Answer:
Explanation:
Hello,
In this case, for the dissociation of calcium fluoride:
The equilibrium expression is:
![Ksp=[Ca^{2+}][F^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2)
In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent 
is computed as follows:
Thus, the molar solubility equals the reaction extent , therefore:
Regards.
KH₂PO₄ hydrolyzes as;
H₂PO₄⁻ + H₂O ↔ H₃PO₄ + OH⁻
Let x amount of H₂PO₄⁻ has reacted with water then,
Kb₁ = [H₃PO₄][OH⁻] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.8-x M
Kb₁ = x² / (0.8 - x)
Given Ka₁ = 7.5 x 10⁻³
so Kb₁ = 1 x 10⁻¹⁴ / (7.5 x 10⁻³) = 1.33 x 10⁻¹²
From this information:
1.33 x 10⁻¹² = x² / 0.8
x = [OH⁻] = 1.03 x 10⁻⁶ M
pOH = - log (1.03 x 10⁻⁶) = 5.99
pH = 14 - pOH = 14 - 5.99 = 8.01
An ionized has consisting of positive ions and free elections in proportion resulting in more or less no overall electric charge
