The Ksp of calcium hydroxide, Ca(OH)2, is 4.68 x 10-6 at 289.15K. Calculate the pH of a saturated solution.
Here's my work:
4.68 x 10-6= 4s^3
s= 0.0105
-log(0.0105)= 1.98
14-1.98=12.02
The answer is supposed to be 12.324, but I'm not sure what I'm doing wrong. Is it because it is not at 25C? Please help!
No, not the T. It could be EXCEPT Ksp is given at the T in the problem so that is ok. The problem here, I think, is that you failed to write the equation and because of that you slipped up on a tiny bit of the problem. Here is the dope.
.......Ca(OH)2 --> Ca^2+ + 2OH^-
I......solid........0........0
C......solid........s........2s
E......solid........s........2s
So you are exactly right and your work is great and s is your answer of 0.0105 M to the digit. BUT (OH^-) is 2s so if you will take the -log(2s) you will get the right answer for OH^- and convert that to pH.
Thanks for showing your work. It really helps us figure what's wrong.
Ah, the joys of chemistry calculations! Let's see if we can figure out what went wrong here.
First, it's important to note that the Ksp value you provided is at 289.15K, not at 25°C (298.15K). So you're correct in suspecting that the temperature difference might be causing the discrepancy in the answer.
To calculate the pH of a saturated solution, we need to consider the dissociation of calcium hydroxide:
Ca(OH)2(s) ⇌ Ca2+(aq) + 2 OH-(aq)
The equilibrium expression for Ksp is given by:
Ksp = [Ca2+][OH-]^2
We are given that Ksp = 4.68 x 10^-6. Since Ca(OH)2 is a strong base, it completely dissociates, leading to [Ca2+] = s and [OH-] = 2s (where s represents the solubility).
Substituting these values into the Ksp expression:
4.68 x 10^-6 = s * (2s)^2
4.68 x 10^-6 = 4s^3
Solving for s gives us s ≈ 0.0105, just as you calculated.
Now, here's where the temperature comes into play. The pH of a saturated solution depends on the concentration of hydroxide ions ([OH-]). At 25°C, we can use the concentration of hydroxide ions to determine the pH, but at 289.15K, we need to consider the ionization constant of water, Kw, which changes with temperature.
At 289.15K, the value of Kw is different from its usual value of 1.0 x 10^-14 at 25°C. However, knowing that Kw = [H+][OH-] at equilibrium, you can use the value of [OH-] (2s) you calculated earlier to find [H+]. From there, you can obtain the pH using the equation pH = -log[H+].
So, to get the correct pH at 289.15K, you need to consider the ionization constant of water at that temperature. That should help you arrive at the correct answer of 12.324.
Your calculation is mostly correct, but there is a small error in the calculation of the concentration of hydroxide ions (OH-) in the solution.
Let's go through the steps again:
1. Write the balanced equation for the dissociation of calcium hydroxide:
Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)
2. Write the expression for the Ksp of calcium hydroxide:
Ksp = [Ca2+][OH-]^2
3. Since calcium hydroxide is a strong base and it is a saturated solution, we can assume that almost all of it dissociates, so [Ca2+] is approximately equal to the concentration of OH- ions.
4. Let's assume the concentration of OH- ions is s. Now we can write the Ksp expression in terms of s:
Ksp = (s)(s^2) = s^3
5. Plug in the Ksp value and solve for s:
4.68 x 10^-6 = s^3
s = (4.68 x 10^-6)^(1/3)
s ≈ 0.0107 M (rounded to four significant figures)
6. Now, calculate the pOH of the solution:
-pOH = log10(0.0107) ≈ 1.97 (rounded to two decimal places)
7. Calculate the pH using the relation pH + pOH = 14:
pH = 14 - 1.97 ≈ 12.03 (rounded to two decimal places)
So, the pH of a saturated solution of calcium hydroxide at 289.15K is approximately 12.03, not 12.324 as you mentioned.
To calculate the pH of a saturated solution of calcium hydroxide (Ca(OH)2), you need to consider the dissociation of calcium hydroxide and the subsequent hydrolysis of the hydroxide ion.
The dissociation of calcium hydroxide can be represented as:
Ca(OH)2 ↔ Ca2+ + 2OH-
The equilibrium expression for this reaction is given by the solubility product constant, Ksp, which is expressed as the product of the concentration of the dissociated ions raised to the power of their respective stoichiometric coefficients:
Ksp = [Ca2+][OH-]^2
In this case, the value of Ksp is given as 4.68 x 10^-6. However, it's important to note that the expression for Ksp is valid only at a specific temperature, in this case, 289.15K.
To find the concentration of hydroxide ions in the saturated solution, we need to solve the equilibrium expression with the given Ksp:
Ksp = [Ca2+][OH-]^2
4.68 x 10^-6 = (x)(2x)^2
4.68 x 10^-6 = 4x^3
Solving the equation, we find:
x ≈ 0.0105
Now that you have the concentration of hydroxide ions in the saturated solution, you can calculate the pOH, which is the negative logarithm (base 10) of the hydroxide ion concentration:
pOH = -log(0.0105) ≈ 1.98
To calculate the pH of the solution, you should subtract the pOH from 14:
pH = 14 - pOH
pH ≈ 14 - 1.98
pH ≈ 12.02
Based on your calculations, it seems you have made an error in rounding off the pOH value. The correct pOH is approximately 1.98, leading to a pH of 12.02, not 12.324.
Therefore, the pH of the saturated solution of calcium hydroxide at 289.15K is approximately 12.02.