Question

A 25.0 mL sample of 0.293 M acetic acid solution is titrated with 37.5 mL of 0.195 M NaOH solution. What is the pH of this solution if K = 1.8 x 10^-5? 

Final Answer: _______ 

Answer 1

Number of moles of acetic acid=Molarity x Volume (L)

=0.293 M x 25.0 mL/1000 mL/L (1 L=1000 mL)

=0.007325 mol

Number of moles of NaOH=Molarity x Volume (L)

=0.195 M x 37.5 mL/1000 mL/L

=0.0073125 mol

Total volume after mixing=25.0 mL+37.5 mL=62.5 mL

=62.5 mL/1000 mL/L=0.0625 L

1 mol NaOH reacts with 1 mol acetic acid

0.0073125 mol NaOH reacts with 0.0073125 mol acetic acid to form 0.0073125 mol CH₃COONa

Remaining mol of acetic acid=0.007325 mol-0.0073125 mol=0.0000125 mol

Concentration of acetic acid in solution=number of moles of acetic acid/Volume of solution=0.0000125 mol/0.0625 L=0.0002 M

1 mol CH₃COONa gives 1 mol CH₃COO^- and 1 mol H⁺

So 0.0073125 mol CH₃COONa gives 0.0073125 mol CH ₃COO^- and 0.0073125 mol H⁺

Concentration of CH₃COO^-=number of moles of acetate/volume of solution (L)

=0.0073125 mol/0.0625 L=0.117 M

pH of buffer solution=pKa+log

=-log Ka+log (0.117 M/0.0002 M)

=-log (1.8x10^-5)+log(585)

=4.74+2.77=7.51

So pH of given solution=7.51