IONIC EQUILIBRIUM 🧪🧑🏻‍🔬

What are electrolyte

What are two types of electrolytes

Any substance which get dissociated into its ion in its solution is called electrolyte

2 type

1 strong electrolyte : 100 percent dissociation into ions

2 weak electrolyte : not 100 percent dissociation into ions

Background details for knowing why ionic equilibrium will be establish only by weak electrolyte

Strong electrolyte don't exist equilibrium

Cause reactants get over

Weak electrolyte reaction will not completel over reactants

Do equilibrium establish hence reversible reaction

It is involving ions so it is ionic equilibrium

Hence weak electrolyte will only establish ionic equilibrium

Ostwalds dilution law

Common ion effect

If a strong containing atleast one common ion is added to a solution of weak electrolyte then degree of weak electrolyte then degree of dissociation of weak electrolyte decreases

Approximations

Concentration of common ion at equilibrium to be considered coming only from strong electrolyte

Equilibrium concentration of weak electrolyte to be considered equal to its initial concentration

Now solve the given question

Just solve it 💅🏻 🫰🏻

Just solve it 💅🏻 🫰🏻

Ionic product of water (Kw)

Water is a very weak electrolyte

Water undergoes self ionisation

Kw is just like expression of eqm constant of dissociation water

At 25 degree celcius 10^-14

Pure water at 25 degree celcius ionizes into equals no. Of OH- AND H +

LIKE 1MOLECULE OF WATER WILL DISSOCIATE INTO 1 OH MINUS AND H PLUS

H PLUS IS OF ACIDIC NATURE AND OH MINUS IS OF BASIC NATURE HENCE PURE WATER US ALWAYS NEUTRAL

OH MINUS AND H PLUS CONCN LIKELY TO SAY ACTIVE MASS IS 10 TO THE POWER MINUS 7

pH scale

It is a way of representing molar concn oh H plus

Is solutions are always neutral on pH 7 ???

No solution are not neutral because of pH it is neutral when [OH-] = [H+]

14=pH +pOH (FOR ANY AQUEOUS SOLUTION SUM OF H AND OH IS ALWAYS EQUALS TO 14 AT 25 DEGREE CELCIUS)

Calculate pH of 0.01M HNO3 solution at 25 degree

Calculate pH of 0.05 M H2SO4 solution at 25 degree celcius

If some ion comes from two electrolyte and their concn difference is more than 10¹ then ignore smaller one

If concentration are equal or differ just by 10¹ then don't ignore any of the concentration

Calculate pH of 10-⁸bM HCl solution at 25 degree celcius

PKw equalsto minuslogKw

At any temperature concentration of proton in water is HPLUS EQUALS TO UNDER ROOT OF Kw

Calculate pH of 10^-12 M HCl solution at 25 degree celcius

Calculate the pH of 0.001M KOH. (aq) solution at 25 degree celcius

Option 2

Ionic product if water increases idpf

1 pressure is reduced

2 Hplus is added

3 OH minus is added

4 Temperature increases

Option 4

Calculate the pH of 0.001 KOH (aq) solution at 25 degree celcius

The ionic product if water is 10^-14. The Hplus ion concentration in o.o1 M NaOH. Solution is

And 10 to the power minus 12

The number of H plus ions in 1ml of a solution having pH equals to 13 is

Ans 6.023 × 10⁷

pOH of 0.002 M HNO3 is

Ans 11 plus log 2

The concentration of an acid is 10 –9 The pOH if the solution is

Ans almost 7

If pH. of solution of NaOH is 12.0 the pH of H2SO4 solution of same molarity will be

Ans 1.7

Answer 0.4g per litre

Calculate [H+] in a solution having pH 3.7

Calculate Hplus concentration in a beer having pH. 2.3

Pehle usko next integer me lao 3.7 ko jo ki milega 4 toh ab usko 10 ki power minus 4 kr de uske baad dekh ki kitna subtract krne par 4 me se 3.7 aayega aur woh hoga 0.3 auryeh aata hai log 2 ke wjh se toh ab answer hogya 2 into 10 to the power minus 4

pH calculation of weak acid and weak base

Calculate the pH of 0.01 NH4OH (Kb = 10–⁶) solution at 25 degree celcius

Answer 10

A solution of pH. 9.0 is 1000 times more basic as a solution The pH of this solution is

The pH of 1 M aqueous solution of a a weak acid HA is 6.0 It's dissociation constankt is

Answer 10-¹²

The concentration of a monoprotic acid is C moles per litre and ionisation constant ka The pH of the solution is

Calculate pH of 0.1 M NH4OH (aq) solution at 25 degree celcius

You may have forgot that NH4OH is a weak electrolyte😐🫡

Calculate pH of 0.1M NH4OH (aq) solution at 25degree celcius

(Ka(NH4+) = 10-⁹ )

You may have forgot that ka or kb is not the concentration it's is equilibrium constant

pH mixture idpf acids

Calculate pH of a solution obtained by mixing 1L of 0.01M HCl with 1L of 0.02 M NaOH solution

Answer 11.7

How many grams of caustic potach required to completely neutralise 12.6 g HNO3

Answer 11.2 g

Caustic potash is KOH.

