Question

1. Find the pH of a 1.5 x 10-5M weak acid (HA) solution with a pKa of 4.5.

Answer 1

Concentration or Molarity of acid HA = [HA] = 1.5 * 10^-5 M

Now, HA is a weak acid it will dissociate in equilibrium as shown below,

HA <====> H+ + A-

Hence , Ka = [H+] * [A-] / [HA]

Where all the concentration terms are equilibrium concentrations.

pKa is given as 4.5 , hence Ka = antilog (- pKa) = antilog (- 4.5) = 3.16 * 10^-5

Now , at equilibrium , [H+] = [A-] and [HA] = 1.5 * 10^-5 ( as it is a weak acid therefore concentration does not change significantly )

Hence equilibrium constant , Ka = [H+]²/ ( 1.5 * 10^-5 )

[H+] = √ ( Ka * 1.5 * 10^-5 ) = √ ( 3.16 * 10^-5 * 1.5 * 10^-5 )

= 2.67 * 10^-5 M

pH = - log [H+] = - log ( 2.67 * 10^-5 ) = 4.57 (answer )

I hope you find it useful. For further clarification please ask in comments.