Question

5 moles of CH3COONa are mixed with 2 mole of HCl in aq solution.The Ph of resulting solution ..Pka=4.47​

Answer 1

The pH of resulting solution is 4.65.

• pH of a solution is nothing but the measure of the concentration of hydrogen ions in the solution.

• It is expressed as the negative logarithm to the concentration of hydrogen ions. pH = -log[H⁺]

• For acids, pH is always lesser than 7, for water it is 7 and for bases, it is more than 7.

Now, we know that the reaction between CH₃COONa and HCl is:

                 CH₃COONa + HCl → CH₃COOH + H₂O

Initial moles of CH₃COONa and HCl are 5 and 2 respectively. So,

Initial :          5 moles       2mole             0           0     [as no products are formed yet]

Final :          (5-2)                 (2-2)              2           2

                   = 3 moles             0                2           2

The solution contains CH₃COONa and CH₃COOH and due to this presence of CH₃⁺ ion, the solution acts as a buffer. Also,

Here, in the buffer sol., salt is CH₃COONa and its final concentration is 3 moles and the acid is CH₃COOH and its concentration is 2 moles.

The expression relating pH and pKa is:

pH = pKa + log $\frac{[salt]}{[acid]}$

Putting the respective concentrations, and the value of pKa as 4.47, we get:

pH = 4.47 + log $\frac{3}{2}$

pH = 4.47 + log(1.5)

pH = 4.47 + 0.17609

pH = 4.65

So, the value of pH is 4.65