Density = mass / volume
9.51 = m / 9.26
9.51 * 9.26 = m
m = 88.0626
PbI(ii) ionization in the solution of PBI(ii) into water is:
<span>PbI</span>₂(solution) <==> Pb₂⁺ + 2I⁻
If the conc. of PbI(ii) in the sol. is xM then the conc. of Lead(ii) will be x M and conc. of iodide will be 2 x M.
Therefore,
<span>Ksp=<span>[Pb</span></span>²⁺][I-]²
Plugging the values:
1.4×10⁻⁸ = x ⋅ (2x)²
1.4×10⁻⁸ = 4x³
x³ = {1.4×10⁻⁸}÷4
x³ = 0.35 x 10⁻⁸
or
x³ = 3.5 x 10⁻⁹
x = 1.51 x 10⁻³
Hence,
Concentration of iodide ions in the solution:
2x = 3.02 x 10⁻³
Answer:
i think multipling 30 to the 40
30 × 40 = 1200ppt
