The solubility of a slightly soluble salt can be greatly affected by the additio
ID: 880784 • Letter: T
Question
The solubility of a slightly soluble salt can be greatly affected by the addition of a soluble salt with a common ion, that is, with one of the ions in the added soluble salt being identical to one of the ions of the slightly soluble salt. The general result of the addition of the common ion is to greatly reduce the solubility of the slightly soluble salt. In other words, the addition of the common ion results in a shift in the equilibrium of the slightly soluble salt.
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 0.120 M NaOH?
Express your answer with the appropriate units.
A buffer is a mixed solution of a weak acid or base, combined with its conjugate. Note that this can be understood essentially as a common-ion problem: The conjugate is a common ion added to an equilibrium system of a weak acid or base. The addition of the conjugate shifts the equilibrium of the system to relieve the stress of the added concentration of the common ion. In a solution consisting of a weak acid or base, the equilibrium shift also results in a pH shift of the system.
It is the presence of the common ion in the system that results in buffering behavior, because both added H+ or OH ions can be neutralized.
Part D
What is the pH change of a 0.240 M solution of citric acid (pKa=4.77) if citrate is added to a concentration of 0.175 M with no change in volume?
Explanation / Answer
Answer –
Part B-
In this part we are given, [NaOH] = 0.120 M
We know Ksp for Mg(OH)2 = 1.8*10-11
We also know NaOH is strong base, so it dissociate completely
[NaOH] = [OH-] = 0.120 M
Ksp expression for the Mg(OH)2 is
Ksp = [Mg2+] [OH-]2
So, = 1.8*10-11 = x * (0.120)2
x = 1.8*10-11 / (0.120)2
= 1.29*10-9 M
So molar solubility of Mg(OH)2 in 0.120 M NaOH is 1.29*10-9 M
Part D –
In this part we are given, [C6H8O7] = 0.240 M , pKa = 4.77, [C6H7O7-] = 0.175 M
Now we need to calculate pH of 0.240 M of citric acid first.
We know it is weak acid, so put ICE chart
C6H8O7 + H2O ----------> H3O+ + C6H7O7-
I 0.240 0 0
C -x +x +x
E 0.240-x +x +x
Now we need to calculate the Ka from given pKa
We know,
pKa – log Ka
som Ka = 10 –pKa
= 1.69*10-5
So, Ka = [H3O+] [C6H7O7-] / [C6H8O7]
1.69*10-5 = x*x / (0.240-x)
We can neglect x in the 0.240-x, because Ka value is too small
So, x2 = 1.69*10-5 * 0.240
= 4.075*10-6
x = 0.00202 M
so, x = [H3O+] = 0.00202 M
pH = -log [H3O+]
= - log 0.00202 M
= 2.69
Now we need to calculate the pH after added citrate , so in the solution there is already citrate so it reacted with acid and formed acid
So, we assume 1 L solution
Moles before added citrate
So moles of citric acid = 0.240 moles
Moles of citrate = 0.00202 moles
Now new moles
Moles of citric acid = 0.240 -0.175 = 0.065 moles
Moles of citrate = 0.00202 +0.175 = 0.415 moles
So [C6H8O7] = 0.065 M and [C6H7O7-] = 0.415 M
Now using the Henderson Hasselbalch equation –
pH = pKa + log [conjugate base] / [acid]
pH = 4.77 + log 0.415/0.065
= 5.58
So change in pH = 5.58 -2.69
= 2.88