH2 and salt CH3NH3+Cl- remained in the solution. so it can form buffer.

and this also half-equivlence point.

Kb of CH3NH2 = 4.2 x 10^-4

pKb = -log (4.2 x 10-4) = 3.4

pOH = pKb + log [salt/base]

        = 3.4 + log (1.25/1.25)

        = 3.4

pH + pOH = 14

pH = 14- pOH

pH = 14-3.4

pH = 10.60 --------------------------(answer)

alizarine yellow R   indicator is suitable because its pH range 10.2 to 12

b)

millimoles of HCl = 25 x 0.1 =2.5

CH3NH2 + HCl ---------------------> CH3NH3+Cl-

2.5              2.5                                0        ---------------------> initial

0                 0                                   2.5 ------------------------> equilibrium

here salt CH3NH3+Cl- remained in the solution. so we have to go for salt hydrolysis

salt concentration (C) = 2.5 / total volume

                                    = 2.5 / (25+25)

                                     =0.05 M

above salt is formed from weak base and strong acid

in the salt hydrolysis pH can be calculated by formula

pH = 7 - 1/2 [pKb + logC]

pH = 7 - 1/2 [3.4 + log (0.05)]

pH = 5.95 -------------------------------------------> answer

methyl red indicator is suitable for this . because methyl red pH range = 4.4 to 6.2