Unformatted text preview: l2(s) and AgCl(s) will form. 20. Why must the dilute HCl added in to the solution in question 19 be cold? PbCl2 is slightly soluble in water. Solubility decreases as temperature decreases. So by using cold water, we ensure that PbCl2(s) precipitates from the solution. 21. Why is it better to use dilute HCl in question 19 dilute rather than use concentrated HCl? We use dilute HCl to prevent the precipitation of basic chlorides and hydroxides. 22. In the group I analysis, what is the basis of separation for AgCl(s) and PbCl2(s). Why is this separation possible? By heating the solution, we separate PbCl2 and AgCl because the solubility of PbCl2 increases dramatically with increasing temperature, so its precipitate dissolves upon heating. However, the solubility of AgCl remains fairly constant, so it remains solid. 23. What color is PbCrO4(s)? Yellow Factors Influencing Solubility (Section 17.5) 24. How many moles of MgF2 (Ksp = 6.4 x 10–9) will dissolve in 0.50 L of 0.20 M NaF? This is an example of the common ion effect. Because some F‐ is already in the solution due to the presence NaF, the solubility of MgF2 will decrease because its solubility equilibrium will shift left away from the “added ion”, in accordance with LeChatelier’s Principle: Mg2+(aq) + 2F‐(aq) MgF2 (s) 5 Dr. Fus CHEM 123 To find the number of moles of MgF2 that will dissolve in 0.50 L of solution, we must calculate the molar solubility of MgF2 in this solution, using an ICE table. **Remember, because NaF is completely soluble in water, the initial concentration of F‐ is NOT 0 M, it is the concentration of F‐ from NaF (in this case, 0.20 M): Mg2+(aq) + 2F‐(aq) MgF2(s) ___ Initial ‐‐‐‐‐
0 M
0.20 M Change ‐‐‐‐‐
+1x M
+2x M
Equilibrium ‐‐‐‐‐
x M
2x+0.20 M Ksp=[ Mg2+] [F‐]2= (x)(2x+0.20)2 This presents a problem because the added 0.20 M makes this a quadratic equation. However, as long as Ksp is low (which it almost always is), the 2x can be ignored: Ksp= (x)(2x+0.20)2= (x)(0.20)2=6.4 x 10‐9 x= 1.6 x 10‐7 M However, this problem asks for moles of MgF2 in 0.5 L of solution, so to get the final answer: 1.6 x 10‐7 mol x 0.50 L = 8.0 x 10‐8 mol 25. Calculate the molar solubility of CrF3 in 0.20 M NaF. Initial Change Equilibrium CrF3(s) ___ Cr3+(aq) + 3F‐(aq) ‐‐‐‐‐
0 M
0.20 M ‐‐‐‐‐
+1x M
+3x M
‐‐‐‐‐
x M
3x+0.20 M Ksp=[ Cr3+] [F‐]3= (x)(3x+0.20)3=(x)(0.20)3=6.6 x 10‐11 x= 8.25 x 10‐9 M 26. What is the molar solubility of MgF2 in 0.40 M F‐? Ksp MgF2 = 6.4 x 10‐9 Initial Change Equilibrium MgF2(s) ___ Mg2+(aq) + 2F‐(aq) ‐‐‐‐‐
0 M
0.40 M ‐‐‐‐‐
+1x M
+2x M
‐‐‐‐‐
x M
2x+0.40 M Ksp=[ Mg2+] [F‐]2=(x)(2x+0.40)2= (x)(0.40)2=6.4 x 10‐9 x= 4.0 x 10‐8 M 27. The solubility product constant for BiI3 is 8.1 x 10‐19. Calculate the molar solubility of BiI3 in 0.20 M Bi(NO3)3. Be careful! Here, the common ion is the cation, unlike in the previous problems: Bi3+(aq) + 3I...
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 Spring '06
 larosa
 Solubility, Dr. Fus

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