When excess solid Mg(OH)2 is shaken with 1.00 L of 1.2 M NH4Cl solution, the resulting saturated solution has pH = 9.30. Calculate the Ksp of Mg(OH)2
I am not sure where to even begin on this problem.
5 answers
3.3x10^-11
how did you get that?
can anyone please explain how to solve this?
To begin the problem, I calculated the [OH-] from the pH.
14-9.3 = 4.7 pOH,
[OH-] = 10^-(4.7) = 1.995x10^-5
Here's where things get difficult. NH4Cl breaks apart in water into NH4+ and Cl-. Therefore we can say the 1.2M is the concentration of NH4+. NH4+ combines with OH to form NH3 and H2O.
NH4+(aq)+OH-(aq) <-> NH3(aq)+H2O(l)
This equation is the equation for finding the Kb of NH3, but backward. Since we know that in order to flip a chemical equilibrium equation, we have to invert the K value,
Kb(NH3)=1.8x10^-5, Kb(NH3,rev)=5.6x10^4
Looking back at the equilibrium expression for NH4+ and OH- combining, we can solve an equilibrium to figure out what the OH had to be in the beginning in order to have excess [OH-] of 1.995x10^-5 M.
NH4+(aq)+OH-(aq) <-> NH3(aq)+H2O(l)
1.2M ? 0
-x -x +x
1.2-x 2.0x10^-5 x
5.6x10^4= x / (2.0x10^-5 x (1.2 - x))
x = 0.63398
So the [OH-] before NH4+ reacted with it was 0.63396 + 2x10^-5 = 0.63398
Now we work backward again and set up an ICE chart for the dissociation of Mg(OH)2.
Mg(OH)2(s) <-> Mg2+(aq)+2 OH-(aq)
N/A 0 0
N/A +x +2x
N/A ? 0.63398
In order to obtain 0.63398 as an answer, the change (x) must be 0.63398 / 2, making [Mg2+]= 0.317.
Ksp(Mg(OH)2)= [Mg2+]x[OH-]^2
Ksp = (0.317) x (2x10^-5)^2 = 1.262x10^-10 for this particular problem.
14-9.3 = 4.7 pOH,
[OH-] = 10^-(4.7) = 1.995x10^-5
Here's where things get difficult. NH4Cl breaks apart in water into NH4+ and Cl-. Therefore we can say the 1.2M is the concentration of NH4+. NH4+ combines with OH to form NH3 and H2O.
NH4+(aq)+OH-(aq) <-> NH3(aq)+H2O(l)
This equation is the equation for finding the Kb of NH3, but backward. Since we know that in order to flip a chemical equilibrium equation, we have to invert the K value,
Kb(NH3)=1.8x10^-5, Kb(NH3,rev)=5.6x10^4
Looking back at the equilibrium expression for NH4+ and OH- combining, we can solve an equilibrium to figure out what the OH had to be in the beginning in order to have excess [OH-] of 1.995x10^-5 M.
NH4+(aq)+OH-(aq) <-> NH3(aq)+H2O(l)
1.2M ? 0
-x -x +x
1.2-x 2.0x10^-5 x
5.6x10^4= x / (2.0x10^-5 x (1.2 - x))
x = 0.63398
So the [OH-] before NH4+ reacted with it was 0.63396 + 2x10^-5 = 0.63398
Now we work backward again and set up an ICE chart for the dissociation of Mg(OH)2.
Mg(OH)2(s) <-> Mg2+(aq)+2 OH-(aq)
N/A 0 0
N/A +x +2x
N/A ? 0.63398
In order to obtain 0.63398 as an answer, the change (x) must be 0.63398 / 2, making [Mg2+]= 0.317.
Ksp(Mg(OH)2)= [Mg2+]x[OH-]^2
Ksp = (0.317) x (2x10^-5)^2 = 1.262x10^-10 for this particular problem.
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