A Ca(OH)2 solution has a pH of 8.0. Determine the values of [H3O+], [OH– ], and [Ca(OH)2] for the solution
To determine the values of [H3O+], [OH–], and [Ca(OH)2] for the Ca(OH)2 solution with a pH of 8.0, we need to use the relationship between pH and pOH, along with the dissociation constant of water (Kw) and the dissociation constant of Ca(OH)2 (Ksp).
1. Calculate pOH:
pOH = 14 - pH = 14 - 8.0 = 6.0
2. Calculate [OH–]:
Using the relationship between pOH and [OH–]:
pOH = -log[OH–]
6.0 = -log[OH–]
To solve for [OH–], take the antilog of both sides:
[OH–] = 10^(-pOH) = 10^(-6.0) = 1 x 10^(-6) M
So, the [OH–] concentration in the Ca(OH)2 solution is 1 x 10^(-6) M.
3. Calculate [H3O+]:
Since the solution has a pH of 8.0, we know that it is basic (pH > 7). In a basic solution, the concentration of [H3O+] is related to [OH–] using the dissociation constant of water (Kw), which is equal to 1.0 x 10^(-14) at 25°C:
Kw = [H3O+][OH–]
1.0 x 10^(-14) = [H3O+](1 x 10^(-6))
To solve for [H3O+], rearrange the equation:
[H3O+] = Kw / [OH–] = (1.0 x 10^(-14)) / (1 x 10^(-6))
[H3O+] = 1 x 10^(-8) M
So, the [H3O+] concentration in the Ca(OH)2 solution is 1 x 10^(-8) M.
4. Calculate [Ca(OH)2]:
To determine the concentration of [Ca(OH)2], we need to consider that Ca(OH)2 is a strong base and fully dissociates in water:
Ca(OH)2 → Ca^2+ + 2OH–
We know that the concentration of [OH–] is 1 x 10^(-6) M, and since Ca(OH)2 dissociates in a 1:2 ratio, the [Ca(OH)2] concentration is equal to half the [OH–] concentration:
[Ca(OH)2] = (1 x 10^(-6)) / 2
[Ca(OH)2] = 5 x 10^(-7) M
So, the [Ca(OH)2] concentration in the Ca(OH)2 solution is 5 x 10^(-7) M.
Therefore, the values of [H3O+], [OH–], and [Ca(OH)2] in the Ca(OH)2 solution with a pH of 8.0 are:
[H3O+] = 1 x 10^(-8) M
[OH–] = 1 x 10^(-6) M
[Ca(OH)2] = 5 x 10^(-7) M
Whoa, let me whip out my chemistry jokes for this one! Alright, so we have a pH of 8.0. That means we have a basic solution on our hands, but not too basic. Now, for the values you're looking for:
1. [H3O+]: Well, in a basic solution, the concentration of H3O+ ions is lower than that of OH- ions. So it'll be somewhere below 1. But, as an obedient bot, I'll need a specific value. How about we dub it the elusive "X"? Yeah, let's go with that. [H3O+] = X.
2. [OH-]: Ah, our star of the show! In this solution, we know that Ca(OH)2 is yielding two OH- ions for every molecule. So the concentration of OH- is double that of Ca(OH)2. The concentration of [OH-] = 2[Ca(OH)2]. Time to break out your calculator!
3. [Ca(OH)2]: Ah, calcium hydroxide. It's the substance that's giving us those precious OH- ions. Now, to calculate its concentration, we can assume that it completely dissociates (which is a bit of a simplification, but hey, we're clowning around here!). So [Ca(OH)2] = half of the concentration of OH-. Let's put that calculator to work again!
Remember, these calculations are just approximations, but don't worry, chemistry is all about the fun of approximations! Just make sure you double-check your numbers with a more precise method if it's needed. Have a blast! 🎉
To determine the values of [H3O+], [OH–], and [Ca(OH)2] for the solution, we first need to use the fact that water autoprotolysis occurs, which is represented by the equation:
2H2O ⇌ H3O+ + OH–
In a neutral solution, the concentration of H3O+ ions (equivalent to H+) is equal to the concentration of OH– ions. However, in this case, the solution has a pH of 8.0, indicating it is slightly basic. Therefore, the concentration of OH– ions is higher than the concentration of H3O+ ions.
Step 1: Determine [H3O+]
Given that the solution has a pH of 8.0, pH is defined as the negative logarithm (base 10) of the concentration of H3O+ ions. Using the equation:
pH = -log[H3O+]
Rearranging the equation gives:
[H3O+] = 10^(-pH)
Substituting pH = 8.0 gives:
[H3O+] = 10^(-8.0)
Calculating this value gives [H3O+] = 1.0 x 10^(-8.0).
Step 2: Determine [OH–]
Since the concentration of OH– ions can be determined using the autoprotolysis equation, we know that the concentration of [H3O+] is equal to the concentration of [OH–] in a neutral solution. However, since this solution is basic, [OH–] is higher than [H3O+].
Therefore, [OH–] will have a concentration greater than 1.0 x 10^(-8.0) M.
Step 3: Determine [Ca(OH)2]
To determine the concentration of [Ca(OH)2], we need the concentration of OH– ions. Since [OH–] is higher than [H3O+], we can calculate the concentration of [Ca(OH)2] using the balanced chemical equation for its dissociation:
Ca(OH)2 ⇌ Ca2+ + 2OH–
The concentration of [Ca2+] will be equal to twice the concentration of [OH–]. Therefore, [Ca(OH)2] can be calculated based on the concentration of [OH–] determined in Step 2.
pH = - log(H^+) = 8 so (H^+) = 1E-8 M
pH + pOH = 14 so pOH = 6 and (OH^-) = 1E-6 M
[Ca(OH)2] = 1/2*(OH^-) = 5E-4 M
I did not try to correct for the H3O^+ and/or OH^- in solution from the water. The numbers shown above are 10 times away from normal (10^-7 for each) so there will be only a very slight change if that were included. I've ignored the difference.