Question

You have a 5.5 Liter container of water. How many grams of NaCN would you have to add to get a pH of 11.6.

Answer 1

base NaOH. OH NacN is a salt of weak acid HN and strong Therefore, it dissociates In water to produce The chemical equation corresponding to dissociation of Na CN in water is: Na CN6w) + H20(l) = HCN (001) + NaOH(a) As Na OH produced in a strong it dissociates completely to produce of ions. Given pH Therefore, por pk w pH Oh, [OH-] = 3.98 x 10 base = 14 - 11.6. = 2.4 -3 M since source of [OH-] us the Naoh produced [HON] - [na 0h] - [OH-] = 3.98X10-3 M at equilibriu Now let c'M NA CN was added, therefore [Na CN at equilibrium = [< -13.98 x103)] Hence, Kg = [Na 04] [Hen] [NaCN C-(3.98 x103) Ka for HON = 6. 2x10 10. Therefore Kp = Kw/ka (3-98x103)? -14 10 10 62x10 5 -1.613X10 5 - 1.613x10 (3.98x103)² . C-13.98x103) - 04, C 0.986 M Annount of Na CN that should be added to the 5.5 l container water = (0.986X5.5) moles X molar, of Naw : 0.986 X5.5 X 49 of wan - 265.739