Question

6. a. Calculate the pH of a solution prepared by diluting 2.1 mL of 2M HCl to a final volume of 100 mL with water.

b. What is the pH of a solution that contains 0.5 M acetic acid and 0.25 M sodium acetate? The pKa for acetic acid is 4.7

Please explain why we do the steps in the calculations also.

Answer 1

Answer – 6. a) Given, M1 = 2.0 M , V1 = 2.1 mL , V2 = 100 mL , M2 = ?

First we need to calculate the M2 using the dilution law

M1V1 = M2V2

M2 = M1V1/V2

     = 2.0 M * 2.1 mL / 100 mL

     = 0.042 M

We know, HCl is strong acid,

So, [HCl] = [H₃O⁺] = 0.042 M

We know pH = - log [H₃O⁺]

                      = - log 0.042 M

                      = 1.38

b) We are given, [acetic acid] = 0.5 M , [acetate] = 0.25 M

pKa = 4.7

we know, Henderson Hasselbalch equation –

pH = pKa + log [acetate] / [acetic acid]

     = 4.7 + log 0.25 /0.50

= 4.4