Calculate the molar solubility of Cr(OH)3 in a solution with a PH of 11.6 knowing that the Ksp(Cr(OH)3)=6.7×10^-31.

Well, well, well, looks like we've got an equilibrium problem on our hands! Let's dive right in.

First, we need to know that the solubility of Cr(OH)3 is determined by the concentration of hydroxide ions (OH-) in the solution. And since we're given the pH, we can find the OH- concentration using the good old pH formula:

pOH = 14 - pH

Now, let's calculate the pOH:

pOH = 14 - 11.6
= 2.4

Next, we convert the pOH to OH- concentration:

[OH-] = 10^(-pOH)
= 10^(-2.4)

Now that we have the concentration of OH-, we can calculate the molar solubility of Cr(OH)3 using the Ksp expression:

Ksp = [Cr3+][OH-]^3

Since the molar solubility of Cr(OH)3 is equal to [Cr3+], let's call it x. And since [OH-] is already given, we substitute the values:

Ksp = x * ([OH-])^3

Rearranging, we get:

x = Ksp / ([OH-])^3

Now plug in the values:

x = (6.7×10^-31) / (10^(-2.4))^3

Hold on, let me get my trusty calculator...

*x takes time to calculate*

Alright, through the magic of calculations (and my trusty calculator), the molar solubility of Cr(OH)3 in a solution with a pH of 11.6 is approximately equal to **2.89 x 10^-12 M**.

Just remember, it's always important to keep your sense of humor even when dealing with molar solubility calculations. Stay clowny!

To calculate the molar solubility of Cr(OH)3, we can use the concept of hydrolysis. The hydrolysis of Cr(OH)3 is given by the equation:

Cr(OH)3 ⇌ Cr^3+ + 3OH^-

From the balanced equation, we can see that when one mole of Cr(OH)3 dissolves, it produces one mole of Cr^3+ and three moles of OH^- ions.

Since Cr(OH)3 is a strong base, it reacts with water to form OH^- ions. This means that the concentration of OH^- ions will be equal to the concentration of the hydroxide ions produced by the dissociation of water, which is determined by the pH of the solution.

The concentration of OH^- ions can be calculated using the equation:

OH^- = 10^(-pOH)

where pOH is the negative logarithm of the hydroxide ion concentration.

Given that pH = 11.6, we can calculate pOH as follows:

pOH = 14 - pH
pOH = 14 - 11.6
pOH = 2.4

Now, let's calculate the concentration of OH^- ions:

OH^- = 10^(-pOH)
OH^- = 10^(-2.4)
OH^- = 4.0 x 10^(-3) M

Since the concentration of OH^- ions is equal to the concentration of Cr^3+ ions, we can use this value to calculate the molar solubility of Cr(OH)3.

The Ksp expression for Cr(OH)3 is as follows:

Ksp = [Cr^3+][OH^-]^3

Given that Ksp = 6.7 x 10^(-31) and [OH^-] = 4.0 x 10^(-3) M, we can substitute these values into the Ksp expression:

6.7 x 10^(-31) = [Cr^3+](4.0 x 10^(-3))^3

Let's solve for [Cr^3+] (which is the molar solubility of Cr(OH)3):

[Cr^3+] = (6.7 x 10^(-31))/(4.0 x 10^(-3))^3
[Cr^3+] ≈ 1.05 x 10^(-23) M

Therefore, the molar solubility of Cr(OH)3 in a solution with a pH of 11.6 is approximately 1.05 x 10^(-23) M.

To calculate the molar solubility of Cr(OH)3 in a solution with a pH of 11.6, we need to consider the concept of hydrolysis.

When a hydroxide (OH-) ion is added to water, it reacts with water to form hydroxide ions and hydroxide ion concentration increases. This increases the pH of the solution.

In the case of Cr(OH)3, it is an ionic compound. When it dissolves in water, it dissociates into Cr3+ ions and OH- ions. The OH- ions contribute to the high pH of the solution.

Since the Ksp (solubility product constant) of Cr(OH)3 is given as 6.7×10^-31, we can write the solubility equilibrium expression as follows:

Ksp = [Cr3+][OH-]^3

Since Cr(OH)3 is a weak base, it reacts with water (H2O) to form OH- ions. This means that:

[Cr3+] = [OH-] (concentration of Cr3+ ions is equal to the concentration of OH- ions)

Thus, we can substitute [OH-] for [Cr3+] in the Ksp expression:

Ksp = [OH-]^4

To calculate the molar solubility of Cr(OH)3, we need to find the concentration of OH- ions in the solution. Since pH is given as 11.6, we can calculate the hydroxide ion concentration ([OH-]) using the pOH formula:

pOH = 14 - pH

pOH = 14 - 11.6 = 2.4

To find the [OH-] concentration, we need to convert pOH to OH- concentration using the formula:

[OH-] = 10^(-pOH)

[OH-] = 10^(-2.4) = 4.0 × 10^(-3) M

Now, we can substitute [OH-] into the Ksp expression and solve for the molar solubility of Cr(OH)3:

Ksp = [OH-]^4
6.7×10^-31 = (4.0 × 10^(-3))^4

Taking the fourth root of both sides:

(6.7×10^-31)^(1/4) = 4.0 × 10^(-3)

Molar solubility of Cr(OH)3 in the given solution is approximately 4.0 × 10^(-3) M.

pH+pOH=14

14-pH=pOH

pOH=14-11.6=2.4

pOH=-log[OH^-]

[OH^-]=10^-[2.4]

[OH^-]=3.98 x 10^-3 M= initial concentration

Cr(OH)3 -------> Cr^3+ + 3OH^-

Ksp=[x][3.98 x 10^-3 M]^3

Ksp/[3.98 x 10^-3 M]^3=x

x=6.7×10^-31/[3.98 x 10^-3 M]^3

Solve for x