[GeneBelcher42648]

Question: 8.3 grams of calcium hydroxide is dissolved in 342 mL of water. Calculate the volume of water required to dilute the solution for it to have a pH of 9.2.

I received this question in a practice test and cannot seem to get an answer even remotely similar to my teacher’s, and upon discussing with my teacher how to get the answer I was still confused on how she got it and am even a bit iffy on if her answer is completely correct. I believe all the information needed is presented but if not let me know.

[Meter]

Can you show your work?

[GeneBelcher42648]

Calcium Hydroxide = Ca(OH)₂ , gmol^-1 = 74.1
8.3/74.1 = 0.112mol
0.112/0.342 = 0.3275M

10^-9.2 = 6.31 × 10^-10 = [h₃O⁺]
10^-14/(6.31 × 10^-10) = 1.585 × 10^-5 = [OH^-]
c₁v₁ = c₂v₂
342 × 0.3275 = (1.585 × 10^-5) × v
v = 112/0.00001585 = 7066246mL
My Answer: 7066L

[Meter]

What you really want to ask yourself is what concentration of hydroxide you need in your solution in order to achieve a pH of 9.2. Hint: since pH + pOH = 14, you might be better off asking what pOH you want and then set up an equation and solve it.

[GeneBelcher42648]

Thanks, I think I get it now after trying again with your hint. Appreciate it.

[Meter]

Thanks, I think I get it now after trying again with your hint. Appreciate it.

I must admit that I am a little skeptical of the result you get from my method, but since pH behaves logarithmically, things can get large quickly.

[Borek]

Ca(OH)₂ Ca²⁺ + 2OH^-

(that's not entirely true, this is an equilibrium reaction, but that's where you went wrong)

[Meter]

Ca(OH)₂ Ca²⁺ + 2OH^-

(that's not entirely true, this is an equilibrium reaction, but that's where you went wrong)

Question: what do you get the added volume to be?

[AWK]

The task is unrealistic for several reasons. Only about 0.5 g of Ca(OH)₂ will be dissolved in 342 ml of water. In the case of calcium hydroxide, the solubility cannot be increased by heating because its solubility decreases as the temperature rises.
If we are going to dilute the suspension, it doesn't matter what amount of hydroxide we use. But if it is a solution, as in the problem, the actual concentration of the solution (and not the suspension) will be about 15 times lower (then you need to know the temperature at which the solution was made and the solubility of the hydroxide at this temperature).

[Borek]

Ca(OH)₂ Ca²⁺ + 2OH^-

(that's not entirely true, this is an equilibrium reaction, but that's where you went wrong)

Question: what do you get the added volume to be?

More or less twice more than what OP got.

[Meter]

Ca(OH)₂ Ca²⁺ + 2OH^-

(that's not entirely true, this is an equilibrium reaction, but that's where you went wrong)

Question: what do you get the added volume to be?

More or less twice more than what OP got.

Strange, I got more or less half (~3000 L). Can I see your work just so I can compare?

Edit: I think I see my mistake. I mistakenly thought that we had 2 hydroxides for every calcium hydroxide, but this isn't the case. But still, correcting for this only gives me a little over 6000 L.

[Borek]

First things first: what concentration of Ca(OH)₂ do we need for pH of 9.2?

[Meter]

First things first: what concentration of Ca(OH)₂ do we need for pH of 9.2?

Since Ca(OH)₂ is a base, we can't use pH = -log([H₃O⁺]), but we can use pOH = -log([OH^-]), and since we know that pH + pOH = 14, we can find that for pH = 9.2 we have pOH = 4.8. Solving for hydroxide concentration, we see that [OH^-] = 10^-4.8, which gives us the required concentration of Ca(OH)₂ for pH = 9.2. Is my reasoning correct here?

We can calculate the amount of Ca(OH)₂ using n = m/M which yields n = 0.11 mol. Using c = n/V we can solve for V and find out the total amount of water needed to have pH = 9.2. Then we simply subtract our initial volume from that number to get how much water we have to add.

Is this correct?

[Borek]

Solving for hydroxide concentration, we see that [OH^-] = 10^-4.8, which gives us the required concentration of Ca(OH)₂ for pH = 9.2. Is my reasoning correct here?

Technically yes, but the way you worded it it is not clear if you took the factor of 2 into account.

Plus, no idea what is the number you got

We can calculate the amount of Ca(OH)₂ using n = m/M which yields n = 0.11 mol. Using c = n/V we can solve for V and find out the total amount of water needed to have pH = 9.2.

OK

Then we simply subtract our initial volume from that number to get how much water we have to add.

While technically correct final volume differs from the initial by several orders of magnitude, so the subtraction can be safely ignored.

[Meter]

Solving for hydroxide concentration, we see that [OH^-] = 10^-4.8, which gives us the required concentration of Ca(OH)₂ for pH = 9.2. Is my reasoning correct here?

Technically yes, but the way you worded it it is not clear if you took the factor of 2 into account.

Plus, no idea what is the number you got

We can calculate the amount of Ca(OH)₂ using n = m/M which yields n = 0.11 mol. Using c = n/V we can solve for V and find out the total amount of water needed to have pH = 9.2.

OK

Then we simply subtract our initial volume from that number to get how much water we have to add.

While technically correct final volume differs from the initial by several orders of magnitude, so the subtraction can be safely ignored.

Since we have 2 moles hydroxides per 1 mole calcium hydroxide, then we relate the concentrations of each as

2 [OH^-] = [Ca(OH)₂]

Which means that we get a factor of 1/2 rather than 2 when we solve for the hydroxide concentration, no?

The number I got is ~3470 L