Molarity of lead ( II ) ion in the original solution if 38.5 mL of lead ( II ) nitrate solution reacts completely with excess sodium iodide solution to yield 0.628 g of precipitate is to be calculated. Concept introduction: Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate. The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature. The reaction of lead ( II ) nitrate and sodium iodide is an example of precipitation reaction. Pb ( NO 3 ) 2 ( a q ) + 2 NaI ( a q ) → PbI 2 ( s ) + 2 NaNO 3 ( a q ) According to the solubility rules, all the common nitrates are soluble but PbI 2 is insoluble in nature.

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Chapter 4, Problem 4.47P

Interpretation Introduction

Interpretation:

Molarity of lead(II) ion in the original solution if 38.5 mL of lead(II)nitrate solution reacts completely with excess sodium iodide solution to yield 0.628 g of precipitate is to be calculated.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature. The reaction of lead(II)nitrate and sodium iodide is an example of precipitation reaction.

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

According to the solubility rules, all the common nitrates are soluble but PbI2 is insoluble in nature.

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