Answer:
The concentration of [Ca²⁺] is 8.47 x 10⁻³ M
Explanation:
We consider the solubility of hydroxyapatite,
Ca₁₀(PO₄)₆(OH)₂ ⇔ 10Ca²⁺ + 6PO₄³⁻ + 2 OH⁻
Assumed that there is <em>a</em> mol of hydoxyapatite disolved in water, yielding <em>10a</em> mol Ca²⁺ of and <em>6a</em> mol of PO₄³⁻
We also have Ksp equation,
Ksp = [Ca²⁺]¹⁰ x [PO₄³⁻]⁶ x [OH⁻]² = 2.34 x 10⁻⁵⁹
⇔ 10a¹⁰ x 6a⁶ x (5.30 x 10⁻⁶)² = 2.24 x 10⁻⁵⁹
⇔ 60a¹⁶ = 2.24 x 10⁻⁵⁹ / 5.30 x 10⁻¹²
⇔ a¹⁶ = 0.007 x 10⁻⁴⁷ = 7 x 10⁻⁵⁰
⇔ a =
= 8.47 x 10⁻⁴
Hence,
[Ca²⁺] = 10<em>a</em> = 8.47 x 10⁻³ M
Answer:
What are we helping you on?
Explanation:
..
The volume of I- is missing in your question by assuming it = 1L
moles I- = molarity * volume
= 0.00237 * 1 L
= 0.00237 mol
[I-] = moles / total volume
= 0.00237 / 1.625L
= 0.00146 M
moles Pb2+ = molarity * volume
= 0.00785 * 0.625 L
= 0.0049 mol
[Pb2+] = 0.0049 / 1.625L
= 0.003 M
when PbI2(s) ↔ Pb2+(aq) + 2I-(aq)
when Q = [Pb2+][I-]^2 and we neglect [PbI2] as it is solid
∴ Q = 0.003 * (0.00146)^2
= 6.4 x 10^-9
by comparing the value of Q with Ksp value we will found that:
Q < Ksp which mean that more solid will dissolve, and this is an unsaturated solution which has ion concentrations < equilibrium concentrations, so the reaction will go forward until achieving equilibrium.