Question

Calculate the Ka of the acid if the pH at the 1/4 point was 4.30. 15.0 mL of 0.100 M NaOH were added to 50.0 mL of the weak acid to reach the 1/4 point.

Answer 1

the no of moles of NAOH = 15 * 0.1 = 1.5 milli moles

no of milli moles of weak acid for (1/4)  = 1.5 =

for 100% moles of weak acid = 1.5 * 4 = 6 milli moles     

given that pH for (1/4 ) point = 4.30 it means that the neutralisation of weak acid has 4.3 pH

we first calculate at that concentration

pH = -log ( H+) = - log (4.3) = 0.63 M

as the H + concentration for 1/4 (25%) of weak acid = 0.63 for (100%) = 0.63 * 4 = 2.52 milli moles

as HA + H2O -------> H+ + A -

initial concentration 6 0 0

equillibrium concentration 6-2.52 2.52 2.52

   = 3.48

Ka = [H+] [A-] / [HA] =   2.52 *   2.52 / 3.48 = 1.82