A solution is 5 mH in each of the following ions: Indicate which of the metal io
ID: 507187 • Letter: A
Question
A solution is 5 mH in each of the following ions: Indicate which of the metal ions would precipitate (or start to precipitate) at each of the following pH values. Indicate your answer with the number of the ion. Use 0 to indicate no precipitate. If more than one precipitate is expected, list the numbers in increasing order and separate them with commas. For example, 3, 4, 5 is ok but 5, 4, 3 is not. pH = 6.00: pH = 10.00: What is the pH to the nearest 0.1 pH unit at which Cu(OH)_2 begins to precipitate? pH =Explanation / Answer
Part A
The pH of the solution is 6.00. Since pH = 6.00, we must have pOH = 14.00 – pH = 14.00 – 6.00 = 8.00 (since pH + pOH = 14).
Again, pOH = -log [OH-]
===> [OH-] = antilog (-pOH) = antilog (-8.00)
===> [OH-] = 1.0*10-8
The concentration of hydroxide ion is 1.0*10-8 M.
Note that all the salts contain divalent metal cation with monovalent OH- ion. The dissociation reaction is
M(OH)2 = M2+ (aq) + 2 OH- (aq)
The reaction quotient is given as Q = [M2+][OH-]2. A precipitate will occur when Q > Ksp. When Q = Ksp, the solution is just saturated and the precipitate begins to form. When Q < Ksp, no precipitate forms.
Construct the following table. Note that the concentration of the metal ions = 5 mM = 5.0*10-3 M.
Number
Ion
Q = [M2+][OH-]2
Whether precipitate forms or not?
1
Mg2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
No, because Q is less than Ksp
2
Cd2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
No, because Q is less than Ksp
3
Co2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
No, because Q is less than Ksp
4
Zn2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
No, because Q is less than Ksp
5
Cu2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
Yes, because Q exceeds Ksp
At pH 6.00, none of the metal ions except Cu2+ will form precipitates. Only Cu2+ will form a precipitate at pH 6.00.
Ans: 5
Part B
The pH of the solution is 10.00. Since pH = 10.00, we must have pOH = 14.00 – pH = 14.00 – 10.00 = 4.00.
Again, pOH = -log [OH-]
===> [OH-] = antilog (-pOH) = antilog (-4.00)
===> [OH-] = 1.0*10-4
The concentration of hydroxide ion is 1.0*10-4 M.
Note that all the salts contain divalent metal cation with monovalent OH- ion. The dissociation reaction is
M(OH)2 = M2+ (aq) + 2 OH- (aq)
Construct the following table. Note that the concentration of the metal ions = 5 mM = 5.0*10-3 M.
Number
Ion
Q = [M2+][OH-]2
Whether precipitate forms or not?
1
Mg2+
(5.0*10-3)*(1.0*10-4)2 = 5.0*10-11
Yes, because Q exceeds Ksp
2
Cd2+
(5.0*10-3)*(1.0*10-4)2 = 5.0*10-11
Yes, because Q exceeds Ksp
3
Co2+
(5.0*10-3)*(1.0*10-4)2 = 5.0*10-11
Yes, because Q exceeds Ksp
4
Zn2+
(5.0*10-3)*(1.0*10-4)2 = 5.0*10-11
Yes, because Q exceeds Ksp
5
Cu2+
(5.0*10-3)*(1.0*10-4)2 = 5.0*10-11
Yes, because Q exceeds Ksp
All the metal ions will form precipitate with OH- at pH 10.00
Ans: 1, 2, 3, 4, 5
Part C
Given [Cu2+] = 5.0*10-3 M. Cu(OH)2 will begin to precipitate when Q = Ksp = 2.2*10-20.
Therefore, 2.2*10-20 = [Cu2+][OH-]2
===> 2.2*10-20 = (5.0*10-3)*[OH-]2
===> [OH-]2 = (2.2*10-20)/(5.0*10-3) = 4.4*10-18
===> [OH-] = 2.098*10-9
Therefore, the concentration of OH- at which Cu(OH)2 just begins to precipitate is 2.098*10-9 M and the corresponding pOH = -log [OH-] = -log (2.098*10-9) = 8.678
Therefore, pH = 14 – pOH = 14 – 8.678 = 5.322 5.3 (ans).
Number
Ion
Q = [M2+][OH-]2
Whether precipitate forms or not?
1
Mg2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
No, because Q is less than Ksp
2
Cd2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
No, because Q is less than Ksp
3
Co2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
No, because Q is less than Ksp
4
Zn2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
No, because Q is less than Ksp
5
Cu2+
(5.0*10-3)*(1.0*10-8)2 = 5.0*10-19
Yes, because Q exceeds Ksp
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.