Weakest Conjugate Base - General Chemistry - Solved Quiz, Exercises of Chemistry

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1. ( 4 points ) Using 1 – 4, organize the species in this group from the strongest acid (1) to weakest acid (4).

H 2 S [Pb(H 2 O) 6 ]2+^ HNO 2 H 2 PO 4 1-

2 4 1 3

2. ( 4 points ) Using 1 – 4, organize the species in this group from which produces the strongest conjugate base (1) to which produces the weakest conjugate base (4).

HSO 4 1-^ HClO [Al(H 2 O) 6 ]3+^ H 2 PO 4 1-

4 1 3 2

3. ( 6 points ) Will each of these salts form an acidic, basic, or neutral solution when dissolved in water?

BASIC Na 2 S ACIDIC NaH 2 PO 4

BASIC NaF NEUTRAL K 2 SO 4

ACIDIC AlCl 3 BASIC LiC 6 H 5 CO 2

4. ( 3 points ) Circle the species in each group of acids that will produce the weakest conjugate base.

HCl HBr HI

H 2 O H 2 Se H 2 S

HMnO 4 HMnO 3 HMnO 2

5. ( 4 points ) Write the appropriate equilibrium constant expression for each of the reactions below.

a. CH 4 (g) + 2 H 2 O (g) ↔ 4 H 2 (g) + CO 2 (g) (write an expression for K p)

b. C (s) + 2 H 2 (g) ↔ CH 4 (g) (write an expression for K c)

2 CH HO

CO

4 H p 4 2

2 2 P P

P P

×

×

K =

2 2

4 c [H ]

[CH ]

(g)

(g) K =

6. ( 8 points ) Select the answer from the column on the right that best matches each description from the column on the left. Each answer can be used, at most, only once.

_ H ___

This equilibrium constant describes a system that is heavily product-favored A. exothermic

__ F __

The equilibrium constant for an reaction will increase as you raise the temperature. B. concentration

__ G __

Removing some of the reactants of a reaction at equilibrium will cause it to shift in this direction

C. K << 1

__ C __ This equilibrium constant describes a system that is strongly reverse-reaction favored D. reactant

__ B __

A change in will not change the size of equilibrium constant of a reaction. E. temperature

__ I ___

A reaction proceeds in this direction when its current value of Q is less than its Keq F. endothermic

__ A ___

The equilibrium constant for an reaction will increase as you lower the temperature. G. reverse

__ D ___ When the reaction quotient is greater than the equilibrium constant a system is -favored

H. K >> 1

I. forward

J. product

7. ( 5 points ) Consider the equilibrium below. Indicate whether each of the stresses listed will cause the system to shift to the right (favor the products), left (favor the reactants), or have no effect.

2 SO 2 (s) + O 2 (g) ↔ 2SO 3 (g) Δ H rxn = -198 kJ /mol

left increase the temperature

right add more oxygen

right remove some of the SO 3 by condensing it

right increase the volume of the reaction chamber

no effect add more sulfur dioxide

8. ( 8 points ) What is the pK a of the conjugate acid of a weak base whose K b = 7.7×10 -6?

14.00 5.11 8.

14.

5.

(7.7 10 )

( )

a

a b

6

b b

= − =

+ =

=

=− ×

=−

−

pK

pK pK

log

pK log K

9. ( 8 points ) The reaction below has a concentration-based equilibrium constant equal to 279 M -1^ at 1000 K. Use this information to determine the value of the pressure- based equilibrium constant for the system.

2 SO 2 (g) + O 2 (g) ↔ 2 SO 3 (g)

10. ( 10 points ) The p H of a 0.133- M solution of an unknown weak acid is 3.652. Determine the pK a of the acid.

HA + H 2 O → A-^ + H 3 O + [ ] 0 0.133 0 0 Δ [ ] - x^ + x^ + x [ ]eq 0.133 2.23×10 -4^ 2.23×10 -

log 6427

374 10

0.

(2.23 10

[HA]

[A ][HO ]

H 3.652 [HO ] 10 223 10

a a

7 4 2 3 a

• H 4 3

pK K.

.

