Question

Why is the change in pH of 3.00 mL of 0.100M pure acetic acid (weak acid that does not ionize fully) greater than the change in pH of a 3.00 mL sample of a .100 M acetic acid + 0.500-gram acetate buffer solution when you add 2 drops (1x10^-4 L) of 3M NaOH? Please show with equations and calculations!

Answer 1

Answer – We are given, [CH₃COOH] = 0.100 M, volume = 3 mL

Acetic acid is weak acid and it does not dissociate completely, so we need to put ICE

    CH₃COOH + H₂O ------> H₃O⁺ + CH₃COO^-

I    0.100 0               0

C    -x                                 +x            +x

E 0.100-x                         +x             +x

Ka = [H₃O⁺] [CH₃COO^-] / [CH₃COOH]

1.8*10^-5 = x*x /(0.100-x)

1.8*10^-5 *(0.100-x) = x²

Now we know the Ka value is too small, so we can neglect x in the 0.100-x

1.8*10^-5*0.100 = x²

x = 0.00134M

x = [H₃O⁺] = 0.00134 M

pH = -log [H₃O⁺]

      = -log 0.00134 M

       = 2.87

Now we need to calculate change in pH after added 2 drops (1x10^-4 L) of 3M NaOH

Moles of acetic acid = 0.100 M * 0.003 L

                                    = 0.0003 moles

Moles of NaOH = 3 M * 1*10^-4 L

                         = 0.0003 moles

Reaction –

    CH₃COOH + NaOH ------> H₂O + CH₃COONa

      0.0003     0.0003                          0.0003

so mole of both acid and base are equal, so there is formation of conjugate baase only

moles of CH₃COO^- = 0.0003 moles

total volume = 3 mL + 1.0*10^-7 mL = 3 mL

[CH₃COO^-] = 0.0003 moles / 0.003 L

                    = 0.100 M

Now ICE chart for conjugate base

    CH₃COO^- + H₂O ------> HO^- + CH₃COOH

I    0.100 0               0

C    -x                              +x            +x

E 0.100-x                         +x             +x

Kb = [HO^-] [CH₃COOH] / [CH₃COO^-]

Kb = 1*10^-14 / 1.8*10^-5 = 5.56*10^-10

5.56*10^-10= x*x /(0.100-x)

5.56*10^-10*(0.100-x) = x²

Now we know the Kb value is too small, so we can neglect x in the 0.100-x

5.56*10^-10*0.100 = x²

x = 7.45*10^-6 M

x = [HO^-] = 7.45*10^-6 M

pOH = -log [HO^-]

      = -log 7.45*10^-6 M

       = 5.12

pH = 14 – 5.12

      = 8.87

so pH change = 8.87 – 2.87

                          = 6

so pH difference is 6

Now we need to calculate pH difference for the buffer solution –

3.00 mL sample of a .100 M acetic acid + 0.500-gram acetate buffer solution

Moles of acetate CH₃COO^- = 0.500 g / 59.045 g.mol^-1

= 0.00847 moles

[CH₃COO^-] = 0.00847 moles / 0.003 L

                   = 2.82 M

Now using the Henderson Hasselbalch equation

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

      = 4.74 + log 2.82 M /0.100 M

      = 4.74 + 1.45

       = 6.19

Now pH after added NaOH

Moles of acetic acid = 0.100 M * 0.003 L

                               = 0.0003 moles

Moles of NaOH = 3 M * 1*10^-4 L

                         = 0.0003 moles

Moles of CH₃COO^- = 0.00847 moles

When we added NaOH moles of base increase and moles of acid decreased

Moles of acetic acid = 0.0003 – 0.0003 moles = 0

Moles of CH₃COO^- = 0.00847 moles + 0.0003 = 0.00877 moles

total volume = 3 mL + 1.0*10^-7 mL = 3 mL

[CH₃COO^-] = 0.00877 moles / 0.003 L

                    = 2.92 M

Now ICE chart for conjugate base

    CH₃COO^- + H₂O ------> HO^- + CH₃COOH

I    2.92 0               0

C    -x                             +x            +x

E 2.92-x                         +x             +x

Kb = [HO^-] [CH₃COOH] / [CH₃COO^-]

Kb = 1*10^-14 / 1.8*10^-5 = 5.56*10^-10

5.56*10^-10= x*x /(2.92-x)

5.56*10^-10*(2.92-x) = x²

Now we know the Kb value is too small, so we can neglect x in the 2.92-x

5.56*10^-10*2.92 = x²

x = 4.02*10^-5 M

x = [HO^-] = 4.02*10^-5 M

pOH = -log [HO^-]

      = -log 4.02*10^-5 M

       = 4.39

pH = 14 – 4.39

      = 9.60

so pH change = 9.60 – 6.19 = 3.41

so pH change of 3.00 mL of 0.100M pure acetic acid is 6 and pH change of a 3.00 mL sample of a .100 M acetic acid + 0.500-gram acetate buffer solution is 3.41. So 3.00 mL of 0.100M pure acetic acid greater than the change in pH of a 3.00 mL sample of a .100 M acetic acid + 0.500-gram acetate buffer solution.