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Chapter 12: Problem 173

What volume of a \(0.245 \mathrm{M}\) solution of \(\mathrm{NaI}\) would you need to add to \(100.0 \mathrm{~mL}\) of a \(0.300 \mathrm{M}\) solution of lead acetate to precipitate out all the lead? Which ions are spectators?

Short Answer

Expert verified
To precipitate out all the lead from a 100.0 mL, 0.300 M solution of lead acetate, you would need to add 245 mL of a 0.245 M solution of sodium iodide (\(NaI\)). The spectator ions that do not participate in the reaction are sodium (Na+) and acetate (CH_3COO-) ions.

Step by step solution

01

Understand the Chemical Reaction

The chemical reaction takes place between sodium iodide (\(NaI\)) and lead acetate, and in this reaction, sodium iodide will react with lead acetate to form lead iodide and sodium acetate. The balanced chemical equation is as follows: \[2NaI(aq) + Pb(CH_3COO)_2(aq) \rightarrow PbI_2(s) + 2NaCH_3COO(aq)\] Notice that the stoichiometry of the reaction is such that 2 moles of \(NaI\) react with 1 mole of \(Pb(CH_3COO)_2\) to form 1 mole of \(PbI_2\).
02

Calculate the moles of Lead Acetate

We first need to calculate the number of moles of lead acetate in the 100.0 mL solution. We can use the concentration (mol/L) = moles/volume (L) equation to find the moles, thus: Moles of lead acetate = Concentration * Volume = 0.300 mol/L * 100.0 mL * (1L/1000mL) = 0.0300 moles
03

Calculate the Volume of Sodium Iodide Needed

Now use stoichiometry to determine the moles of \(NaI\) and then use the molarity to figure out the volume. Since the ratio is 2:1 (NaI to lead acetate), we need 2 * 0.0300 = 0.0600 moles of \(NaI\). The volume, V, of NaI solution is calculated using the formula: V = moles/concentration = 0.0600 moles / 0.245 mol/L = 0.245 L or 245 mL
04

Identify the Spectator Ions

Spectator ions are the ions that do not participate in the reaction. This means they are present in the same form on both the reactant and product sides of the equation. From the balanced chemical equation, we see that sodium (Na+) and acetate (CH_3COO-) ions are on both sides of the equation and do not change during the reaction. Therefore, the spectator ions are Na+ and CH_3COO-.

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