CHM 1046 Quiz 4 1. A precipitate will form when a solution containing a cation a
ID: 1045638 • Letter: C
Question
CHM 1046 Quiz 4 1. A precipitate will form when a solution containing a cation and a solution containing an anion aze sixed if the exceeda the value of the Answer 2. The pt ot a saturated solution of Mg (0)2 vhose Ksp is 7.1-10 12, Ansver 3. The solubility product for chomium(I) fluoride is Ksp -6.6x10 11. Using this information, what is the molar solubility of chromiun(IID luoride? Answer 4. Calculate the concentration of chloride ions in a saturated lead (II) chloride solution. The Kap 2.4 x 104 tor lead(II) chloride. Answer: 5. A student mixes 50.00 mi. of 0.500 M NaCl with 75.00 nl of 0.200 M AgNO3. Will a precipitate form? What is the identity of the precipitate? Consult Table 17.1 in your text. Answer: 6. What is the molar solubility of Ca (08)2 in a 0.300 M NaOH solution? The Kap of Ca (OH)2is 6.5 x 10. Ansver:Explanation / Answer
1. A precipitate from a solution is formed when the value of Qsp exceeds the value of Ksp.
2. Ksp Mg(OH)2 = [Mg2+][OH-]^2 = 7.1 x 10^-12
let x amount of salt is in solution, than x amount of Mg2+ and 2x amount of OH- would be in solution
7.1 x 10^-12 = (x)(2x)^2
x = 1.21 x 10^-4 M
So,
[OH-] = 2 x 1.21 x 10^-4 = 2.42 x 10^-4 M
[H+] = 1 x 10^-14/2.42 x 10^-4 = 4.13 x 10^-11 M
pH = -log[H+] = -log(4.13 x 10^-4) = 10.38
4. let x amount of PbCl2 is in solution
than x amount of Pb2+ and 2x amount of Cl- would be in solution
Ksp = [Pb2+][Cl-]^2
2.4 x 10^-4 = (x)(2x)^2
x = 0.039 M
concentration [Cl-] = 2 x 0.039 = 0.08 M
5. Qsp AgCl = [Ag+][Cl-]
= (0.5 M x 50 ml/125 ml)(0.2 M x 75 ml/125 ml)
= 0.024
Ksp of AgCl = 1.8 x 10^-10
So,
Qsp > Ksp, therefore solution is saturated. Precipitate would form.
6. Ksp for Ca(OH)2 = [Ca2+][OH-]^2
with,
[OH-] = 0.3 M
molar solubility = [Ca2+] = Ksp/[OH-]^2
= 6.5 x 10^-6/(0.3)^2
= 7.22 x 10^-5 M
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