Page 76 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

60 Chapter 5
5. Ksp = (~)(2x)~ 1.1 x + 4x3 = 1.1 x 10-l8. Therefore, x
=
= 6.5 x 10-7 M.
6. [Hg;'] = x = 6.5 x M, and [Cl-] = 2x = 2(6.5 x low7)
= 1.3 x M.
7. From the reaction: 1[Hg2C12] = l[Hgi+] = 2[C1-] + molar
solubility of [Hg2C12] = 6.5 x M, since 1[Hg2C12] E
1 [Hd+I.
Answer: Molar solubility of HgzCl2 = 6.5 x 10- 7M

Example No. 2: Determine the solubility of Zn(CN)2 at 298 K in the
presence of 0.15 M KCN, given that Ksp Zn(CN)2 is 8.0 x

Solution:

1. Solubility product question!
2. Zn(CN)2(,) Zn2+ + 2CN-(,q)-heterogeneous equilibrium
(two phases).
3. ~sp = [zn2 + (aq)l[c~-(aq~~.

4. ZNCN)~(~) + zn2+(aq> + 2CN-(aq)
Initial conc. 0 0.15
Change +X + 2x
Final conc. +X (0.15 + 2x)
8
5. Ksp = (x)(0.15 + 2~)~ 10-l2. Assume 2x << 0.15 =+-
x
=
(0.15)2x = 8 x 10-l2, and therefore x = 3.6 x lo-'' M. Hence,
(3.6 x 10-10)/0.15% = 2.4 x which is <5%, i.e. the
assumption made was justified.
6. [Zn2+caq,] = 3.6 x lo-'' M and [CN-(aq)] = (0.15 + 2x), i.e.
approximately 0.15 M.
7. From the reaction, l[Zn(CN),(,)] = l[Zn2'] = 2[CN-] * solubi-
lity of [Zn(CN)2] = 3.6 x lo-'' M, since l[Zn(CN)2(s)] =
1 [zn2 + I.
Answer: Molar solubility of Zn(CN)2 = 3.6 x lO-'OM

SUMMARY
Figure 5.3 summarises the five types of problems on aqueous solution
equilibrium.