Answer:
0.0099M = [F⁻]
Explanation:
<em>For BaF2, Ksp = 1.7x10⁻⁶</em>
<em />
When BaF₂ is in solution, the equilibrium between the solid and the dissociated ions occurs as follows:
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)
Where Ksp = 1.7x10⁻⁶ is defined as:
1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
<em>Where [] are equilibrium concentrations of each ion in solution.</em>
<em />
That means you will add F⁻ until its concentration exceeds:
1.7x10⁻⁶ = [0.0173] [F⁻]²
9.827x10⁻⁵ = [F⁻]²
<h3>0.0099M = [F⁻]</h3><h3 />
When more F⁻ is added, BaF₂ begins its precipitation.
Answer:
13.5 mL of solution.
Explanation:
Given:
the solution is of 2% concentration.
The dosage required is 10mg/kg
The weight of dog is 27 kg
Solution:
The mass of barbiturate required for the 27 kg dog = 27 X 10 mg = 270 mg
As the solution is 2%, so there is 2g of barbiturate is dissolved in 100mL of solution
The mass of barbiturate required = 0.270 g
For 2 g we will need 100mL of solution
For 1 g we will need = 50mL of solution
for 0.270g we will need = 50 X 0.270 mL of solution = 13.5 mL of solution.