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Question

# What happens when aqueous lead nitrate solution reacts with potassium iodide solution? Explain the type of reaction and balance the given equation

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Solution

## 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 Pb2+, 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(NO3)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 PbI2. Finally, in potassium nitrate, the charges balance in another 1:1 ratio, giving a formula of KNO3 . The symbol equation is as follows: Pb(NO3)2+KI→PbI2+KNO3 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(NO3)2+KI→PbI2+2KNO3 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(NO3)2+2KI→PbI2+2KNO3

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