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?
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
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?
Because the pH is above 7, we convert it to pOH.
pOH = 14 - 10.30
Hence, the pOH value is 3.70.
We use the pOH to get the
Let us write the chemical equation in equilibrium of ions.:
Notice that based on comparison of the coefficients, then
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
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