Question # 1

pKb value of NH₄OH is 4.74. If the concentration of NH₄OH is 1 molar containing 0.1 molar NH₄Cl, then pH of this buffer will be

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3.74

10.26

4.74

9.26

K_bvalue of NH₄OH is 1.81 x 10^-5and its conjugate acid has K_a= 5.7 x 10^-10pKb of the base is 4.74, pKa of its conjugate acid is

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-4.74

4.74

10

9.26

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Pressure change

Temperature change

Concentration change

Catalyst

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In the particular reaction for the valueK_{c 1 x}10^-25which statement is correct :

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Almost forward reaction is completed.

Amount of reactant is negligible as compared to product.

Amount of product is negligible as compared to reactant.

Amount of product is equal to amount of reactant.

pH of water is 7, if 0.01 M NaOH is added, than its pH is

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12

14

zero

10

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4 mole per dm³

2 mole per dm³

0.33 mole per dm³

0.67 mole per dm³

1 mole of N₂and 2 moles of H₂are allowed to react in a 1 dm³vessel. At equilibrium 0.8 mole of NH₃is formed. The concentration of H₂in the vessel is

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0.6 mole

0.8 mole

0.2 mole

0.4 mole

A large value of K_cmeans that at equilibrium :

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Less reactant and more products.

Reactants and product in same amounts.

More reactants and less products.

None of above.

K_avalue of HF acid is 6.7 x 10^-15the acid is a

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Weak acid

Moderately strong acid

Strong acid

Very weak acid

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8

4

9

3

In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant will

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Also be doubled

Be halved

Becomes one fourth

Remains the same

Addition of solid NaHCO₃in water causes ionization of NaCHO₃its K_a= 4.7 x 10^-1. Then this solution has character

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Acidic

Very weakly basic

Alkaline

Neutral

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The value of K_pfalls with a rise in temperature

The value of K_pfalls with increasing pressure

Adding V₂O₅catalyst increase the equilibrium yield of sulphur trioxide equilibrium yield of sulphur trioxide

The value of K_pis equal to K_c

N₂O₄⇌ 2NO₂

For the above reaction, which of the Following expression of K_c correct :

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Kc = [ N₂O₄]/[ NO₂]²

Kc = [ N₂O₄]/ [NO₂]

Kc = [ N₂O]²/[ N₂O₄]

Kc = [ N₂O₄]/[ N O₂]

The solubility of KCIO₃salt in water is decreased by adding

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NaCIO₃

NaCI

KCIO₄

KCI

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Forward reaction is favoured

Backward reaction is favoured

No effect

None of the above

1.1 mol of A is mixed with 2.2 mol of B and the mixture is kept in on litre flask till the equilibrium is reached. At equilibrium, 0.2 mol of C is formed. If the equilibrium reaction is A+2B 2C+D, the value of equilibrium constant is

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0.002

0.004

0.001

0.003

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High temperature and low pressure

Low temperature and low pressure

Low temperature and high pressure

High temperature and high pressure

Base buffer solution can be prepared by mixing

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Weak acid and its salt

Strong acid and its salt with weak base

Weak base and its salt with strong acid

Strong base and its salt with weak acid

Buffers having pH less than 7 are made

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Mixture of weak acid + salt of it with strong base

Mixture of weak acid + salt of it with weak base

Mixture of weak base + salt of it with strong acid

Mixture of weak base + salt of it with weak base

pH and pKa of the buffer are related by Henderson equation which is

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For which system does the equilibrium constant, kc has units of (concentration) ?

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N₂+3H_2-------2NH₃

H₂+L₂---------2HL

2NO₂----------N₂O₄

2HF--------H2+F₂

At certain temperature, 50% of HI is dissociated into H₂and I₂the equilibrium constant is

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1.0

3.0

0.5

0.25

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450°C

250°C

850°C

1000°C

For the above reaction the relationship b/w k_c and k_p will be :

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K_p = K_cRT

Kp = K_c(RT)_-1

K_p = K_c(RT)_-2

K_c = K_p

N₂ + O₂ ⇌2NO

The unit of K_c for tis reaction will be:

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No unit

mol^-2 dm^-3

mol^-1 dm^-3

mol^-2 dm⁺³

The solubility product of AgCl is 2.0 x 10^-3 mol² dm^-6 , The maximum concentration of Ag ion in the solution is :

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2.0 x 10^-10 mol dm^-3

1.41 x 10^-10 mol dm^-3

1.0x 10^-10 mol dm^-3

4.0x 10^-10 mol dm^-3

A buffer of a 0.09 molar acetic acid and 0.11 molar sodium acetate has pH = 4.83. If 0.01 mole NaOH in 1 dm³of the buffer solution is added, then pH of the buffer becomes

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4.74

4.92

5.0

4.0

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KC = KP

Kp = KcRT

Kp = kc(RT)^-2

Kp = Kc(RT)^-1

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High temperature

Low temperature

Low pressure

High pressure

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