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If 38.5 mL of lead (II) nitrate solution reacts completely with excess sodium iodide to yield 0.0628 g or precipitate, what is the molarity of lead (II_

If 38.5 mL of lead (II) nitrate solution reacts completely with excess sodium iodide to yield 0.0628 g or precipitate, what is the molarity of lead (II_ ion in the original solution.. . I was wonderin
The balanced equation that describes the reaction between lead (II) nitrate and sodium iodide to produce lead (II) iodide and sodium nitrate is expressed Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3. All nitrates are soluble so lead (II) iodide is the precipitate. given 0.0628 g PbI2 is equal to 1.3623x10^-4 moles PbI2. This is equal to 1.3623x10^-4 moles also of Pb(NO3)2. Hence the molarity of the solution is 1.3623x10^-4 moles per 0.0385 L or 0.003538 M.  ...

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