Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.

For 260.0 mL of a buffer solution that is 0.265 M in CH3CH2NH2 and 0.235 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.

Answer 1

initially

mol of CH3CH2NH2 = MV = 0.265*0.260 = 0.0689

mol of CH3CH2NH3Cl= MV = 0.235*0.260 = 0.0611

initial pH

pOH = pKb + log(CH3CH2NH3Cl/CH3CH2NH2 )

pKb = 3.193

pOH = 3.193+ log(0.235/0.265)

pH = 14-pOH = 14-3.14082 = 10.85918

b)

after 0.005 mol of base

mol of CH3CH2NH2 = 0.0689 + 0.005 = 0.0739

mol of CH3CH2NH3Cl= MV = 0.0611 - 0.005 = 0.0561

pOH = pKb + log(CH3CH2NH3Cl/CH3CH2NH2 )

pKb = 3.193

pOH = 3.193 + log(0.0561/0.0739)

pH = 14-pOH = 14-3.0733

pH = 10.9267