1. Use the given Ksp for PbCl2 (s) value (Ksp = 2x10-5), calculate the minimum c

Question

1. Use the given Ksp for PbCl2 (s) value (Ksp = 2x10-5), calculate the minimum chloride ion concentration [Cl-] required to initiate precipitation of lead(II) chloride from a solution that initially contains 0.02M Pb2+?

2. A.)What volume of 6M HCl solution must be added to a 2mL of solution initially containing no Cl- ions so that the final Cl- ion concentration will be 0.03M afterwards (2pts). B.) If one drop is approximately 0.05mL, how many drops will that be equivalent to?

3. H2S(aq) + 2H2O 2H3O+ (aq) + S2-(aq) Ka = 1x10-22 Predict how [S2-] concentration may vary if the pH of the solution is decreased (in another word please answer if [S2-] will increase or decrease or unchanged if pH is decreased). Explain your answers. (2pts) Calculate the hydrogen ion concentration you should keep at in order to control [S2-] concentration to be less than 2x10-23M. (2pts). You may assume that the concentration of H2S(aq) in solution is 0.1M.

Explanation / Answer

1.

PbCl2 = Pb2+(aq) + 2 Cl-(aq)

The solubility product of PbCl2 is Ksp = 2x10-5

Given that [Pb2+] = 0.02M

we know,

Ksp = [Pb2+] [Cl-]2

2x10-5= 0.02 [Cl-]2

[Cl-] = 0.032 M

Hence the minimum chloride ion concentration [Cl-] required to initiate precipitation of lead(II) chloride is 0.032M

2.

a)From the definition of molarity,

the amount of [Pb2+] moles present = 0.02/1000*2 = 4x10-5mols

we know that, for every mole of [Pb2+] present, there are 2 moles of [Cl-] present.

Hence to start precipitation, we need, 2x4x10-5= 8x10-5mols of [Cl-].

HCl(aq) = H+(aq) + Cl-(aq)

Since HCl is 6M solution, There are 6 moles of HCl present in 1000ml

so the volume needed to form 8x10-5 mols of HCl = 1000x8x10-5/ 6 = 0.013ml

b) If one drop is approximately 0.05mL of HCl, the volume required in above experiment will be less than one drop.

3.

a)pH of the solution is defined as negative log of [H+] concentration.

If the pH of the solution decreases, the concentration of [S2-] increases.

b) Given that Ka = 1x10-22

From the equation

H2S(aq) + 2H2O = 2H3O+(aq) + S2-(aq)

we can write, Ka = [H3O+] [S2-] / [H2S]

Substituting the values, we get

1x10-22= [H3O+] * 2x10-23/ 0.1

[H3O+] = [H+] = 0.5M

The Hydrogen ion concentration should be kept at 0.5 M.

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