Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>
Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48
K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span>
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M)
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075
NH3 = 0.44 M
Answer:
Explanation:
The solution contain 0.01 M concentration of Ba²⁺
0.01M concentration of Ca²⁺
Ksp ( solubility constant) for BaSO₄ = 1.07 × 10⁻¹⁰
Ksp for CaSO₄ = 7.10 × 10⁻⁵
(BaSO₄) = (Ba²⁺) (SO₄²⁻)
1.07 × 10⁻¹⁰ = 0.01 M (SO₄²⁻)
1.07 × 10⁻¹⁰ / 0.01 = ( SO₄²⁻)
1.07 × 10⁻⁸ M = ( SO₄²⁻)
so the minimum of concentration of concentration sulfate needed is 1.07 × 10⁻⁸ M
For CaSO₄
CaSO₄ = ( Ca²⁺) ( SO₄²⁻)
7.10 × 10⁻⁵ = 0.01 (SO₄²⁻)
(SO₄²⁻) = 7.10 × 10⁻⁵ / 0.01 = 7.10 × 10⁻³ M
so BaSO₄ will precipitate first since its cation (0.01 M Ba²⁺) required a less concentration of SO₄²⁻ (1.07 × 10⁻⁸ M ) compared to CaSO₄
b) The minimum concentration of SO₄²⁻ that will trigger the precipitation of the cation ( 0.01 M Ba²⁺) that precipitates first is 1.07 × 10⁻⁸ M
Answer:
An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings
