Mg (OH)_2 is a sparingly soluble salt with a solubility product constant, K_sp,
ID: 1014225 • Letter: M
Question
Mg (OH)_2 is a sparingly soluble salt with a solubility product constant, K_sp, of 5.61 Times 10^-11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal waste water streams. Calculate the ratio of solubility of Mg(OH)_2 dissolved in pure H_2O to Mg(OH)_2 dissolved in a 0.180 mol L^-1 NaOH solution. Express your answer numerically to three significant figures. The common-ion effect and buffer systems 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 behaviour, because both added H^+ or OH^- ions can be neutralized. What is the pH change of a 0.200 mol L^-1 solution of citric acid (pK_a = 4.77) if citrate is added to a concentration of 0.130 mol L^-1 with no change in volume? Express the difference in pH numerically to two decimal places.Explanation / Answer
Part A
The solubility of sparingly substance of ionic substance is decreases in a solution of another ionic compound when the two substance have the ion in common.
Whenever the process in equilibrium addition of more OH- reducing the concentration of Mg2+
We can Express the solublity product of Mg(OH)2 in H2O
Ks=[Mg2+][OH-]2= S X S =S2
Ks=5.61X10^-11
5.61X10^-11 =S2
Since S=Ks =5.61X10^-11 = 0.000007489 or 7.489X 10^-6
We can Express the solublity product of Mg(OH)2 in NAOH
Ks=[Mg2+][OH-]2= S X (concentration of OH- ion +2S)2
= S X ( 0.18 +2S)2
Most of the practical case S is so large compare with 2S term. So 2S term can be negleated and the equation becomes
Ks = SX (0.18)2
S = Ks / (0.18)2
=5.61X10^-11 /(0.18)2
= 1.7314 10-9(mol/lit)2
Since the ratio of solubility product is
= Mg2+ in H2O/Mg2+ in NAOH
=1.7314 10-9(mol/lit)2/ 0.000007489
Sratio =2.3 X 10^5 (mol/lit)2
Part B
pH= Pka + log (salt) / (acid)
here
pH= Pka + log (concentration of Citrate) /(concentration of Citric acid)
Pka=4.77
concentration of Citrate =0.13 mol/lit
concentration of Citric acid = 0.2 mol/lit
therefore
pH= 4.77+ log (0.13) /(0.2)
= 4.77 +0.65
pH = 5.42
pH of citric acid= -log (concentration of H+ ion)
=-log(0.2)
=1.93
change in pH = pH of citric acid by adding citrate ion - pH of citric acid without adding citrate ion
= 5.42-1.93
= 3.49
Part A
The solubility of sparingly substance of ionic substance is decreases in a solution of another ionic compound when the two substance have the ion in common.
Whenever the process in equilibrium addition of more OH- reducing the concentration of Mg2+
We can Express the solublity product of Mg(OH)2 in H2O
Ks=[Mg2+][OH-]2= S X S =S2
Ks=5.61X10^-11
5.61X10^-11 =S2
Since S=Ks =5.61X10^-11 = 0.000007489 or 7.489X 10^-6
We can Express the solublity product of Mg(OH)2 in NAOH
Ks=[Mg2+][OH-]2= S X (concentration of OH- ion +2S)2
= S X ( 0.18 +2S)2
Most of the practical case S is so large compare with 2S term. So 2S term can be negleated and the equation becomes
Ks = SX (0.18)2
S = Ks / (0.18)2
=5.61X10^-11 /(0.18)2
= 1.7314 10-9(mol/lit)2
Since the ratio of solubility product is
= Mg2+ in H2O/Mg2+ in NAOH
=1.7314 10-9(mol/lit)2/ 0.000007489
Sratio =2.3 X 10^5 (mol/lit)2
Part B
pH= Pka + log (salt) / (acid)
here
pH= Pka + log (concentration of Citrate) /(concentration of Citric acid)
Pka=4.77
concentration of Citrate =0.13 mol/lit
concentration of Citric acid = 0.2 mol/lit
therefore
pH= 4.77+ log (0.13) /(0.2)
= 4.77 +0.65
pH = 5.42
pH of citric acid= -log (concentration of H+ ion)
=-log(0.2)
=1.93
change in pH = pH of citric acid by adding citrate ion - pH of citric acid without adding citrate ion
= 5.42-1.93
= 3.49