Question

Calculate pH of a salt solution. What is the pH of a 0.172 M aqueous solution of sodium cyanide, NaCN? (K for HCN = 4.0x1010) pH

Answer 1

Comment in case of any doubt.

, уоu hot Aь котч Ка - Ч -S O 2.SXO 4.0 xirja = 2.5x105 He know that Kg = kw = 10 So, kb for NacN = 10 Ha of MON LOS The base dissociation equilibrium of Nach is : ICE Table No CN (92) + H2Old) . Hen(aq) + OH-(ar), K₂=2.500 I: 0.172 ci -a - toe to E: (0-172-01) 2.5X10 = K = CHCNJLON CHCNJCOH 10) tou) [Nach] (0.172-5) I since I 2.5x105 is very small and [nacalinitial is significant, we can make an assumption that 0.172 >>n. [Note: If a comes out to be comparable to 0.172, then iterations would be performed / so, from is above 2.5750 x ² = Jo.172 42.5x18s * *= 2.074 x103M= CHCN] = [OH-] since a has come out to be << 0.172, so no need of iterations. Thus, [hot] = 10 [OHT pH = -log[ H20t] = -log (4.822 X1O4) = 3:32] If number of significant figures are also to be kept in mind then enter pH=3.3 (upto a sig. digists as ka is only in a sig. digits). oly 1014 3074103 = 4.822no%