Question

A beaker with 105 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.30 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

Given:

CH3COOH                

Molarity          0.1       M

volume buffer 0.105   L

CH3COO-                 

Molarity          0.1       M

Calculation of mol of acetic acid

Mol = Molarity x volume in L

n CH3COOH = 0.1 M x 0.105 L

                        =0.0105 mol acetic acid

Calculation moles of conjugate base

n CH3COO- = 0.1 M x 0.105 L

                      = 0.0105 mol

Both have equal moles and when we add HCl to this buffer then some moles of conjugate base will be reacted with H+ of HCl and form acetic acid.

Lets show the reaction

Reaction:

            HCl + CH3COO- ---- > CH3COOH + Cl-

This reaction shows that number of moles of HCl reacted = number of moles of acetic acid formed and number of moles of HCl reacted =number of moles of conjugate base decreased.

Calculation of moles of HCl

n HCl = 0.0043 L x 0.40 M

            = 1.72 E-3 mol

Calculation of resultant moles of acid and its conjugate base.

Mol acetic acid = 0.0105 (original) + 1.72 E-3 ( from HCl)

                        = 1.22E-02 mol

Mol acetate ion = 0.0105 (original ) – 1.72 E-3 (from HCl)

                        = 8.78E-03 mol

Equilibrium concentration of acetic acid

[CH3COOH]= mol acid / volume of solution in L

            = 1.22E-02 mol / (0.105+0.0043) L

            = 1.12E-1 M

Equilibrium concentration of conjugate base (CH3COO-)

[CH3COO-] = 8.78E-03 mol / (0.105+0.0043) L

                        =8.03E-02

We know got concentrations of both acid and its conjugate base of the buffer solution.

We can use Henderson Hasselbalch equation

pH = pka + Log ( [ CH3COO-]/ [CH3COOH] )

pH = 4.740+log(8.03E-02/1.12E-01)

pH = 4.60

pH change is = 5.0 -4.60 = 0.40

Ans : pH change will be 0.40