Question

a. Calculate the pH of a buffer solution that contains 0.50 M CH3COOH and 0.70 M NaCH3COO

b. What is the mximum number of moles of H+ that can be absorbed by 100 mL of buffer given in the previous question

c. Write the net ionic equation for the reaction of H+ wiht the buffer component

Answer 1

Answer –

a)

We are given, [CH3COOH] = 0.50 < , [CH3COONa] = 0.70 M

We know , pKa for the CH3COOH = 4.75

We can use Henderson Hasselbalch equation –

pH = pKa + log [conjugat base] / [acid]

pH = pKa + log [CH3COO^-] / [CH3COOH]

     = 4.75 + log 0.70 / 0.50

     = 4.75+ 0.146

     = 4.90

b) For calculating the maximum number of moles of H+ that can be absorbed by 100 mL of buffer, we need to calculate the how much conjugate base in the solution after the reaction

Volume of solution = 100 mL

From the part A , we can calculate CH3COOH is dissociate,

So, [H⁺] =x

pH = -log [H⁺]

so, [H⁺] = 10^-pH

               = 10^-4.90

              = 1.27*10^-5 M

So, x = [H⁺] = [CH3COO^-] = 1.27*10^-5 M

So there is formed [CH3COO^-] = 1.27*10^-5 M

Moles of CH3COO^-]= 1.27*10^-5 M *0.100 L

                                 = 1.27*10^-6 mol

So 1.27*10^-6 mol of H⁺ that can be absorbed by 100 mL of buffer.

c) Net ionic reaction –

CH3COO^-(aq) + H⁺(aq) <-----> CH3COOH(aq)