Question

Calculate the theoretical pH of the resulting solution when 0.25 mL of 0.100 M NaOH is added to 2.00 mL of an acetic acid/sodium acetate buffer (Ch3COOH/CH3CooNa - 1 ml of 0.1 M acetic acid and 1 ml of 0.10 M sodium acetate). Enter your answer in the box and show your work below. (pKa of CH3COOH = 4.74)

Answer 1

Sol :-

As,

Number of moles = Molarity x Volume of solution in L

So,

Number of moles of NaOH = Molarity of NaOH x Volume of solution in L

= 0.100 M x 0.00025 L

= 0.000025 mol

Similarly,

Number of moles of each CH₃COOH and CH₃COONa = Molarity x Volume of solution in L

= 0.100 M x 0.001 L

= 0.0001 mol

ICF table is :

...............CH₃COOH (aq).................+...............NaOH (aq) <--------------------> CH₃COONa (aq) .............+..........H₂O (l)

Initial (I)...0.0001 mol.......................................... 0.000025 mol............................ 0.0001 mol................................

Change (C).....- 0.000025 mol..............................- 0.000025 mol..........................+ 0.000025 mol................................

Final (F)............0.000075 mol....................................0.0 mol............................................0.000125 mol............................

Using Henderson-Hasselbalch equation :

pH = pK_a + log [Salt]/[Acid]

pH = pK_a + log [CH₃COONa]/[CH₃COOH]

pH = 4.74 + log 0.000125 / 0.000075

pH = 4.74 + 0.2218

pH = 4.96

Hence, pH of the resulting solution = 4.96