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Unformatted text preview: M Change ‐‐‐‐‐
+x M 22 Dr. Fus CHEM 123 Ksp=[ Mn2+] [S2‐]= (x)(x)= x2= 2.3 x 10‐13 x= 4.8 x 10‐7M *The molar solubilities of Zn(OH)2 and Pb(OH)2 were calculated in statement C, and they are both higher than that of MnS. Thus MnS does NOT have the highest molar solubility in water, making this statement FALSE. e) From themolar solubilities calculated in statement C, we can calculate the concentration of Pb2+ in each solution. Since the coefficient of Pb2+ in both equilibria is one, the molar solubility equals [Pb2+]. Pb(OH)2 has the higher molar solubility, and therefore also the higher [Pb2+], making this statement FALSE. Qualitative Analysis Group II & III (Section 17.7 & “Isolation and Characterization of Metal Ions: Exploitation of Differences in Solubility” Lab) 72. A solution contains 0.015 M Cu2+ and 0.015 M Ni2+. The solution is saturated with H2S (0.10M) and adjusted to pH = 2.00. Which of the metal sulfides will precipitate? Ksp of NiS is 3 x 10–20, Ksp of CuS is 6 x 10–37. For these type of problems, we need to consider two different equilibria that influence the solubility of these metal sulfides: MS(s) M2+(aq) + S2‐(aq) 2‐(aq) + H O (l) HS‐(aq) + OH‐(aq) S
2 M2+(aq) + HS‐(aq) + OH‐(aq) MS(s) + H2O (l) From this overall equilibrium, we can set up an expression for Q: Q=[M2+][HS‐][OH‐] [M2+] is given in the problem. We can calculate [OH‐] from the given pH: pH=‐log[H+]=2.00 [H+]=1.0 x 10‐2 M +][OH‐]=1.0 x 10‐14 (always true) [H
(1.0 x 10‐2 M) +][OH‐]=1.0 x 10‐14 [OH‐]=1.0 x 10‐12 M ‐], we need to consider yet another equilibrium, that of the weak acid H S: To get [HS
2 H+(aq) + HS‐(a...
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- Spring '06