Question

23.

A solution is prepared by adding 0.10 mole of Ni(NH3). Cl2 to 0.50 L of 3.2 M NH3. Calculate (Ni (NH3)&2+) and (Ni2+in this solution. Koveralt for Ni (NH3)22+ is 5.5 x 108. That is, 5.5 x 108 = Ni (NH3) 2+1 [Ni2+] [NH3] for the overall reaction Ni2+ (aq) + 6NH3(aq) = Ni (NH3). 2+ (aq) (Ni (NH3), 2+] = 0 [Ni2+] = C M

Answer 1

0.10 mol Initial concentrati on of Ni* =[N(NH3), a,] = 500= 0.2 M Equation representing the formation of (N( complex is as follows; N* (aq) + 6NH; (aq)=[Ni(NH3)** (aq) Formati on constant of [Ni(NH3))*=5.5108 ICE table: N* (aq) + 6NH, (aq) = [Ni(NH3).]** (aq) 0.2 3.2 Initial (M): Change (M) Equilibrium (M): - 6x 3.2-6x 0.2-x [Ni(NH3).]** [N7** ][ NH,T 5.5*10=- (0.2- x)(3.2-6x) Since x is so small, we can neglect 6x 3.2-6x 3.2 x 5.5x 10 (0.2-x)(3-2) x=0.19999999999966133724559494491117 From ICE table [Ni**] = (0.2 – x)M = (0.2-0.19999999999966133724559494491117) M = 3.4x10-3 M [N (NH,)*]=(x)M = 0.19999999999966133724559494491117 M