Question

What is the maximum volume of 0.250 M HCl that can be added to a cetain buffer system before the capcity is exceeded if that buffer system consist of 500 mL of 0.300 M NaC₂H₃O₂ and 0.100 M HC₂H₃O₂?

(correct answer is 600 mL)

Answer 1

[NaC2H3O2] = 0.300 M

[HC2H3O2] = 0.100 M

pKa of acetic acid = 4.756

pH = pKa + log [salt] / [acid]

= 4.756 + log (0.300 / 0.100)

= 4.756 + log 3

= 5.2331

[HCl] = 0.250 M

Let X litre of HCl is added to the buffer.

Final pH of buffer = 5.2331 + 1

= 6.2331

Total [H+] in buffer = 10 ^{- 6.2331}

= 5.846x10^-7

Total volume of buffer = X + 0.50 L

Total moles of H+ ions in buffer = (5.846x10^-7)*(X + 0.50)

= X*5.846x10^-7 + 2.923x10^-7

X*0.250 - 0.50*0.300 = X*5.846x10^-7 + 2.923x10^-7

0.250*X - 0.15 = X*5.846x10^-7 + 2.923x10^-7

0.250*X = 0.150

X = 0.6

So, volume of HCl added = 0.6 L = 600 mL