Question

A beaker with 175 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.20 mL of a 0.400 MHCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

We have Henderson equation for buffer

pH = pka + log [ acetate ]/[acetic acid]

5 = 4.74 + log [acetate ]/[acetic acid]

[acetate] = 1.82 [acetic acid] .........(1)

given [acetic acid] + [acetate] = 0.1 ............(2)   ( i.e concentartion of buffer 0.1M)

using (1) in (2) we get   [acetic acid] + 1.82 [acetic acid] = 0.1

[acteic acid] = 0.03546 , acetic acid moles = Mx V = 0.03546 x 175/1000 = 0.0062057

[acetate ] =1.82 x 0.03546 = 0.06454 , acetate moles = M x V = 0.06454 x 175/1000 = 0.0112944

Now H+ added = M x V of Hcl = 0.4 x 8.2/1000 = 0.00328

CH3COO-(aq) + H+ (aq) = CH3COOH (aq) is the equation

hence H+ addes to acetate and converts it acetci acid.

Now acetate moles after H+ reaction = 0.0112944 -0.00328 = 0.008

Moles of acetic acid = 0.0062057 + 0.00328 = 0.009486

Now total vol = 175 + 8.2 = 183.2 ml = 0.1832 L

now pH = pka + log [acetate]/[acetic acid]   with new conectrations

pH = 4.74 + log ( 0.008/0.1832) / ( 0.009486/0.1832)

   = 4.74 + log ( 0.008/0.009486)

= 4.67

pH change = 5-4.64 = 0.334