This exercise will help you review molarity which allows the qualification of ch

Question

This exercise will help you review molarity which allows the qualification of chemical reactions in solution, specifically the two precipitation reactions under study. Reaction I In this lab you will mix 25 mL of 0.05 M lead nitrate with 1.4 mL of 0.025 M sodium carbonate. The precipitation reactions occur quickly. After the reaction occurs you will filter the solution to remove the precipitate. You will then test the remaining solution for excess lead ion and for excess carbonate ion. Imagine that you mix the two volumes and then freeze frame the reaction does not proceed. 1. Identify all the ionic species present in the solution before the reaction takes places. 2. Compute the molarity and the number of moles of each ionic species present before the reaction takes place. Now let the reaction go to completion and 3. Identify all the ionic species present in the solution. 4. Compute the molarity and the number of moles of each ionic species present. 5. Clearly predict the results of the experiment by determining which reactant is limiting and which is in excess. Reaction II Repeat the above analysis for reaction II where 25 mL of a 0.05 M lead nitrate solution is mixed with 1.4 mL of 0.25 M potassium iodide solution. Clearly predict the results of the experiment by following steps 1 to 5.

Explanation / Answer

Both lead nitrate and sodium carbonates are inorganic compounds. On dissolution both gives clear solutions and those salts completely dissosiate into their constituent ions.

1. Pb(II), NO3-, Na+, CO3-2 ions are present in the solution before the reaction to occur.

2. On combining both these volumes of 25 mL of 0.05 M lead nitrate (1.25 mol of Pb(II), 2.5 mol of NO3- ions [mol = molarity X volume]) and 1.4 mL of 0.025 M sodium carbonate (0.7 mol of Na+, 0.035 mol of CO3-2 ions) results in generation of total volume of 26.4 mL which contains 1.25 mol of Pb(II), 2.5 mol of NO3-, 0.7 mol of Na+ and 0.035 mol of CO3-2 ions before the reaction to takes place.

then the molarities of each ion in the solution (before the reaction):

Pb(II) molarity = mol/volume = 1.25/ 26.4 = 0.0473 M

NO3- molarity = mol/volume = 2.5/ 26.4 = 0.094 M

Na+ molarity = mol/volume = 0.7/ 26.4 = 0.0265 M

CO3-2 molarity = mol/volume = 0.035/ 26.4 = 0.0013 M

3. After completion of the reaction, the solution mixture now contains 0.035 mol of PbCO3, excess Pb(II) ions & Na+ and NO3- spectator ions.

4. After reaction completion, solution contains 1.215 mol of Pb(II), and the remaining two spectator ions and their molarities do not alter.

Na+ molarity = mol/volume = 0.7/ 26.4 = 0.0265 M; NO3- molarity = mol/volume = 2.5/ 26.4 = 0.094 M

remaining Pb(II) molarity = 1.215/ 26.4 = 0.046 M.

5. Sodium carbonate is the limiting reactant and lead nitrate is the excess reactant.

For the secnd reaction:

1. before the reaction, solution mixture contains Pb(II), NO3-, K+, and I- ions.

2. Before the reaction, total volume of the reaction mixture is 26.4 mL and it contains 1.25 mol of Pb(II), 2.5 mol of NO3-, 0.35 mol of K+, and 0.35 mol of I- ions

molarities of individual ions,

molarity of Pb(II) = mol/ volume = 1.25/ 26.4 = 0.0473 M

molarity of NO3- = 2.5/ 26.4 = 0.094 M

molarity of K+ = 0.35/ 26.4 = 0.0132 M

molarity of I- = 0.35/ 26.4 = 0.0132 M

3. after the reaction completion, it contains 0.175 mol of PbI2, excess Pb(II), spectator ions K+ and NO3- ions

4. After reaction completion, remaining Pb(II) mol = 1.075 mol and its molarity = 1.075/ 26.4 = 0.040 M; spectator ions do not alter their mols/ molarities.

5. lead nitrate is the excess reactant and potassium iodide is the limiting reactant

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