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adelina 88 [10]
3 years ago
9

Solid NaCl is added slowly to a solution containing 0.10M AgNO3 and 0.20M Pb(NO3)2. Ksp AgCl = 1.8 E-10 Ksp PbCl2 = 1.6 E-5 Writ

e a net ionic equation and corresponding Ksp expression for the dissolution of solid i. Silver Chloride (AgCl) ii. lead (II) chloride (PbCl2) How do I do this?
Chemistry
1 answer:
ZanzabumX [31]3 years ago
8 0

Answer:

Ag+(aq) + Cl-(aq) <--->AgCl(s)

Ksp = [Ag+] [Cl-]

aii) Pb2+(aq) + 2Cl-(aq) <--->PbCl2(s)

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

Explanation:

In this question, we are to write the net ionic equation and corresponding Ksp expression for the salts in the question.

We proceed as follows;

Ag+(aq) + Cl-(aq) <--->AgCl(s)

Ksp = [Ag+] [Cl-]

aii) Pb+2(aq) + 2Cl-(aq) <--->PbCl2(s)

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

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<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

Relation of K_p\text{ with }K_c is given by the formula:

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K_p = equilibrium constant in terms of partial pressure = ?

K_c = equilibrium constant in terms of concentration = 6.5\times 10^4

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T = temperature = 35^oC=[35+273]K=308K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43

K_p is the constant of a certain reaction at equilibrium while Q_p is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of Q_p for above equation follows:

Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}

We are given:

p_{NOCl}=1.76atm

p_{NO}=1.01atm

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Putting values in above equation, we get:

Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23

We are given:

K_p=1583.43

There are 3 conditions:

  • When K_{p}>Q_p; the reaction is product favored.
  • When K_{p}; the reaction is reactant favored.
  • When K_{p}=Q_p; the reaction is in equilibrium

As, K_p>Q_p, the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

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