At 22 °C an excess amount of a generic metal hydroxide M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.49. What is the Ksp of the salt at 22 °C?

Question

At 22 °C an excess amount of a generic metal hydroxide M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.49. What is the Ksp of the salt at 22 °C?

Is the ksp=1.38*10^-11?

Answer 1

M(OH)2---> M^+ + 2OH

14-pH=pOH

14-10.49=3.51

pOH=-log[OH]

10^(-3.51)=OH concentration

OH concentration is twice as much as M from the reaction.

Ksp=products/reactants=[OH Concentration]^2[OH concentration/2]

I calculate 1.48 x 10^-11

Answer 2

I agree with 1.48E-11

Answer 3

To determine the Ksp of a salt, we need to use the pH of the equilibrium solution. However, the given information in the question is not sufficient to directly calculate the Ksp value. We will need to perform some calculations to obtain the Ksp.

First, let's convert the pH of the solution into the concentration of hydroxide ions ([OH-]).

The pH of a solution is related to the concentration of hydroxide ions by the equation:

pOH = 14 - pH

pOH represents the negative logarithm of the hydroxide ion concentration.

Therefore, the pOH of the solution is:

pOH = 14 - 10.49 = 3.51

Next, we need to convert the pOH into the concentration of hydroxide ions ([OH-]).

The relationship between pOH and [OH-] is given by the equation:

[OH-] = 10^(-pOH)

Therefore, the concentration of hydroxide ions ([OH-]) in the solution is:

[OH-] = 10^(-3.51)

Now, let's assume that the concentration of the metal ions (M2+) is also equal to the concentration of hydroxide ions ([M2+] ≈ [OH-]). This assumption is based on the balanced chemical equation for the dissociation of the metal hydroxide M(OH)2:

M(OH)2 --> M2+ + 2OH-

Now, let's substitute the concentration of hydroxide ions ([OH-]) into the equation:

[OH-]^2 ≈ [M2+][OH-]

Since we can assume [M2+] ≈ [OH-], we can rewrite the equation as:

[OH-]^2 ≈ [OH-]^2

This means that the concentration of hydroxide ions squared is approximately equal to itself, which is always true.

This implies that the equilibrium constant (Ksp) for the dissolution of the metal hydroxide is:

Ksp = [M2+][OH-]^2

Since we assume [M2+] ≈ [OH-], we can rewrite the equation as:

Ksp = [OH-]^3

Finally, plug in the value of [OH-] = 10^(-3.51) into the equation to calculate the Ksp:

Ksp = (10^(-3.51))^3

Evaluating this expression gives:

Ksp ≈ 2.07 x 10^(-11)

Therefore, the Ksp of the generic metal hydroxide M(OH)2 at 22 °C is approximately 2.07 x 10^(-11), not 1.38 x 10^(-11) as mentioned in your question.