Question

Consider the following solutions:
Solution 1: 0.100 L of 0.25 M NaCH₃CO₂ + 0.050 L of 0.25 M HCl
Solution 2: 0.100 L of 0.25 M HCH₃CO₂ + 0.050 L of 0.25 M NaOH

What is the pH of Solution 1 after adding 4.4 mmol of strong base?

What is the pH of Solution 2 after adding 4.4 mmol of strong base?

Answer 1

Solution 1: 0.100 L of 0.25 M NaCH3CO2 + 0.050 L of 0.25 M HCl

Solution 2: 0.100 L of 0.25 M HCH3CO2 + 0.050 L of 0.25 M NaOH

What is the pH of Solution 1 after adding 4.4 mmol of strong base?

pH = pka + log(NaCH3CO2 - HCl / CH3COOH)

   pka of CH3COOH = 4.74

No of mol of NaCH3CO2 = 0.1*0.25 = 0.025 mol

No of mol of HCl = HCH3CO2 = 0.05*0.25 = 0.0125 mol

pH = 4.74 + log((0.025-0.0125)/0.0125) = 4.74

after addition of 4.4 mmol of strong base

    pH = 4.74 + log((0.0125+4.4*10^-3)/(0.0125-4.4*10^-3))

       = 5.06

What is the pH of Solution 2 after adding 4.4 mmol of strong base?

pH = pka + log(NaCH3CO2 / CH3COOH-NaOH)

   pka of CH3COOH = 4.74

No of mol of NaOH = NaCH3CO2 = 0.05*0.25 = 0.0125 mol

No of mol of HCH3CO2 = 0.1*0.25 = 0.025 mol

pH = 4.74 + log(0.0125/(0.025-0.0125)) = 4.74

after addition of 4.4 mmol of strong base

    pH = 4.74 + log((0.0125+4.4*10^-3)/(0.0125-4.4*10^-3))

       = 5.06