1. Suppose 50.00 mL of 2.0 × 10–4 M Fe(NO3)3 is added to 50.00 mL of 2.0 ×10-6 M

Question

1. Suppose 50.00 mL of 2.0 × 10–4 M Fe(NO3)3 is added to 50.00 mL of 2.0 ×10-6 M KIO3. Which of the following statements is true? For Fe(IO3)3, Ksp = 1.0 × 10–14.
A) A precipitate forms because Qc > Ksp.
B) A precipitate forms because Qc < Ksp.

C) No precipitate forms because Qc < Ksp.

D) No precipitate forms because Qc = Ksp.

E) No precipitate forms because Qc > Ksp.

2. For which of the following reactions is S° > 0 at 25°C?

A) 2H2(g) + O2(g) 2H2O(g)
B) 2ClBr(g) Cl2(g) + Br2(g)
C) I2(g) I2(s)

D) 2NO(g) + O2(g) 2NO2(g)

E) NH4HS(s) NH3(g) + H2S(g

3. What is E of the following cell reaction at 25°C? Cu(s) | Cu2+(0.017 M) || Ag(s), (Ag+ = 0.18M)

E°cell = 0.460 V.

A) 0.468V

B) 0.282 V

C) 0.460 V

D) 0.490 V

E) 0.479V

Explanation / Answer

Q1 Suppose 50.00 mL of 2.0 × 10–4 M Fe(NO3)3 is added to 50.00 mL of 2.0 ×10-6 M KIO3. Which of the following statements is true? For Fe(IO3)3, Ksp = 1.0 × 10–14.

Solution :-

Using the given concentration values lets calculate the Qc

Qc equation is as follows

Qc=[Fe^3+][IO3^-]3

Lets put the values in the formula after adding the solution total volume will become 100 ml therefore the initial concentrations will halved.

[Fe^3+] = 2.0E-4 / 2 = 1.0E-4 M

[IO3^-] = 2.0E-6 / 2 = 1.0E-6 M

Qc = [1.0E-4 ] [1.0E-6]2

Qc = 1E-16

Therefore Qc is smaller than Kc means reaction will proceed in forward direction.

Therefore correct answer is

No precipitate forms because Qc < Ksp.

Q2 solution

For the reaction

NH4HS(s) NH3(g) + H2S(g)

Delta S > 0at 25 c

Because here solid is changing to the gases.Gases have the greatest entropy than liquid and solid.

Therefore the reaction shown in the option E will have entropy greater than 0

Q3 solution :-

Cu(s) | Cu2+(0.017 M) || Ag(s), (Ag+ = 0.18M)                       E°cell = 0.460 V.

Formula to calculate the E of cell is as follows

E cell = Eo cell – (0.0592/n)logQ

Where log Q = [product ]/[reactant]

N=number of electrons transferred.

Lets put the values in the formula

E = 0.460 V – (0.0592/2)log [0.017/0.18]

E= 0.460 V – (-0.0303)

E= 0.490 V

Therefore correct answer is option ‘D’

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