A saturated solution of lead(II) iodide, PbI2 has an iodide concentration of 3.0 x 10^-3 mol/L. a) What is the molar solubility of PbI2?
b) Determine the solubility constant, Ksp, for lead(II) iodide.
c) Does the molar solubility of lead (II) iodide increase, decrease, or remain unchanged with the addition of potassium iodide to the solution? EXPLAIN.

a) 1.5 x 10^-3 mol/L

b) 1.35×10^-8

c) decrease

Explanation:

The solubility of lead II iodide is given by the equation;

PbI2(s) > Pb^2+(aq) + 2I^-

By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L

Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L

Ksp= [Pb^2+] [2I^-]^2 =

Let the molar solubility of each ion be x, therefore;

Ksp= 4x^3

Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8

Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.

answer is: nuclear fusion produces elements that are heavier than helium.

nuclei combine to form a heavier nucleus, releasing energy.  

for example, fusion of hydrogen atoms at the sun.  

this fusion nuclear reaction is in the inner core of the sun and hydrogen (atomic number 1) is conveted into helium (atomic number 2).  

the inner core is the central region of the sun and all solar energy is produced in the core by nuclear fusion.