Question

1. A buffer is made by adding 0.150 mol of CH3COOH and 0.250 mol of NaCH3COO to enough water to make 1.00 L of solution.

a. Calculate the pH of the buffer solution.

b. Calculate the pH of this solution after 35.0 mL of 0.50M NaOH is added to the buffer solution.

c. Calculate the pH of this solution after 35.0 mL of 5.0 M NaOH is added to the buffer solution.

Answer 1

Answer – We are given, moles of CH₃COOH = 0.150 moles

Moles of NaCH₃COO = 0.250 moles , volume = 1.0 L

So, [CH₃COOH] = 0.150 moles / 1.0 L = 0.150 M

[CH₃COO^-] = 0.250 mol / 1.0 L = 0.250 M

We know the pKa value for the CH₃COOH and it is 4.75

Now we need to use Henderson Hasselbalch equation

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

      = 4.75 + log 0.250 / 0.150

      = 4.97

b) pH after added 35.0 mL of 0.50M NaOH

moles of NaOH = 0.50 M * 0.035 L = 0.0175 moles

when we added NaOH the moles of acid decrease and its conjugate base increase

moles of CH₃COOH = 0.150 moles – 0.0175 mole = 0.133 moles

moles of CH₃COO^- = 0.250 moles + 0.0175 moles = 0.266 moles

total volume = 1000 +35 = 1035 mL = 1.035 L

so, [CH₃COOH] = 0.133 moles / 1.035 L = 0.128 M

[CH₃COO^-] = 0.266 mol / 1.035 L = 0.258 M

pH = pKa + log [CH₃COO^-] / [CH₃COOH]

      = 4.75 + log 0.258 / 0.128

      = 5.06

c) pH after 35.0 mL of 5.0 M NaOH

moles of NaOH = 5.0 M * 0.035 L = 0.175 moles

when we added NaOH the moles of acid decrease and its conjugate base increase. The all acid gets reacted and converted to base

moles of CH₃COO^- = 0.250 moles + 0.175 moles = 0.425 moles

total volume = 1000 +35 = 1035 mL = 1.035 L

[CH₃COO^-] = 0.425 mol / 1.035 L = 0.411 M

We need to use the ICE

    CH₃COO^- + H₂O <----> CH₃COOH + OH^-

I    0.411                                   0             0

C     -x                                     +x              +x

E   0.411-x                               +x           +x

We know , Ka = 1.8*10^-5

So, Kb = 1.0*10^-14/ 1.8*10^-5 = 5.55*10^-10

5.55*10^-10 = x*x /0.411-x

We can neglect x in the 0.411-x, since Kb value is too small

So, x² =   5.55*10^-10 *0.411

     x = 1.51*10^-5 M

, so, [OH^-] = 1.51*10^-5 M

pOH = -log [OH^-]

      = - log 1.51*10^-5 M

      = 4.82

So, pH = 14 – 4.82

             = 14-4.82

             = 9.18