Question

2. Calculate the pH at the equivalence point when 30 mL of 0.206 M CH₃CH₂NH₂, is titrated with 0.559 M HBr.

Answer 1

CH3CH2NH2 + HBr ---> CH3CH2NH2·HBr

no fo moles of CH3CH2NH2 = 0.206M x 0.03 L = 0.00618 moles

from the balanced equation one equilent of CH3CH2NH2 required one equilent of HBr

0.00618 moles of CH3CH2NH2 required 0.00618 moles of CH3CH2NH2

and 0.00618 moles of salt will form

volume of HBr = 0.00618 mol / 0.559 mol/L

= 0.01105 L

= 11.05 mL

total volume = 30 + 11.5 = 41.5 mL

concentration of CH3CH2NH2·HBr = 0.00618 / 0.0415 = 0.15 M

Kb = 4.3 x 10^-4  from this calculate the Ka using Ka x Kb = 10^-14

Ka of CH3CH2NH2 = 2.3 x 10^-11

from this pKa = -logKa = -log(2.3 x 10^-11 ) = 10.64

now use the direct formula

pH = 1/2 (pKa -logC)

pH = 1/2 (10.64 - log0.15)

pH = 5.73