Asked by Jeff
Lead(II) nitrate solution reacts with potassium iodide solution to produce lead(II) iodide and potassium nitrate.
The answer in the book is:
Pb(NO3)2(aq) + 2 KI(aq) ---> 2KNO3(aq) + PbI2(s)
I understand how to write and balance the equation. However, I am confused about how to determine the states of matter. Why are some aqueous and one solid? Does it have something to do with the solubility rules? Thanks for your help.
The answer in the book is:
Pb(NO3)2(aq) + 2 KI(aq) ---> 2KNO3(aq) + PbI2(s)
I understand how to write and balance the equation. However, I am confused about how to determine the states of matter. Why are some aqueous and one solid? Does it have something to do with the solubility rules? Thanks for your help.
Answers
Answered by
DrBob222
yes.
PbI2 is not soluble in water; therefore, it is written as PbI2(s).
KNO3 is soluble in water; therefore, in solution it is written as KNO3(aq).
Pb(NO3)2(aq) means you have a solution of Pb(NO3)2 and not solid Pb(NO3)2.
PbI2 is not soluble in water; therefore, it is written as PbI2(s).
KNO3 is soluble in water; therefore, in solution it is written as KNO3(aq).
Pb(NO3)2(aq) means you have a solution of Pb(NO3)2 and not solid Pb(NO3)2.