Question

8. Calculate the pH and % ionization of 0.0200 M HF (Ka = 6.8x104)? (1! 9. What is the pH of a buffer that is 0.300 M HF (Ka = 6.8x104) and 0.500 M KF?

Answer 1

® Hf is a weak Acid. And the dissociation of HF can be shown as:- . Hf ~ Hot po quitial c o o If @is the degree of dissociation dhen. at eam coca ca ca The equililibrium constant Ka can be shown as - Ka- [147] [60] . (HF) cohere, Square Beacketts shows the molar concentration the corresponding species. it Kaa (C^) (Cd) c-ca Ka a 4 (ca) ella) e 씀 e can be neglected in comparison to 1 år denominator, as for weak acid, dissociation is wery less. alka z CQ2 T to formula #1. and conch of Ch®] 2 cx

.: [HO] 2 CTO op []. Seke to formula nom uzation - Raſka and Kaz 6.80104 1 2 = 16.8*10-4 c. 0.0200 M v0.0.200 Ke 0.1843 % ionization 2 18.43% + Ausser, OPH is the negative doguithem of H@ ion concentration pH = -log Ho] - = -log Jeka I-dog 10.02) (6.8710") PH: 2.43 to Answer 1 Hf is a weak Acid and Kf is a selt having common & ion with Hf. This a Buffer solution is formed. and pH of Buffer solution is calculated by asing Hasselbaich Henderson equation : – pH ₂ pka + log ( conjugate Base] u [weakfaid] - pka -log Ka - pka a bag (6.8x10") = 131677 [weak Acid) = 0.300 M [ conjugate base], 0.500M. pH = 3.167 + dog 10.50 [ pH: 3.39 7. Aunser. [0.300)