# 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

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 10.30. what is the ksp of the salt at 22 °c?

The balanced reaction equation:
M(OH)2 ↔ M^2+ + 2(OH)^-

and when the Ksp = [M^2+][OH-]^2
when PH + POH = 14
∴ POH = 14-10.3= 3.7
and when POH = - ㏒ [OH-]
3.7 = -㏒[OH-]
∴[OH] = 2x10^-4
and when [M^2+] = 1/2[OH-]
∴[M^2+] = (2x10^-4) / 2 = 0.0001 M
So, by substitution in Ksp formula:
∴Ksp = (0.0001 * (2x10^-4)^2 = 4x10^-12

1.985 x 10⁻¹²

Further explanation

Given:

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 10.30.

Question:

What is the Ksp of the salt at 22°C?

The Process:

Step-1

Because the pH is above 7, we convert it to pOH.

pOH = 14 - 10.30

Hence, the pOH value is 3.70.

Step-2

We use the pOH to get the

Therefore,

Step-3

Let us write the chemical equation in equilibrium of ions.:

Notice that based on comparison of the coefficients, then

Step-4

The Ksp expression:

Let's calculate the Ksp value.

Thus, the Ksp is

Learn more  Write the equilibrium constant for the reaction  Calculating the pH value of weak base About electrolyte and nonelectrolyte solutions

Keywords: Ksp, equilibrium, pH, pOH, metal hydroxide, M(OH)₂, pure water, the chemical equation, ions,

2.29x10⁻¹² is Ksp of the salt

Explanation:

The Ksp of the metal hydroxide is:

M(OH)₂(s) ⇄ M²⁺ + 2OH⁻

Ksp = [M²⁺] [OH⁻]²

As you can see in the reaction, 2 moles of OH⁻ are produced per mole of M²⁺. It is possible to find [OH⁻] with pH, thus:

pOH = 14- pH

pOH = 14 - 10.22

pOH = 3.78

pOH = -log[OH⁻]

1.66x10⁻⁴ = [OH⁻]

And [M²⁺] is the half of [OH⁻], [M²⁺] = 8.30x10⁻⁵

Replacing in Ksp formula:

Ksp = [8.30x10⁻⁵] [1.66x10⁻⁴]²

Ksp = 2.29x10⁻¹² is Ksp of the salt

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