You mix together aqueous solutions of 100.0 mL of 0.05 M sodium nitrate and 10.0

Question

You mix together aqueous solutions of 100.0 mL of 0.05 M sodium nitrate and 10.0 mL of potassium phosphate and the reaction proceeds to completion. 1. What is the balanced chemical equation for this reaction? Label all states. 2. What is the concentration of phosphate ion in solution after the reaction? You mix together 50.0 mL of 0.005 M lead(II) nitrate and 20 mL of YY M ammonium iodide and the reaction proceeds to completion. 1. What is the balanced chemical equation for this reaction? Label all states. 2. What is the concentration of iodide ions in solution after the reaction? This will take more than 1 calculation.

Explanation / Answer

First question :

1]

3NaNO3(aq) + K3PO4(aq) ------> Na3PO4(aq) + 3KNO3(aq)

2]

Intital concentration is 0.05 M

Final concnetration = Initial concentration*Volume / Total volume = 0.05*10 / 110 = 0.004545 M

second question :

1]

Pb(NO3)2 (aq) + 2 NH4I (aq) > PbI2 (s) + 2NH4NO3 (aq)

2]

Moles = molarity*V in L = 50*0.005 = 0.25 millimoles of Pb(NO3)2

Moles of NH4I = 20*0.005 = 0.1 millimoles

1 mole of Pb(NO3)2 reacts with 2 moles of NH4I

0.1 millimoles of NH4I reacts with 0.1/2 = 0.05 millimoles of Pb(NO3)2

Moles of Pb(NO3)2 left = 0.25 - 0.05 = 0.2 milimoles

[Pb+2] = moles / total volume in L = 0.2 / 50+20 = 2.857*10^-3 M

Ksp of PbI2 is 1.4 x 10^-8

PbI2 ---> Pb+2 + 2I-

Ksp = [Pb+2] [I-]^2

1.4 x 10^-8 = [2.857*10^-3] * [I-]^2

[I-] = 2.213*10^-3 M

so the concentration of I- = 2.213*10^-3 M

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