Part A a solution that is 0.18 M in HCHO2 and 0.12 M in NaCHO2 Part B a solution
ID: 843626 • Letter: P
Question
Part A
a solution that is 0.18M in HCHO2 and 0.12M in NaCHO2
Part B
a solution that is 0.14M in NH3 and 0.21M in NH4Cl
Express your answer numerically to two decimal places. Use a minus ( ? ) sign if the pH has decreased.
Part A
What is the solubility of in pure water?
Part B
What is the solubility of in a 0.202 solution of ?
Part A: Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.6110?11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams.
Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in pure H2O?
Part B: Calculate the molar solubility in NaOH
Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.150M NaOH?
Part C: What is the pH change of a 0.300M solution of citric acid (pKa=4.77) if citrate is added to a concentration of 0.175M with no change in volume?
Express the difference in pH numerically to two decimal places.
Explanation / Answer
ACID-BASE EQUILIBRIA
655
17.104
a.
[H
3
O
+
]
?
0.150 M
b.
The
0.150
-M
H
2
SeO
4
ioni
zes to 0.150
-M
H
3
O
+
and 0.1
50-M
HSeO
4
-
.
A
ssem
b
le t
h
e
usu
a
l table, a
nd su
bstitute
into the K
a2
e
quilibri
um-constant expression for
H
2
SeO
4
.
Solve the resulting qua
drat
ic equ
ation.
Set up the table.
Con
c
. (M
)
HSeO
4
-
+
H
2
O
H
2).
M(NO3)2 ==> [M2+] + 2 [NO3-]
0.202 M ==> 0.202 M
M(OH)2 ==> [M2+] + 2[OH-]
5.05*10^-18 ===> s + [2s]^2
5.05*10^-18 ===> 0.202 + [2s]^2
5.05*10^-18 = 0.202 * 4s^2
4s^2 = 25*10^-18
s^2 = 6.25*10^-18
s = 2.5*10^-9
So, the solubility is 2.5*10^-9
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Ksp = [Mg] [OH]2 You know that the Mg concentration is x and the Oh concentration is 2x, so you have:
[x] [2x]2 = 5.61*10^-11 which is 4x^3 = 5.61* 10^-11 Solving for x gives you
x = 2.41 *10^-4 .....and that's the solubility of Mg(OH)2 in moles per liter in pure water.
Now if we add Mg(OH)2 to a solution of 1.5 molar OH- ions, the Ksp for Mg(OH)2 is still the same, but we have to plug in the 1,5 molar concentration for the OH ion concentration. Since the concentration of the OH ions from the NaOH is so much greater than the additional concentration the OH ions from the dissolved Mg(OH)2, we can ignore the OH ions from the Mg(OH)2 and we have
Ksp = [Mg] [1.5]2 or 5.61*10^-11 = [x] [1.5]2.......if you work this out and solve for x you should get x =2.49*10^-11 and the ratio of solubilities is
2.41*10^-4 divided by 2.49*10^-11 and that comes out to be
9.66*10^6 What it amounts to is that Mg(OH)2 is ~ 10 million times more soluble in water than it is in a 1.5 molar NaOH solution. I hope you followed this, Kobe, you should check my math; I'm pretty sure it's okay but I make mistakes, and it would be good for you to run through the math for practice anyway.
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Mix appropriate volumes of stock and add an equal volume of distilled water to make a final 0.1 M S