Question

For each solution, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

B.) 520.0 mL of a buffer solution that is 0.120 M in HC2H3O2 and 0.100 M in NaC2H3O2

pHinitial =

pHfinal =

C.) 520.0 mL of a buffer solution that is 0.150 M in CH3CH2NH2 and 0.130 M in CH3CH2NH3Cl

pHinitial =

pHfinal =

Answer 1

B.) 520.0 mL of a buffer solution that is 0.120 M in HC2H3O2 and 0.100 M in NaC2H3O2

moles of HC2H3O2 = 0.120 x 520 / 1000 = 0.0624

moles of NaC2H3O2 = 0.100 x 520 / 1000 = 0.052

pKa = 4.74

pHinitial = pKa + log [NaC2H3O2 / HC2H3O2]

pH = 4.74 + log (0.052 / 0.0624)

pH = 4.66

on addition of C moles strong acid HCl

C = 0.02

pHfinal = pKa + log [NaC2H3O2 -C / HC2H3O2 +C]

pH = 4.74 + log (0.052 -0.02 / 0.0624 + 0.02)

pH = 4.33

C)

moles of base = 0.150 x 520 / 1000 = 0.078

moles of salt = 0.130 x 520 / 1000 = 0.0676

pOH = pKb + log [salt / base]

pOH = 3.2 + log [0.0676 / 0.078]

pOH = 3.14

pH + pOH = 14

pH initial = 10.86

pH final :

on addition of C moles HCl

pOH = pKb + log [salt +C / base-C]

pOH = 3.2 + log [0.0676+0.020 / 0.078-0.020]

pOH = 3.38

pH + pOH = 14

pH final = 10.62