Cliffs AP Chemistry, 3rd Edition

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(e) Given: 0.125 g AgBr(s) with AgNO 3 (s).

KspAgOH = 1.52 × 10 –8

Restatement: Will AgOH precipitate?

Step 1:Write the equation in equilibrium for the dissociation of AgOH.

AgOH s()DAg aq()+OH aq-()

Step 2:Calculate the concentration of the ions present.

.'

.

..
.

Ag liter sol n

g AgBr

g AgBr

mole AgBr

mole AgBr

mole Ag

M

OH M

0 065987

0 125

187 772

1

1

1

0 0101

0 012

==##

=

+

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7

7

A

A

Step 3:Solve for the ion product, Q.

Q=[Ag+][OH–] = (0.0101)(0.012) = 1.32 × 10−4

KspAgOH = 1.52 × 10−8

Since Q > Ksp, AgOH will precipitate

Part A: Question 2

2. Given: H 2 O + 2.51 g SbCl 3 →1.906 g SbxOyClz
   0.802 g gas, 97.20% Cl, and 2.75% H

(a) Restatement: Simplest formula for gas.

If you had 100. grams of the gas, 97.20 grams would be due to the weight of chlorine

atoms, and 2.75 grams would be due to the weight of hydrogen atoms.

97.20 grams Cl / (1 mole Cl / 35.453 g / mole) = 2.742 moles Cl

2.75 grams H / (1 mole H / 1.00794 g / mole) = 2.73 moles H

Because this is essentially a 1:1 molar ratio, the empirical formula of the gas is HCl.

(b) Restatement: Fraction of chlorine in the solid product and in the gas phase.

1.The mass of chlorine in the original compound:

2.51 g of SbCl 3 ×106.36 g Cl / 228.10 g SbCl 3 = 1.17 g Cl

2.Fraction of chlorine in the gas.

According to the Law of Conservation of Mass, if you have 1.17 grams of chlorine in the

original compound, you must account for 1.17 grams of chlorine on the product side.

Answers and Explanations for the Practice Test