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.
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.
Well, it seems like we have a chemical comedy unfolding here! Let's dive right in and solve this puzzle, shall we?
To find the Ksp of Mg(OH)2, we need to use some chemical calculations. The first thing we need to do is figure out the concentration of OH- ions in the saturated solution.
Since the pH is given to be 9.30, we can use the equation pH = -log[H+]. We know that the concentration of OH- is equal to [H+], so we can rewrite the equation as:
pOH = -log[OH-]
Now, to calculate pOH, we need to subtract the given pH from 14:
pOH = 14 - 9.30
pOH ≈ 4.70
Now that we have pOH, we can use it to calculate the concentration of OH-. Since pOH is the negative logarithm of the OH- concentration, we can write:
OH- = 10^(-pOH)
OH- = 10^(-4.70)
OH- ≈ 2.99 x 10^(-5) M
Since Mg(OH)2 is a strong electrolyte and dissociates completely, we can assume that the concentration of Mg2+ ions is equal to twice the concentration of OH- ions:
[Mg2+] ≈ 2 × 2.99 x 10^(-5) M
Now, we can substitute this value of [Mg2+] into the expression for the Ksp of Mg(OH)2 to solve for Ksp:
Ksp = [Mg2+] × [OH-]^2
Ksp ≈ (2 × 2.99 x 10^(-5))^2
Ksp ≈ 8.95 x 10^(-9)
And there you have it - the Ksp of Mg(OH)2 is approximately 8.95 x 10^(-9). Voilà!
Remember, chemistry can be a bit tricky, so it's important to approach it with a sense of humor. If you have any more questions, feel free to ask!
To calculate the Ksp (solubility product constant) of Mg(OH)2, we first need to understand the equation that represents the dissolution of this compound in water. The balanced chemical equation is:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
The solubility product expression is then defined as follows:
Ksp = [Mg2+][OH-]^2
However, in this problem, we are given the pH of the saturated solution formed, rather than the actual concentrations of Mg2+ and OH-. So, we need to use the given pH to determine the concentration of OH- ions in the solution.
The concentration of OH- ions in a solution can be calculated using the following formula:
pOH = -log[OH-]
To obtain the concentration of OH-, we need to calculate the pOH, which is the negative logarithm of the hydroxide ion concentration.
Given that the pH of the saturated solution is 9.30, we can calculate the pOH:
pOH = 14 - pH
pOH = 14 - 9.30
pOH = 4.70
Now that we have the pOH, we can find the concentration of OH- ions using the formula:
[OH-] = 10^(-pOH)
[OH-] = 10^(-4.70)
[OH-] = 2.99 x 10^(-5) M
Now, we can plug the concentration of OH- into the solubility product expression and solve for Ksp:
Ksp = [Mg2+][OH-]^2
Ksp = (x)(2.99 x 10^-5)^2
However, since excess solid Mg(OH)2 is present, it can be assumed that all of the Mg(OH)2 dissolves and provides an equal concentration of Mg2+ ions in the solution. Hence, we can substitute [Mg2+] with 2[OH-] in the Ksp expression:
Ksp = (2[OH-])([OH-])^2
Ksp = 2([OH-])^3
Ksp = 2(2.99 x 10^-5)^3
Ksp = 5.37 x 10^-16
Therefore, the solubility product constant (Ksp) for Mg(OH)2 is 5.37 x 10^-16.