Question

calculate the initial pH and the pH after each 2ml addition of 0.10M NaOH.

i) initial pH

a) 2ml of 0.10M of NaOH has added to 50ml of 0.50M HC2H3O2/0.050M NaC2H3O2

b) 8ml of 0.10M of NaOH has added to 50ml of 0.50M HC2H3O2/0.050M NaC2H3O2

c) 10ml of 0.10M of NaOH has added to 50ml of 0.50M HC2H3O2/0.050M NaC2H3O2

Answer 1

(1). Sol :-

Since, HC₂H₃O₂ is a weak base and NaC₂H₃O₂ is its salt of strong base. They form a acidic buffer solution and we know that pH of buffer solution can be calculated by using Henderson-Hasselbalch equation :

pH = pK_a + log [NaC₂H₃O₂] / [HC₂H₃O₂]

pH = 4.76 + log 0.050 M / 0.50 M

pH = 4.76 + log 0.1

pH = 4.76 - 1.0

pH = 3.76

Hence, Initial pH = 3.76

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(a). 2ml of 0.10M of NaOH has added to 50ml of 0.50M HC2H3O2/0.050M NaC2H3O2

No. of moles of NaOH = Molarity x Volume in L = 0.10 M x 0.002 L = 0.0002 mol

No. of moles of HC₂H₃O₂ = 0.50 M x 0.050 L = 0.025 mol

No. of moles of NaC₂H₃O₂ = 0.050 M x 0.050 L = 0.0025 mol

ICF table is :

..........................HC₂H₃O₂ .............+...............NaOH ------------------------> NaC₂H₃O₂ ..........+............H₂O

Initial (I)..............0.025 mol...............................0.0002 mol..........................0.0025 mol...........................

Change (C)..........- 0.0002 mol.........................- 0.0002 mol........................+ 0.0002 mol..........................

Equilibrium (E).......0.0248 mol..........................0.0 mol.................................0.0027 mol........................

Again by using Henderson-Hasselbalch equation, we have

pH = pK_a + log [NaC₂H₃O₂] / [HC₂H₃O₂]

pH = 4.76 + log 0.0027 / 0.0248 M

pH = 4.76 + log 0.108871

pH = 4.76 - 0.963

pH = 3.80