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

Answer 1

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

Answer 2

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.

$\boxed{ \ pH + pOH = 14 \ } \rightarrow \boxed{ \ pOH = 14 - pH \ }$

pOH = 14 - 10.30

Hence, the pOH value is 3.70.

Step-2

We use the pOH to get the $[OH^-].$

$\boxed{ \ pOH = -log[OH^-] \ } \rightarrow \boxed{ \ [OH^-] = 10^{-pOH} \ }$

$\boxed{ \ [OH^-] = 10^{-3.70} \ }$

Therefore, $\boxed{ \ [OH^-] = 1.995 \times 10^{-4} \ molar \ }$

Step-3

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

$\boxed{ \ Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^- \ }$

Notice that based on comparison of the coefficients, then $\boxed{ \ [Mg^{2+}] = \frac{1}{2}[OH^-] \ }$

• $\boxed{ \ [OH^-] = 1.995 \times 10^{-4} \ M \ }$
• $\boxed{ \ [Mg^{2+}] = \frac{1}{2} \times 1.995 \times 10^{-4} = 9.975 \times 10^{-5} \ M \ }$

Step-4

The Ksp expression:

$\boxed{ \ K_{sp} = [Mg^{2+}][OH^-]^2 \ }$

Let's calculate the Ksp value.

$\boxed{ \ K_{sp} = [1.995 \times 10^{-4}][9.975 \times 10^{-5}]^2 \ }$

Thus, the Ksp is $\boxed{ \ 1.985 \times 10^{-12} \ }$

Learn more  

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3. About electrolyte and nonelectrolyte solutions https://brainly.com/question/5404753

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