)

K

p p. M

=− =

= ×

×

= =

= ⇒ = = ×

−

− + −

• −

3. 40

(279)(0.08 21 1000 K)

( ) where Δ 1

• 1 molK atmL

gas

Δ P c gas

=

= ⋅

= =−

⋅ ⋅

K K RT n n

11. ( 10 points ) The reaction below is prepared at 12 °C with an initial pressure of NO of 0.824 atm and an initial pressure of Br 2 of 0.658 atm. At equilibrium the pressure of the NOBr is measured to be 0.506 atm. Determine Kp for the reaction.

2 NO( g ) + Br 2 ( g ) ↔ 2 NOBr( g )

P 0 0.824 0.658 0 Δ P - 2x^ + x^ + 2x Peq 0.824 – 2x 0.658 - x 2x P (^) eq 0.318 atm 0.405 atm 0.506 atm

2x = 0.506 ֜ x = 0.

The equilibrium constant is found:

ܭ௣ ܲൌ ே௢஻௥

ܲଶ

ேை

ଶ ܲ·

஻௥మ

ൌ

ሺ0.506ሻଶ

ሺ0.318ሻଶ^ ሺ0.405ሻ

ൌ 6.

12. ( 10 points ) Determine the p H of a solution prepared by dissolving 8.82 grams of lithium hydrogen carbonate, LiHCO 3 , in enough water to make 1.00 L of solution.

[ ]

M

M

1 0.

3

HCO

0.

1.00L

67.96g

1 mol 3 8.82gLiHCO

3 LiCO

⎥^ =

⎦

⎤

⎢⎣

∴⎡^ −

=

×

=

− (^) + ←⎯→ + 1 − 2 2 3

1 HCO 3 HO H CO OH [ ] 0 0.130 0 0 Δ [ ] - x + x + x [ ]eq 0.130 x x

[ ]

[ ]

H 9.

OH [OH ] 4.

x 5.59 10 [OH ]

2.4 10 HCO 0.

HCO [OH ]

1

5 1

8 2 1 3

1 2 3 b

=

=− =

= × =

= = = ×

−

− −

− −

−

p

p log

x K

13. ( 10 points ) For the following reaction, Kc = 0.592 at 200 K.

COCl 2 ( g ) ↔ CO( g) + Cl 2 ( g )

If the reaction mixture initially contains a COCl 2 concentration of 0.400 M , a CO concentration of 0.500 M , and a Cl 2 concentration of 0.600 M , what will be the equilibrium concentration of each species?

COCl 2 ( g ) ↔ CO( g) + Cl 2 ( g ) [ ] 0 0.400 0.500 0. Δ [ ] - x + x + x [ ]eq 0.400- x 0.500+ x 0.600+ x [ ]eq 0.438 M 0.462 M 0.562 M

0.592 ൌ ܭ௖ ൌ

ܱܥሾ ݈ܥሿሾ ଶ ሿ

݈ܥܱܥሾ ଶ ሿ

ൌ

ሺ0.500 ൅ ݔሻሺ0.600 ൅ ݔሻ

ሺ0.400 െ ݔሻ

0.592 ൌ

0.300 ൅ 1.1 ݔ൅ ݔ ଶ

0.400 െ ݔ

0.237 െ 0.592 ݔൌ 0.300 ൅ 1.1 ݔ൅ ݔ ଶ

0 ൌ ݔ ଶ^ ൅ 1.692 ݔ൅ 0.

x = -0.0381, -1.654 (reject -1.654)

14. ( 10 points ) Determine the [OH-], [H 3 O+], p H, and p OH for a solution that has a potassium hydroxide, KOH, concentration of 4.8×10-4^ M.

[OH - ] = [ KOH] = 4.8×10-4^ M

p OH = -log[OH-^ ] = 3.

ܪሾܱଷ ା^ ሿ ൈ ሾܪܱି ሿ ൌ ܭ௪

ܪሾܱଷ ା^ ሿ ൌ

ܭ௪

ିܪܱሾ ሿ

ൌ

1.0 ൈ 10ିଵସ

4.8 ൈ 10ିସ^

ൌ 2.1 ൈ 10ିଵଵ^ ܯ

p H = -log[H 3 O+] = 10. (or 14 – p OH = 14 – 3.32 = 10.68)