If 62.3 ml of lead(ii) nitrate solution reacts completely with excess sodium iodide solution to yield 0.862 g of precipitate, what is the molarity of lead(ii) ion in the original solution?

The molarity of the lead(II) ion in the original solution is 0.03M

Explanation:

Step 1: The balanced reaction

Pb(NO3)2(aq) + 2 NaI aq) → PbI2(s) + 2 NaNO3 (aq)

Pb2+ + 2I- →PbI2

This means that for 1 mole Pb2+ consumed, there is needed 2 moles of I- to produce 1 mole of PbI2

Step 2: Calculate moles of PbI2

Moles of PbI2 = 0.862g / 461.01 g/mole

moles of PbI2 = 0.00187 moles

Step 3: calculate moles of Pb2+

For 1 mole of PbI2 produced we need 1 mole of Pb2+

This means for 0.00187 moles of PbI2 we need 0.00187 moles of Pb2+

Step 4: Calculate the molarity of the Pb2+ ion

Molarity of Pb2+ = moles of Pb2+ / volume =

Molarity of Pb2+ = 0.00187 moles / 62.3*10^-3

Molarity of Pb2+ ion = 0.03M

The molarity of the lead(II) ion in the original solution is 0.03M

a c d and f

just finished the assignment

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.