Question

A beaker with 145 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 5.00 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

Henderson equation is

pH = pKa + log ( [A-]/[HA])

5 = 4.74 + log ([A-]/[HA])

log([A-]/[HA]) = 5-4.74 = 0.26

[A-]/[HA] = 1×10^0.26

= 1.82

[A-] = 1.82 [HA]

  

[A-] + [HA] =0.1M

2.82[HA] = 0.1M

[HA] = 0.0354M

[A-] = 0.1 - 0.035 = 0.0646M

Added HCl react with Acetate (A-) and consumed

No of mole of HCl = (0.490mol/1000ml)×5ml = 0.00245mol

0.00245mol HCl react with 0.00245 mol of Acetate(A-)

Initial mole of A- = (0.0646mole/1000ml)×145ml = 0.009367

After reaction with HCl ,mol of A- = 0.009367-0.00245=0.006917

Final volume = 150ml

Final [A-] = (0.006917mol/150ml)×1000ml= 0.0461M

Now,applying the Henderson equation

pH = 4.74 + log(0.0461/0.0354)

= 4.74 + 0.11

= 4.85