Question

A 20.0 mL of 0.36 M solution of the salt NaA has a pH of 8.40.          

a) Calculate the pK_a value of the acid HA. (Correct to 3 sig. fig)

b) Calculate the pH of a solution containing 0.4 M HA and 0.6 M NaA. (Correct to 3 sig. fig)

Answer 1

Here and pH = 8.40 pH = -log [4+] 8.40 = -log 4+ (ht=10-8.4 = 3-98410 -7M. [04-] = kw Kw = 1810-14 [ht] 1710-14 = 2.51x10-6 M 3.98x10-9 now He Ma A salt dissociaks as NaA - Hydrolysis of 4- A- + ho HA + OH- Initially 0.36 at queitica 0.30-2 st a u Natt A- [[NA] = 0.36M] K = CHA] Low - (0 (2) now ( [04 - u=2.51810-6 [A-] (0.36-2) (2.51 X106) 2 - = 1.75 36 310-) . (0.36- 2.87X106) Ky = - = 5.70 X10-4 1.75 X 10-11 now pka = -log ka = -log (5.70 x 10-y) pka = 3.24 A

(b) now [HA) = 0.4m (MaA] 20.0m 2 T RS NGA - Mat + A- so [A-] = 0.6M ANT Mob pu= pka + log [A-] 1. THA pH = 3.24 + log (0.6) = 3.24 + 0.176 pH = 3.42 Aus