Solubility問題

ken

2013-03-11 4:12 am

1.
0.0044g of Pb(IO3)2 ,dissolved in 200cm3 of water to give a saturated solution.

Calculate the molar solubiliy of Pb(IO3)2.
Calculate the solubility product ,ksp of Pb(IO3)2.

2.
0.331g of BaF2 , dissolved in 250mL of water to give a saturated solution.

Determine the molar solubility of BaF2
Write the solubility equilibrium equation , and find the solubility product , ksp , of BaF2

回答 (1)

andrew

2013-03-11 9:41 am

✔ 最佳答案

1.
Molar mass of Pb(IO3)2 = 207.2 + 126.9x2 + 16x6 = 557 g/mol
No. of moles of Pb(IO3)2 = 0.0044/557 mol
Molar solubility of Pb(IO3)2 = (0.0044/557) / (200/1000)= 3.95 x 10⁻⁵ M

Pb(IO3)2(s) ⇌ Pb²⁺(aq) + 2IO3⁻(aq)
In the saturated solution :
[Pb²⁺] = 3.95 x 10⁻⁵ M
[IO3⁻] = (3.95 x 10⁻⁵) x 2 = 7.9 x 10⁻⁵ M

Ksp = [Pb²⁺] [IO3⁻]² = (3.95 x 10⁻⁵) x (7.9 x 10⁻⁵)² = 2.47 x 10⁻¹³ M³

=====
2.
1.
Molar mass of BaF2 = 137.3 + 19x2 = 175.3 g/mol
No. of moles of BaF2 = 0.331/175.3 mol
Molar solubility of Pb(IO3)2 = (0.331/175.3) / (250/1000)= 7.55 x 10⁻³ M

BaF2(s) ⇌ Ba²⁺(aq) + 2F⁻(aq)
In the saturated solution :
[Ba²⁺] = 7.55 x 10⁻³ M
[F⁻] = (7.55 x 10⁻³) x 2 = 1.51 x 10⁻² M

Ksp = [Ba²⁺] [F⁻]² = (7.55 x 10⁻³) x (1.51 x 10⁻²)² = 1.72 x 10⁻⁶ M³

參考: andrew

👍 0 👎 0