Question

A buffer solution is prepared by mixing 35mL of 0.18M CHCOOH (acetic acid) and 25mL of 0.23M NaCH₃COO (sodium acetate). Fine the change in pH when 5.0 mL of 0.12M HCl is added.

Been stuck on this one for a while. If you can show step by step that would help so much!

Answer 1

Sol:-

Step1:-Calculate pH of given acidic buffer solution :-

We know

Molarity = Number of moles of solute / Volume of the solution in litre

therefore Number of moles of solute = Molarity x Volume of the solution in litre

Now Number of moles of CH₃COOH = 0.18 M x 0.035 L = 0.0063 mol

also

Number of moles of CH₃COONa = 0.23 M x 0.025 L = 0.00575 mol

Total volume = 35 mL + 25 mL = 60 mL = 0.060 L

therefore concentration of CH₃COOH i.e [ CH₃COOH] = 0.0063 mol / 0.060 L = 0.105 mol/L OR 0.105 M

similarly [CH₃COO^-] = 0.00575 mol / 0.060 L = 0.0958 mol/L OR 0.0958 M

also pK_a of CH₃COOH = 4.76

Now from Henderson-Hasselbalch equation , we have

pH = pK_a + log [Conjugate base] / [ Conjugate acid ]

pH = 4.76 + log [CH₃COO^-] / [ CH₃COOH]

pH = 4.76 + log 0.0958 M / 0.105 M

pH = 4.76 + log 0.91238

pH = 4.76 + ( - 0.0398 )

pH = 4.76 - 0.0398

pH = 4.72

Step 2:- Calculate pH of buffer after the addition of HCl :-

Number of moles of HCl added = 0.12 M x 0.005 L = 0.0006 mol

Now reaction of HCl with conjugate base i.e CH₃COO^- is :

           CH₃COO^-(aq) +   HCl (aq) -------------> CH₃COOH (aq) +   Cl^-(aq)

I           0.00575 mol        0.0006 mol                    0.0063 mol             -

C         - 0.0006 mol        - 0.0006 mol                + 0.0006 mol            -

F           0.00515 mol           0 mol                           0.0069 mol         -

again from Henderson-Hasselbalch equation we have

pH = 4.76 + log [CH₃COO^-] / [ CH₃COOH]

pH = 4.76 + log 0.00515 M / 0.0069 M

pH = 4.76 + log 0.7464

pH = 4.76 + ( - 0.1270 )

pH = 4.76 - 0.1270

pH = 4.63

Now Change or decrease in pH value after the addition of HCl = 4.72 - 4.63 = 0.09

Hence Change in pH = 0.09