Ionic Equlibrium Test

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Ionic Equlibrium Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

K_sp of Ca(OH)₂ is 4.0 × 10^-6. At what minimum pH, Ca²⁺ ions start precipitating in 0.01 M CaCl₂?

A
12 – log 2

B
12 + log 2

C
8 – log 2

D
8 + log 2

Solution

K_sp[Ca(OH)₂] = 4.0 × 10^-6 = [Ca²⁺] [OH^-]²

= 0.01 × [OH^-]²⇒ [OH^-] = 2 × 10^-2;

pOH = - log 2 × 10^-2 = 2 - log 2; pH = 14-(2-log2)

= 12 + 2 log 2
Question 2
1 / -0

In which of the following, the buffer action ceases to act?

A
100 ml of 0.1 M CH₃COOH + 25 ml of 0.1 M NaOH

B
100 ml of 0.1 MCH₃COOH + 50 ml of 0.3 M NaOH

C
100 ml of 0.1 M CH₃COOH + 80 ml of 0.1 M NaOH

D
100 ml of 0.1 M CH₃COOH + 50 ml of 0.1M NaOH

Solution

CH₃COOH + NaOH → CH₃COONa + H₂O;

milli mole of CH₃COOH = 100 × 0.1 = 10;

milli mole of NaOH = 50 × 0.3 = 15;

CH₃COOH is completely consumed and NaOH is left in excess.
Question 3
1 / -0

On addition of (NH₄)₂SO₄ to 0.1 M NH₃

A
OH^– ion concentration will decrease

B
pH will decrease

C
pH will increase

D
H₃O⁺ ion concentration will decrease

Solution

Due to common ion effect of NH₄⁺ ionisation of NH₃ is suppressed. As a result [OH^-] decreases and so also pH.
Question 4
1 / -0

Pick out the acid-base conjugate pairs of the following:

A
C₅H₅NH⁺ and C₅H₅N

B
N₂H₆²⁺ and N₂H₅⁺

C
H₃O⁺andOH^–

D
C₂H₅OH and C₂H₅O^–

Solution

In acid-base conjugate pair, the acid and base differ by a proton.

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Re-Attempt Weekly Quiz Competition

Ionic Equlibrium Test - 3

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Ionic Equlibrium Test - 3

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Question 1

12 – log 2

12 + log 2

8 – log 2

8 + log 2

Solution

Question 2

100 ml of 0.1 M CH3COOH + 25 ml of 0.1 M NaOH

100 ml of 0.1 MCH3COOH + 50 ml of 0.3 M NaOH

100 ml of 0.1 M CH3COOH + 80 ml of 0.1 M NaOH

100 ml of 0.1 M CH3COOH + 50 ml of 0.1M NaOH

Solution

Question 3

OH– ion concentration will decrease

pH will decrease

pH will increase

H3O+ ion concentration will decrease

Solution

Question 4

C5H5NH+ and C5H5N

N2H62+ and N2H5+

H3O+andOH–

C2H5OH and C2H5O–

Solution

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100 ml of 0.1 M CH₃COOH + 25 ml of 0.1 M NaOH

100 ml of 0.1 MCH₃COOH + 50 ml of 0.3 M NaOH

100 ml of 0.1 M CH₃COOH + 80 ml of 0.1 M NaOH

100 ml of 0.1 M CH₃COOH + 50 ml of 0.1M NaOH

OH^– ion concentration will decrease

H₃O⁺ ion concentration will decrease

C₅H₅NH⁺ and C₅H₅N

N₂H₆²⁺ and N₂H₅⁺

H₃O⁺andOH^–

C₂H₅OH and C₂H₅O^–