Question

37. You have 500.0 mL of a buffer solution that is 0.30 M HF and 0.50 M KF. Ka for HF is 7.1 x 10-4 a) Calculate the pH of the buffer. b) Calculate the pH of the buffer after adding 0.020 moles of HCI. c) Calculate the pH of the buffer after adding 0.030 moles KOH. - -Loel.

Answer 1

@ pH = pka + log (KES i. pka = 3.15 Ka = 7.1x164 pka = - log Ka = -log (7.1x1024) = -flog.7-1 + log 16"} =-f0-85-4} = 3.15 [RF] = 0.5m [HF] = 0.3M ..pH = pka + log IKI = 3-15 + log (9:3) = 3.15 + dog (.66) = 3.15 +0.220 = 3.37

6 when 0.020 moles of Hel added to the bukten solution, then HF + H, K + , decreased by increased by 0.02 mole 0.020 mole Initially the mole of HE in 500me of solution was 0.5% x 0.3 mole x = 0.15 mode. the mole of KF in 500 ml of solution was = 0.54x0.5mall = 0.25 mol

[RF] = (0.25-0.02) = 0.23 mol Aften addition of Hel [1HF) = (0.15 +0.02) mole = 0.17 mole 2. pH = pka + log TAFELL = 3.15 + log tro = 3:15 tog 1:35 = 3.15 + 0.130 = 3.28

☺ when 0.03 mole of Koh added to the butten solution then HF + OH - -> KF +H₂O decreased by increased by 0.03 mole 0.03 mole So aften addition of KOH [HF] = (0.15-0.03) mole [KF] =(0.25 +0.03) mole = 0.12 mole = 0.28 mole pH = pka + dog The FREE = 3-15 + dogy Fotolog = 3:15 +0.367 = 5.41