Question

A 0.053 M solution of an acid (HA) has a pH of 4.63. What is the pKa?

Answer 1

bH=4.63 pH = -log (ht] [H+] = lo - H [ht] = 10-4.63 2.358 10 5 M ICE Table concentration (M) I HA (99) H+ (aq) + A-() 0.053 с txi +30 E 30 0,053-X ka = [H+] [A-] CHA) [++] = 20 = 2.35 X10-571 = x= 2.35 810-5M (A-] = x= 2.35 x 10-5M CHA) = 0.053-X = (0.053 - 2.3 Xlo-5 )n = 0.052977 M 2.35 x 10-5 X 2.35 8 10-5 1.048 10-8 0.052977 ka = bka= -log ka = -log (1:04 X10-8)= 7.98 pka = 7.98 | Answer