Question

For 540.0mL of a buffer solution that is 0.165M in CH3CH2NH2 and 0.155M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020mol of HCl.

Answer 1

we know that

Ch3CH2NH2 and CH3CH2NH3Cl form a basic buffer

for a basic buffer

pOH = pKb + log [ conjugate acid / base]

so

pOH = pKb + log [CH3CH2NH3+ / CH3CH2NH2]

pKb for CH3CH2NH2 is 3.19

using given values

pOH = 3.19 + log [ 0.155/0.165]

pOH = 3.1628

we know that

pH = 14 - pOH

so

pH = 14 - 3.1628

pH = 10.8372

so the initial pH is 10.8372

now

0.02 mol of HCl is added

we know that

moles = molarity x volume(L)

given

540 ml of buffer solution

moles of CH3CH2NH3+ = 0.155 x 0.54 = 0.0837

moles of CH3CH2NH2 = 0.165 x 0.54 = 0.0891

now the reaction is

CH3CH2NH2 + HCl ---> CH3CH2NH3Cl

from the above reaction

moles of CH3CH2NH2 reacted = moles of HCl added = 0.02

moles of CH3CH2NH2 remaining = 0.0891 - 0.02 = 0.0691

moles of CH3CH2NH3+ formed = moles of CH3CH2NH2 reacted = 0.02

new moles of CH3CH2NH3+ = 0.0837 + 0.02 = 0.1037

now

pOH = pKb + log [CH3CH2NH3+ / CH3CH2NH2]

we kow that

molarity = moles / volume

as the final volume is same for both , they cancel out

so the ratio of molarities = ratio of moles

so

we get

pOH = 3.19 + log [ 0.1037 / 0.0691]

pOH = 3.3663

pH = 14 - pOH

pH = 14 - 3.3663

pH = 10.6337

so

the final pH is 10.6337