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Test: Ionic Equilibrium - 1 - Chemistry MCQ


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20 Questions MCQ Test Topicwise Question Bank for IIT JAM/CSIR/GATE Chemistry - Test: Ionic Equilibrium - 1

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Test: Ionic Equilibrium - 1 - Question 1

1 c.c. of 0.1N HCl is added to 99 CC solution of NaCl. The pH of the resulting solution will be

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 1

Volume of resulting solution = 100 ml

Test: Ionic Equilibrium - 1 - Question 2

10 ml of  is mixed with 40 ml of . The pH of the resulting solution is

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 2


[H+] = 10–2
pH = 2.

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Test: Ionic Equilibrium - 1 - Question 3

The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is:

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 3

Concentration of solution =C=1M
Dissociation is α=0.01
After dissociation, the concentration of H+ ions will be Cα=1×0.01=10−2
pH=−log[H+]=−log(10−2) = +2

Test: Ionic Equilibrium - 1 - Question 4

If K1 & K2 be first and second ionisation constant of H3PO4 and K1 >> K2 which is incorrect.

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 4




Test: Ionic Equilibrium - 1 - Question 5

What is the percentage hydrolysis of NaCN in N/80 solution when the dissociation constant for HCN is 1.3 × 10-9 and Kw = 1.0 × 10-14

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 5

Weak acid + Strong base : Salt


Test: Ionic Equilibrium - 1 - Question 6

Which of the following solution will have pH close to 1.0 ?

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 6

75mL of M/5HCl + 25mL of M/5 NaOH 
Meq of HCl = 15 (∴ Mol. wt. and Eq. wt. of HCl or NaOH are same) 
Meq of NaOH = 5 Hence, in solution 10meq of HCl remains in 100mL solution. 
So, concentration of HCl in mixture = 10/100M = M/10 
In M/10 HCl, [H+] = 10-1 M
pH = -log[H+] = -log10-1 = 1

Test: Ionic Equilibrium - 1 - Question 7

If equilibrium constant of

CH3COOH + H2 CH3COO- + H3O+

Is 1.8 × 10-5, equilibrium constant for

CH3COOH + OH-  CH3COO- + H2O is

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 7

Ka = 1.8 × 10–5
CH3COO + H2 CH3COOH + OH-
Given reaction is reverse of above

Test: Ionic Equilibrium - 1 - Question 8

A solution with pH 2.0 is more acidic than the one with pH 6.0 by a factor of :

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 8

[H+]1 = 10–2 ; [H+]2 = 10–6

Test: Ionic Equilibrium - 1 - Question 9

The first and second dissociation constants of an acid H2A are 1.0 × 10-5 and 5.0 × 10-10 respectively.

The overall dissociation constant of the acid will be:

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 9



Test: Ionic Equilibrium - 1 - Question 10

An aqueous solution contains 0.01 M RNH2 (Kb= 2 × 10-6) & 10-4 M NaOH.

The concentration of OH- is nearly:

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 10



x2  +  10–4 x – 2 × 10–8 = 0
x = 10–4
[OH–] = x + 10–4
= 2 × 10–4

Test: Ionic Equilibrium - 1 - Question 11

The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be:

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 11

Salt of weak acid & weak base

Test: Ionic Equilibrium - 1 - Question 12

How many gm of solid NaOH must be added to 100 ml of a buffer solution which is 0.1 M each w.r.t. Acid HA and salt Na+ A- to make the pH of solution 5.5. Given pKa(HA) = 5 (Use antilog (0.5)= 3.16)

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 12



Suppose x m. mole NaOH was added


Test: Ionic Equilibrium - 1 - Question 13

If Ksp for HgSO4 is 6.4 × 10-5, then solubility of this substance in mole per mis

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 13


Ksp = S2
⇒ 6.4 × 10–5 = S2
⇒ S = 8 × 10–3 mole/L
S = 8 × 10–3 × 103 mole/m3
⇒ S = 8 mole/m3

Test: Ionic Equilibrium - 1 - Question 14

pH of saturated solution of silver salt of monobasic acid HA is found to be 9.

Find the Ksp of sparingly soluble salt Ag A(s).

Given : Ka(HA) = 10-10

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 14




Ksp  = S (S–X) = 11 × 10–6 × 10–6
= 1.1 × 10–11

Test: Ionic Equilibrium - 1 - Question 15

When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8 × 10-10) will occur only with:

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 15

Test: Ionic Equilibrium - 1 - Question 16

 If the value of the solubility product for AgBr is 4.0 x 10-12 at 25°C, calculate the solubility of AgBr(s) in water.

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 16

Ksp = 4.0 x 10-12 
For AB type salt, Solubility = Ksp1/2
= (4.0 x 10-12)1/2
= 2 x 10-6

Test: Ionic Equilibrium - 1 - Question 17

 For HF, pKa = 3.45. What is the pH of an aqueous buffer solution that is 0.1M HF (aq) and 0.300 M KF (aq)?

Detailed Solution for Test: Ionic Equilibrium - 1 - Question 17

Given, pKa = 3.45
Concentration of HF = 0.1 M, concentration of KF = 0.300 M
For acidic buffer;
pH = pKa + log [salt of weak acid]/[weak acid]
= 3.45 + log0.3/0.1
= 3.45 + 0.48
= 3.93

*Answer can only contain numeric values
Test: Ionic Equilibrium - 1 - Question 18

What will be the value of pH of 0.01 mol dm–3 CH3COOH (Ka = 1.74 × 10–5)?


Detailed Solution for Test: Ionic Equilibrium - 1 - Question 18

*Answer can only contain numeric values
Test: Ionic Equilibrium - 1 - Question 19

For A- + H2O ⇔ HA + OH-, Kb = 1 x 10-12

Thus, pKa of HA + H2O ⇔ H3O + A- is ........


Detailed Solution for Test: Ionic Equilibrium - 1 - Question 19

The correct answer is 2
Here
we know [OH-][H+]=10-14 at 25 degree Celsius
so pka+pkb=14
hence answer is 2

*Answer can only contain numeric values
Test: Ionic Equilibrium - 1 - Question 20

How many of the following are Lewis bases?


Detailed Solution for Test: Ionic Equilibrium - 1 - Question 20

Total number of compounds acting as Lewis base in the given example is 6. H+ and BF3 are electron deficient so they can't act as Lewis base while FeCl3 acts as Lewis acid so all the other compounds except these three are Lewis base.

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