What is the complete balanced equation for a reaction when lead II nitrate is added to potassium phosphate?

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In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.

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We must first convert from a word equation to a symbol equation:

#"Lead (II) Nitrate + Potassium Iodide "->" Lead (II) Iodide + Potassium Nitrate"#

The lead (II) ion is represented as #"Pb"^(2+)#, whilst the nitrate ion is #"NO"_3^-#. To balance the charges, we require two nitrate ions per lead (II) ion, and so lead (II) nitrate is #"Pb(NO"_3")"_2# .

The potassium ion is #"K"^+# and the iodide ion is #"I"^-#. The two charges balance in a #1 : 1# ratio, so potassium iodide is simply #"KI"#.

In lead (II) iodide, the charges balance in a #1 : 2# ratio, so the formula is #"PbI"_2#.

Finally, in potassium nitrate, the charges balance in another #1 : 1# ratio, giving a formula of #"KNO"_3# .

The symbol equation is as follows:

#"Pb(NO"_3")"_2 + "KI" -> "PbI"_2 + "KNO"_3#

The most obvious change we must make, when balancing this equation, is to increase the number of nitrate ions on the right hand side of the equation. We can do this by placing a coefficient of #2# before the potassium nitrate:

#"Pb(NO"_3")"_2 + "KI" -> "PbI"_2 + 2"KNO"_3#

In doing this we have upset the balance of potassium ions on each side of the equation. Again, we can fix this: we must simply place another coefficient of #2#, this time before the potassium iodide:

#"Pb(NO"_3")"_2 + 2"KI" -> "PbI"_2 + 2"KNO"_3#

Checking over the equations once more, you will notice that we initially had 1 iodide ion on the right hand side, but 2 on the left. However, we already dealt with this in balancing our potassium ions. Now, our equation is balanced.

And that's it! One last thing to add is that you may have noticed the irregularity in iodide ions rather than nitrate ions. In this case, you would have arrived at the same answer simply by working backwards.