You place 1.234 g of solid Ca(OH)2 in 1.00 L of pure water at 25°C and it partially dissolves. The pH of the solution is 12.40. Estimate the Ksp for Ca(OH)2.

Ca(OH)2 ==> Ca^2+ + 2OH^-

pH = 12.40
pOH = 14-12.40
pOH = -log(OH^-)
Ksp = (Ca^2+)(OH^-)^2.
You know (OH^-) from above, (Ca^2+) will be 1/2 of that, substitute into the Ksp expression and solve for Ksp.

To estimate the Ksp (solubility product constant) for Ca(OH)2, also known as calcium hydroxide, we can use the pH of the solution and the concentration of hydroxide ions (OH-) present.

Step 1: Determine the concentration of hydroxide ions (OH-)
Since we know the pH of the solution is 12.40, we can use the equation:
pOH = 14 - pH
pOH = 14 - 12.40
pOH = 1.60

To convert pOH to OH- concentration, we use the equation:
pOH = -log[OH-]
1.60 = -log[OH-]

Now we can solve for [OH-]:
[OH-] = 10^(-pOH)
[OH-] = 10^(-1.60)
[OH-] = 0.0251 M

Step 2: Calculate the concentration of calcium ions (Ca2+)
Since calcium hydroxide is a strong electrolyte that dissociates completely, the concentration of Ca2+ ions will be twice the concentration of OH- ions.

[Ca2+] = 2 * [OH-]
[Ca2+] = 2 * 0.0251 M
[Ca2+] = 0.0502 M

Step 3: Calculate the solubility product constant (Ksp)
The Ksp expression for Ca(OH)2 is:
Ksp = [Ca2+][OH-]^2

Substitute the calculated concentrations into the equation:
Ksp = (0.0502 M)(0.0251 M)^2
Ksp = 3.16 x 10^(-5) (rounded to the appropriate number of significant figures)

Therefore, the estimated Ksp for Ca(OH)2 is 3.16 x 10^(-5).

To estimate the Ksp (solubility product constant) for Ca(OH)2, we need to first determine the concentration of hydroxide ions (OH-) in the solution.

Given that the pH of the solution is 12.40, we know that the concentration of hydroxide ions can be calculated using the formula: [OH-] = 10^(-pH)

Substituting the pH value into the formula, we get: [OH-] = 10^(-12.40)

Now, since Ca(OH)2 dissociates into one calcium ion (Ca2+) and two hydroxide ions (OH-), we can assume that the concentration of calcium ions is double the concentration of hydroxide ions.

So, the concentration of calcium ions (Ca2+) can be estimated as: [Ca2+] ≈ 2 * [OH-]

Next, we need to calculate the molar concentration (M) of calcium ions ([Ca2+]).

Given that 1.234 g of Ca(OH)2 was added to 1.00 L of water, we can calculate the number of moles of Ca(OH)2 using its molar mass. The molar mass of Ca(OH)2 is approximately 74.096 g/mol.

Number of moles of Ca(OH)2 = mass / molar mass = 1.234 g / 74.096 g/mol

Now, since one Ca(OH)2 molecule yields one Ca2+ ion when it dissolves, the initial concentration of Ca2+ ions can be calculated as: Initial [Ca2+] = (moles of Ca(OH)2) / (1.00 L)

Finally, we can estimate the Ksp of Ca(OH)2 by multiplying the concentrations of the calcium ions and hydroxide ions: Ksp ≈ [Ca2+] * ([OH-]^2)

Substituting the calculated values, we have:
Ksp ≈ (Initial [Ca2+]) * ([OH-]^2)

Plug in the values and calculate to estimate the Ksp for Ca(OH)2.

Well, I have to give it to you - you're really testing my chemistry skills here! But fear not, because when it comes to Ksp, I can still clown around!

So, let's solve this problem step by step. First, we know that Ca(OH)2 partially dissolves in water, meaning it undergoes dissociation. And since Ca(OH)2 is a strong base, it'll produce OH- ions.

The equation for the dissociation of Ca(OH)2 is:

Ca(OH)2 (s) ⇌ Ca2+ (aq) + 2OH- (aq)

Now for the pH. pH is a measure of the concentration of H+ ions in a solution. Since this is a basic solution, we need to find the concentration of OH- ions and then do some calculations to figure out the H+ concentration.

But instead of boring calculations, let's try a humorous approach! Imagine the OH- ions feeling grumpy because they're outnumbered. So, they complain to the H+ ions, saying, "Hey, H+, can you do us a solid and create equilibrium?"

That's when the H+ ions reply, "Sorry, I'm a bit acidic. I can't handle basic situations like this."

Now, I know you're thinking, "Clown Bot, get to the point!" So, Estimating the concen-tration of [OH-] ions using the formula:

pOH = 14 - pH
pOH = 14 - 12.40
pOH = 1.60

10^(-pOH) = [OH-]
10^(-1.60) = [OH-]
[OH-] = 0.0251 M

Since 1 molecule of Ca(OH)2 produces 2 OH- ions, we have:

[Ca(OH)2] = 0.0124 mol/L
[OH-] = 2[Ca(OH)2] = 2(0.0124) = 0.0248 M

Now, let's have a dramatic pause! *Drumrolls*

So, the Ksp for Ca(OH)2 can be calculated by multiplying the concentration of Ca2+ ions with the concentration of OH- ions raised to the power of their stoichiometric coefficient, which is 2.

Ksp = [Ca2+][OH-]^2
Ksp = (x)(0.0248)^2

But wait... 'x' is the concentration of Ca2+ ions. *dramatic music plays*

Now, our dear friend 'x' doesn't want to reveal itself to us. It's playing hide and seek, so let's solve for 'x' using the equation for the equilibrium constant expression. Duh-duh-duh!

So, Ksp = 4x^3 = x(0.0248)^2
4x^3 = 0.0000615x
4x^2 = 0.0000615
x = 0.00309 M

And there you have it! The estimated Ksp for Ca(OH)2 is approximately 0.00309. So, remember, when in doubt, mix a little humor with your chemistry!