Question

your lab mate at grad achool prepared 500ml of a 10mM buffer x with a pH of 8.3. The pKa for this buffer is 8.1. You will find out he ate your lunch from the fridge so you decide you need to mess him up. You add 0.3ml of 6M NaOH to his buffer and put if back on his bench. What is the new pH of his buffer?

Answer 1

For buffer,

pH = pKa + log(base/acid)

8.3 = 8.1 + log(base/acid)

1.585 = base/acid

[base] = 1.585[acid]

Initial moles of buffer = 10 mM x 0.5 L = 5 mmols

[base] + [acid] = 5 mmols

1.585[acid] + [acid] = 5 mmol

[acid] = 1.934 mmol

[base] = 5 - 1.934 = 3.066 mmol

Added NaOH = 0.3 ml of 6 M

moles of NaOH = 6 x 0.3 = 1.8 mmol

Final moles of acid = 1.934 - 1.8 = 0.134 mmol

Final moles of base = 3.066 + 1.8 = 4.866 mmol

molarity of [acid] in buffer = 0.134/0.5 = 0.268 mM

molarity of [base] in buffer = 4.866/0.5 = 9.732 mM

Final pH of buffer,

pH = 8.3 + log(9.732/0.268)

     = 9.66