NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound (composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water, therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq) Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ .......... 0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb = 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN- ] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = - log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2 Solution NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound (composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water, therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq) Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ .......... 0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb = 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN- ] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = - log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2.