Answer is: the maximum concentration of Pb²⁺ is 6.8·10⁻³ M.
Chemical reaction 1: PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq).
Chemical reaction 2: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Ksp(PbCl₂) = 1.7·10⁻⁵.
c(NaCl) = c(Cl⁻) = 0.0500 M.
Ksp(PbCl₂) = c(Pb²⁺) · c(Cl⁻)².
c(Pb²⁺) = Ksp(PbCl₂) ÷ c(Cl⁻)².
c(Pb²⁺) = 1.7·10⁻⁵ M³ ÷ (0.0500 M)².
c(Pb²⁺) = 0.000017 M³ ÷ 0.0025 M².
c(Pb²⁺) = 0.0068 M = 6.8·10⁻³ M.
Answer:
Solution's mass = 200.055 g
[PbSO₄] = 275 ppm
Explanation:
Solute mass = 0.055 g of lead(II) sulfate
Solvent mass = 200 g of water
Solution mass = Solvent mass + Solution mass
0.055 g + 200 g = 200.055 g
ppm = μg of solute / g of solution
We convert the mass of solute from g to μg
0.055 g . 1×10⁶ μg/ 1g = 5.5×10⁴μg
5.5×10⁴μg / 200.055 g = 275 ppm
ppm can also be determined as mg of solute / kg of solution
It is important that the relation is 1×10⁻⁶
Let's verify: 0.055 g = 55 mg
200.055 g = 0.200055 kg
55 mg / 0.200055 kg = 275 ppm
Cells have limitations, if the cell is stronger it targets other cells , when a cell is weak it possibly mean it’s disease.
