Question

What is the solubility of cr(oh)₃ at a ph of 11.00? (ksp cr(oh)₃ is 6.70 × 10⁻³¹)

Answer 1

The solubility of Cr(OH)3 at pH 11.00 can be calculated from its Ksp value and the OH- ion concentration at that pH. The solubility is approximately 6.70 × 10−22 mol/L.

Step-by-step explanation:

To calculate the solubility of Cr(OH)3 at a pH of 11.00 using its Ksp (6.70 × 10−31), we need to determine the concentration of OH− ions at that pH level first. We know that pH + pOH = 14, so pOH = 3. Using pOH, we can find that the concentration of OH− ions is 10−3 M (because pOH = −log[OH−]). We can then set up the equilibrium expression for the solubility product:

Ksp = [Cr3+][OH−]3

Assuming that the solubility of Cr(OH)3 is 's' mol/L, [Cr3+] = s and [OH−]3 = (10−3 − s)3. Plugging these into the Ksp expression, we get:

6.70 × 10−31 = s(10−3 − s)3

We can ignore 's' in the (10−3 − s)3 term given its likely small value compared to 10−3. Therefore, the expression simplifies to:

s = ∼6.70 × 10−31 / (10−9) ∼ − 6.70 × 10−22

The calculated solubility of Cr(OH)3 is approximately 6.70 × 10−22 mol/L at pH 11.00.

Answer 2

The solubility of Cr(OH)3 at a pH of 11.00 can be calculated using its solubility product constant (Ksp) and the hydroxide ion concentration. The expression [Cr(OH)3] = (Ksp / [OH-]^3)^(1/3) can be used to determine the solubility. Substituting the given values into the expression will give the solubility of Cr(OH)3 at a pH of 11.00.

Step-by-step explanation:

The solubility of a compound can be determined from its solubility product constant (Ksp). In this case, the compound is Cr(OH)3 and its Ksp is given as 6.70 × 10-31. To determine the solubility at a pH of 11.00, we need to consider the hydroxide ion concentration in the solution.

The pH of a solution is a measure of its acidity or basicity. At a pH of 11.00, the solution is basic. This means that the hydroxide ion concentration ([OH-]) is high. The solubility of Cr(OH)3 is dependent on the hydroxide ion concentration.

To calculate the solubility, we need to use the Ksp expression for the compound:

Ksp = [Cr(OH)3]3[OH-]3

We can rearrange the expression to solve for the solubility ([Cr(OH)3]):

[Cr(OH)3] = (Ksp / [OH-]3)1/3

Substituting the given values, we find:

[Cr(OH)3] = (6.70 × 10-31 / [OH-]3)1/3

Therefore, the solubility of Cr(OH)3 at a pH of 11.00 can be calculated using the expression above.