Question

Q1)Determine the [H₃O⁺] in a 0.256 M HClO solution. The K_a of HClO is 2.8  10^–9.

Q2)Find the percent ionization of a 0.373 M HF solution. The K_a for HF is 3.4  10^–5.

thanks!

Answer 1

Не ощ) = clo (aq) + H+ (aq) guitial: 0.256 final: (01256x) - Let's assume x to be small x ka is small. as 0.256-x ~ 0.250 Ke = (ceo] [ht] = Chceo] xxx (0.256-x) 2.8 x 10-9 0.256 sc ² = 0.256 X 2.8 x 10-9 - = 7:168 x 10-10 - se = [H* ] = /7:168 xlo-lo M our assumption is correct 2.6773 x 10-5 M In acidic medium, it is same as Hot Hence, [H₂o+] = 2.67 x 10-5M upto 2 decimal places

G H E (aq) = hit coul + F Caps 0.373 (0.373-x) x x I will be smell Let's assume a as be ka is small small. ... 0.373-x = 0.373 ka = [H + ][fr] HEJ - x 0.373 3.4 x 10-> x² = = x = 0.373 x 3.4 x 10-5 12.682 x 10-6 3.561 x 10-3 M os t Cour assumption is corect I % ionization = x x 100 0.373 3.561 X 10-3 X 100 0:373 0.95469_Y. 0.95 %. Upto 2 = decimal places) mee dercent ionization of - HE Upto 2 = 0.95% decimal places