Asaturated solution of lead(ii) chloride, pbcl2, was prepared by dissolving solid pbcl2 in water. the concentration of pb2+ ion in the solution was found to be 1.62×10−2 m . calculate ksp for pbcl2.

Ksp for PbCl2 is 1.7 *10^-5

Explanation:

Step 1: Data given

The concentration of Pb2+ ion in the solution was found to be 1.62 *10^−2 M

Step 2: The balanced equation

PbCl2 ⇔Pb^2+ +2Cl -

Step 2: ICE chart

The initial concentration of Pb2+ is 0 M

There will react X M  and 2X of Cl-

At the equillibrium there is X M  of Pb^2+ and 2X M of Cl-

The concentration of Pb2+ ion in the solution was found to be  1.62 *10^−2  M

Step 3: Calculate Ksp

Since PbCl2 is solid, it doesn't aply for Ksp

Ksp = [Pb^2+][Cl-]²

Ksp = X*(2X)²

Ksp = 4X³

  ⇒ X = 1.62 *10^-2 M

Ksp = 4*( 1.62 *10^-2)³

Ksp =1.7 *10^-5

Ksp for PbCl2 is 1.7 *10^-5

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.