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
The answers is: 375 moles NaOH
You are given two beakers, distilled water, two hot plates, two thermometers and salt. These materials are enough in order to test the effect of salt in the boiling point water. To do this, you set up two beakers. In one of the beakers, you add pure distilled water and nothing else. For the other beaker, you put a solution of salt and water. You place these beakers on separate hot plates and place inside the beakers the thermometers. You heat these substances until they boil and then you measure the boiling points of the substances. You would observe that the boiling point of the solution would have a higher boiling point than the pure liquid.
