Question

18. You are asked to prepare a pH = 8.60 buffer starting from 500 mL of 0.10 M solution of hydrocyanic acid, HCN, and excess sodium cyanide, NACN. (Ka for HCN = 4.9 x 10-10) a) What is the pH of the hydrocyanic acid solution prior to adding NACN? b) How many grams of NaCN should be added to prepare the buffer solution?

Answer 1

(18) Given Ka = 409 x10-10 (a) for very weak aids, degree of icnisation (a) can be neglected in comparison with I For HCN we have ka 5 HUN 2 H+ + CN Initial C 6 0 change -ca +(x + coc Equilibrium C Co-a) cat corect : Ka 2 (H*I (W"] CHONJ Ka.cact (-x) 5 Ka = cx se44), 1- и -1 . 1-a Ka= cх² a = Ska .. [++] = cx = ſkaic : [HJ= 10.4.9810-10 X 0.40 [H* J = 7 X10-26 s we can not neglect : pH: ~ log (7x600+ 1x1097) Labin contribution of Ht due to water as it close to #x10-6]. pH = 5.15 b) pha 2 - log ka = -69 (4.9x10'0) = 9.31 Moles of aud present = Molarity & volume in а о• 10 Их Бого м х L 10M2 = 0.05 mol

pH of buffer la combination of weak acid base) is given by Henderson Hasselbalch and equation, E its pH A s it the road in 3,74:-p pka + log(salt). [acid] i => 3.60 = 9.31 + tog et ) [Salyj w ? cocida - Antilog 69.3148160) = 0:195 www.bapa [Avid 7 → [solt J = 0.195 * 0.05 mol. = 9.75800mor, Molar mass of Nacre 23712 +14 = 49 molta g/mol. & Mass of Nach needed = 9.75 110 g mol X99 g/mol 3 0-4778 9.