Chemistry Test 82

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Chemistry Test 82

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Weekly Quiz Competition

Question 1
4 / -1

A certain weak acid has a dissociation constant of 10^–4. The equilibrium constant for its reaction with a strong base is :-

A
10¹⁰

B
10¹⁴

C
10^–10

D
10^–4

Solution

K_eq = ka/kw = 10¹⁰
Question 2
4 / -1

Which of the following pairs constitutes buffer solution?

A
HNO₃ and NH₄NO₃

B
HCl and KCl

C
HNO₂ and NaNO₂

D
NaOH and NaCl

Solution

A weak acid and its salt with strong base constitutes buffer.
Question 3
4 / -1

It is found that at 25°C 0.1 M solution of four sodium salts NaA, NaB, NaC and NaD have the pH values 7, 9, 10, 11 respectively. Which one of the corresponding acids is strongest?

A
HB

B
HA

C
HC

D
HD

Solution

Aqueous solution of a salt which is obtained from strong acid and strong base is neutral. Hence HA is strong acid in among other acids.
Question 4
4 / -1

On adding 0.1 M solution each of Ag⁺, Ba⁺², Ca⁺² ions in a Na₂SO₄ solution, species first precipited is (K_sp for BaSO₄ = 10^–11, K_sp for CaCO₃ = 10^–6,K_sp for Ag₂SO₄ = 10^–5).

A
Ag₂SO₄

B
BaSO₄

C
CaCO₄

D
All of these compounds will precipitate simultaneously

Solution

Compound with lowest K_sp value, will precipitate first.
Question 5
4 / -1

If pK_b of a base is 5, then K_a of its conjugate acid will be

A
10⁵

B
10^–5

C
10^–9

D
10^–14

Solution

pk_a + pk_b= 14 (for conjugate pairs)

pk_a = 14–pk_b = 14 – 5 = 9

ka = 10^–pka = 10–9
Question 6
4 / -1

Which of the following is the correct quadratic form of Ostwald's dilution law equation for AB type electrolyte :

A
α²C + αK – K = 0

B
α²C – αK – K = 0

C
α²C – αK + K = 0

D
α²C + αK + K = 0

Solution

According to Ostwald's dilution law,

K = Cα²/ 1 - α

or K – Kα = Cα² or Cα² + Kα – K = 0

or α²C + αK – K = 0

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Chemistry Test 82

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Chemistry Test 82

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Question 1

1010

1014

10–10

10–4

Solution

Question 2

HNO3 and NH4NO3

HCl and KCl

HNO2 and NaNO2

NaOH and NaCl

Solution

Question 3

HB

HA

HC

HD

Solution

Question 4

Ag2SO4

BaSO4

CaCO4

All of these compounds will precipitate simultaneously

Solution

Question 5

105

10–5

10–9

10–14

Solution

Question 6

α2C + αK – K = 0

α2C – αK – K = 0

α2C – αK + K = 0

α2C + αK + K = 0

Solution

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10¹⁰

10¹⁴

10^–10

10^–4

HNO₃ and NH₄NO₃

HNO₂ and NaNO₂

Ag₂SO₄

BaSO₄

CaCO₄

10⁵

10^–5

10^–9

10^–14

α²C + αK – K = 0

α²C – αK – K = 0

α²C – αK + K = 0

α²C + αK + K = 0