CHEM 171 1. You mx 200.0 mL of a 0.200 M Lead (D Nitrate with 300.0 mL ofa 0.250

Question

CHEM 171

1. You mx 200.0 mL of a 0.200 M Lead (D Nitrate with 300.0 mL ofa 0.250 M Sodium Iodide and a yellow Lead (IT) Iodide precipitates. (Pbl2 461.00 g/mol Pb(NO3)2(a) 2 Nal(ag)Pbl) + 2 NaNO3(aq) How many grams ofLead (I) Iodide can theoretically form? (think lmiting reactant) What are [Pb2+], [Na+], [1-], and [NO3"] after precipitation is complete (total end volume = 500.0 mL)? a. b. = Limiting Ion rams Pblh M ] M = [NO3-] 2. You titrate 20.0 mL ofa 0.150M HPO4 with a 0.200M NaOH. What vohume ofNaOH was used?

Explanation / Answer

1a) The balanced chemical equation is given.

Pb(NO3)2 (aq) + 2 NaI (aq) -------> PbI2 (s) + 2 NaNO3 (aq)

As per the stoichiometric equation,

1 mole Pb(NO3)2 = 2 moles NaI = 1 mole PbI2.

Moles of Pb(NO3)2 added = (200.0 mL)*(1 L/1000 mL)*(0.200 M)*(1 mol/L/1 M) = 0.04 mole.

Moles of NaI added = (300.0 mL)*(1 L/1000 mL)*(0.250 M)*(1 mol/L/1 M) = 0.075 mole.

Find out the limiting reactant, starting with Pb(NO3)2 and then NaI.

Pb(NO3)2: (0.04 mole Pb(NO3)2)*(2 mole NaI/1 mole Pb(NO3)2) = 0.08 mole NaI.

We do not need to work out the limiting reactant with NaI. Clealry, 0.04 mole Pb(NO3)2 will require 0.08 mole NaI and we have 0.075 mole NaI. Therefore, NaI is the limiting reactant and the yield of the product is governed by the moles of NaI (ans). We will report I- as the limiting ion (ans).

Moles of PbI2 produced = (0.075 mole NaI)*(1 mole PbI2/2 mole NaI) = 0.0375 mole PbI2.

Mass of PbI2 that can be formed theoretically = (moles of PbI2)*(molar mass of PbI2) = (0.0375 mole)*(461.00 g/mol) = 17.2875 g (ans).

b) Consider the dissociation of the aqueous reactants.

Pb(NO3)2 (aq) --------> Pb2+ (aq) + 2 NO3- (aq)

As per the stoichiometric equation,

1 mole Pb(NO3)2 = 1 mole Pb2+ = 2 moles NO3-.

Therefore, 0.04 mole Pb(NO3)2 = 0.04 mole Pb2+ = 0.08 mole NO3-.

NaI (aq) --------> Na+ (aq) + I- (aq)

As per the stoichiometric equation,

1 mole NaI = 1 mole Na+ = 1 mole I-.

Therefore, 0.075 mole NaI = 0.075 mole Na+ = 0.075 mole I-.

PbI2 is sparingly soluble in water and dissociates slightly. The dissociation equation is

PbI2 (s) <=====> Pb2+ (aq) + 2 I- (aq)

As per the stoichiometric dissociation,

1 mole Pb2+ = 2 moles I-.

Since NaI is the limiting reactant (I- is the limiting ion), I- reacts completely. The moles of Pb2+ that reacts with I- = (0.075 mole I-)*(1 mole Pb2+/2 mole I-) = 0.0375 mole Pb2+ and the moles of Pb2+ left unreacted = (0.04 – 0.0375) mole = 0.0025 mole.

Since I- reacts completely to form the precipitate, we must have [I-] = 0.00 M (we ignore the solubility product here for simplicity) (ans).

The total volume of the solution = 500.00 mL = (500.00 mL)*(1 L/1000 mL) = 0.500 L.

[Pb2+] = (moles of unreacted Pb2+)/(volume of solution in L) = (0.0025 mole)/(0.500 L) = 0.0050 mol/L = 0.0050 M (ans).

[Na+] = (0.075 mole)/(0.500 L) = 0.15 mol/L = 0.15 M (ans).

[NO3-] = (0.08 mole)/(0.500 L) = 0.16 mol/L = 0.16 M (ans).

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