Question

For 460.0 mL of a buffer solution that is 0.150 M in CH3CH2NH2 and 0.135 M in CH3CH2NH3Cl. Calculate the inital pH and the final pH after adding 0.020 mol of HCl.

Express your answer using two decimal places separated by a comma.

Answer 1

this is a buffer so

CH3CH2NH3Cl -->CH3CH2NH3+

CH3CH2NH2

initial pH

pKb = 3.35 from data and text

pOH = pKb + log(CH3CH2NH3+/CH3CH2NH2)

pOH = 3.35 + log(0.135/0.15) = 3.3042425

pH = 14-pOH = 14-3.3042425 = 10.69575

after addition of HCl

mmol of base initially = MV = 460*0.15 = 69 mmol of base

mmol of conjugate = MV = 460*0.135 = 62.1 mmol of conjugate

after adding HCl...

mmol = 0.02*10^3 = 20 mmol

base will react , therefore decreases, and form conjguate, which increases

then

mmol of base initially= 69 -20 = 48

mmol of conjugate = 62.1 +20 = 42.1

then

pOH = pKb + log(CH3CH2NH3+/CH3CH2NH2)

pOH = 3.35 + log(42.1/48) = 3.293

pH = 14-pOH = 14-3.293= 10.707