- Organic Chemistry
- Aldehydes and Ketones
- Alkyl Halides, Alcohols and ethers
- Amines and other nitrogen compounds
- Aromatic Chemistry
- Carbohydrates, Amino acids, protein, Vitamin and Fat
- Carboxylic acids and its derivatives
- Chemistry in daily life
- General Mechanism in organic compounds
- Hydrocarbons
- Nomenclature and isomerism

# 28 - Chemical and Ionic Equilibrium Questions Answers

**If ionic product of water is K _{w} = 10^{-6} at 4°C, then a solution with pH = 7.5 at 4°C will**

**(i) turns blue litmus red **

**(ii) turns red litmus blue**

**(ii) Be neutral t****o litmus**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

question is not appropriate because for given K_{w} , [H^{+}] will be 10^{-3 }and max. pH will be 6 and the question is asking about a solution of pH 7.5 which is out of limit

but if it is considered as a correct question then answer will be (ii)

**N _{2} + O_{2} = 2NO . Equilibrium constant K_{c} = 2 . Degree of dissociation is**

**(i) 1/1-****√2 (ii) ****1/1+****√2 (iii) 2****/1-****√2**

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**Joshi sir comment**

** N _{2} + O_{2} = 2NO **

1 1 0

1-x 1-x 2x

so by formula K_{c} = (2x)^{2} / (1-x)(1-x)

now solve for x

**In the equilibrium SO _{2}Cl_{2} = SO_{2} + Cl_{2} at 2000K and 10 atm pressure , % Cl_{2} = % SO_{2} = 40 (by volume) . Then what is the value of K_{p}?**

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**Joshi sir comment**

**SO _{2}Cl_{2} = SO_{2} + Cl_{2}**

_{100 0 0}

_{20 40 40}

_{2 atm 4 atm 4 atm}

**now calculate K _{p} **

**0.2 mole of N _{2} and 0.6 mole of H_{2} react to give NH_{3} and 40 % of reactant mixture is decreased , according to the equation, **

**N _{2} (g) + 3H_{2} (g) = 2NH_{3} (g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is**

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**Joshi sir comment**

**N _{2} (g) + 3H_{2} (g) = 2NH_{3} (g)**

**0.2 0.6 0**

**0.2-x 0.6-3x 2x**

according to the given condition 4x = 0.8*40/100 = 0.32

so x = 0.08

now solve

1 mole of A , 1.5 mole of B and 2 moles of C are taken in a vessel volume one litre. At equilibrium concentration of C is 0.5 mole/L. Equilibrium constant for the reaction

**A (g) + B (g) = C (g)**

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**Joshi sir comment**

**A (g) + B (g) = C (g)**

** 1 1.5 2**

**1-x 1.5-x 2+x**

**according to the given condition 2+x = 0.5 so x = -1.5**

**now calculate K _{c}**

The K_{sp} for a sparingly soluble Ag_{2}CrO_{4} is 4 x 10^{-12}. The molar solubility of the salt is

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**Joshi sir comment**

Ag_{2}CrO_{4 }-------------> 2Ag + CrO_{4}

s 0 0

2s s

so K_{sp }= (2s)^{2} s = 4s^{3}

now solve

A 0.1N solution of sodium bicarbonate has a pH value of

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**Joshi sir comment**

Please submit complete question, K_{a} should be given with percentage of ionisation

pH of 0.1 M NH_{4}Cl solution is

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**Joshi sir comment**

For calculating pH of a salt, only concentration is not sufficient

**pH of 10 ^{2} M HCl is**

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**Joshi sir comment**

This is a strong acid so consider 100 percent ionisation

we get [H^{+}] = 10^{2}

it is more than 1 so pH = 0

**100 c.c. of N/10 NaOH solution is mixed with 100 c.c. of N/5 HCl solution and the whole volume is made to 1 litre. The pH of the resulting solution will be**

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**Joshi sir comment**

mili gm eq of NaOH = NV = 10

mili gm eq of HCl = 20

total mili gm eq = 20-10 = 10 of HCl

so N = 10/200 = 0.05

so pH = log[1/0.05]