At wt if K is 39

pH of mixture of weak acid

Calculate (HPLUS ) (HCOOMINUS) APand (OCN MINUS) in a solution that contains 0.1 M HCOOH (Ka equals to 2.4 into 10 to the minus 4 ) and 0.1 M ( Ka equals to 4 into 10 to the power minus four )

Answer

For H PLUS. 8 into 10to the power minus 3

3 into 10 to the power minus 3

5 to the power minus 3

You may have forgot when we add two reactions then their equilibrium constant get multiplied to give equilibrium constant of the resultant reaction

Salt hydrolysis

Reverse reaction of neutralisation is called salt hydrolysis

It is the breaking up of salt in presence to regenerate respective acid or base

Only acid or only base or both can be regenerated

Types of salts

Salts are of 4 types

Salt of strong acid and strong base

NaCl KNO3. Na2SO4

Salt of strong acid and weak base

NH4Cl (NH4)2SO4.

Salt of weak acid and strong base

NaCN CH3COONa

Salt of weak acid and weak base

CH3COONH4. NH4CN

Hydrolysis of salt of SA and SB

Hydrolysis of WB and SA

Hydrolysis of SB and WA

Learn the examples of strong weak acids strong bases and weak bases and then only try to solve the given. questions

Here are the formatted questions.

The aqueous solution of HCOONa, C6H5NH3+Cl-, and KCN are respectively:

Option 1: Acidic, Acidic, Basic

Option 2: Acidic, Basic, Neutral

Option 3: Basic, Neutral, Neutral

Option 4: Basic, Acidic, Basic

Which of the following salts on being dissolved in water gives pH greater than 7 at 25 degrees Celsius?

Option 1: NH4CN

Option 2: NH4Cl

Option 3: KNO3

Option 4: KCN

Aqueous solution of sodium cyanide is:

Option 1: Acidic

Option 2: Amphoteric

Option 3: Basic

Option 4: Neutral

The pH of 0.1M solution of the following salts increase in the order:

Option 1: NaCl < NH4Cl < NaCN < HCl

Option 2: HCl < NH4Cl < NaCl < NaCN

Option 3: NaCN < NH4Cl < NaCl < HCl

Option 4: HCl < NaCl < NaCN < NH4Cl

The salt solution of strong acid and weak base FeCl3 is:

Option 1: Acidic

Option 2: Basic

Option 3: Neutral

Option 4: None of this

pH calculation if strong acid and strong base f

No hydrolysis

H2O

pH of salt of weak base and strong acid

If we dissolve a salt whose pH is basic into water then it is 100 percent sure that the weak one is acid and the strong one is base and VICE VERSA

pH calculation of salt of weak acid and strong base

### Questions on Salt Hydrolysis

Question 1. A 0.1 M sodium acetate solution is prepared. The of acetic acid is . Which of the following statements is correct?

* The degree of hydrolysis is .

* The concentration of ions is .

* The pH is approximately 8.88.

* All of the above.

Ans 4

Question 2. HX is a weak acid with . It reacts with caustic soda (NaOH) to form the salt NaX, having a concentration of 0.1 M. What is the degree of hydrolysis of NaX?

* 0.01%

* 0.0001%

* 0.1%

* 0.5%

Answer 1

Question 3. What is the pH of a 0.001 M sodium acetate solution? Given that the of acetic acid () is .

* Approximately 11

* Approximately 6.5

* Approximately 14

* Approximately 8

Question 4. The pH of a 0.1 M NaCN solution is 11. What is the percentage of hydrolysi

s?

* 0.1%

* 1%

* 0.01%

* 10%

Answer 1

Buffer solution

Questions related to buffer solution

The solution of string acid and weak base is (FeCl3)

1 acidic

2basic

3 neutral

4 none

Mechanism of buffer action

Solve these questions

A )For preparing a buffer solution of pH 6 by mixing sodium acetate and acetic acid, the ratio of the concentrations of salt and acid should be (Ka = 10⁻⁵).

Options:

(1) 1:10

(2) 10:1

(3) 1:100

(4) 100:1

Ans 2

Question: Two buffer solutions A and B, each made with acetic acid and sodium acetate, differ in their pH by one unit. Solution A has a salt-to-acid ratio of X:Y, while solution B has a salt-to-acid ratio of Y:X. If X > Y, then the value of X/Y is:

Options:

(1) 10,000

(2) 3.17

(3) 6.61

(4) 2.10

Ans 2

Question 2: How many moles of HCOONa must be added to 1 liter of 0.1 M HCOOH to prepare a buffer solution with a pH of 3.4, given that for HCOOH is ?

Options:

(1) 0.01 mol

(2) 0.05 mol

(3) 0.1 mol

(4) 0.2 mol

Ans 0.05

Question 3: To 1 liter of a solution containing 0.1 mole each of NH₃ and NH₄Cl, 0.05 mole of HCl is added. The change in pOH will be:

Given: for NH₃ = 4.74

Options:

(1) +0.30

(2) −0.30

(3) +0.48

(4) −0.48

Ans solve it and get it

pH of basic buffer solution

Question 4: The pH of a buffer solution containing NH4OH and NH4Cl is given by:

Options:

(1) pH = pKb

(2) pH = 1/2 pKb − 1/2 log (Salt/Base)

(3) pH = 14 − pKb − log (Salt/Base)

(4) pH = pOH − pKb + (Salt/Base)

Ans solve it and get it

Question 6: 0.1 mole of CH₃NH₂ (Kb = 4.5 × 10⁻⁴) is mixed with 0.08 mole of HCl, and the solution is diluted to 1 liter. The H⁺ ion concentration in the solution will be:

Options:

(1) 1.6 × 10⁻¹¹

(2) 8 × 10⁻¹¹

(3) 5 × 10⁻⁵

(4) 8 × 10⁻²

Ans solve it and get it

Buffer capacity

It is equal to no. Of moles of acid required to be added into 1L buffer solution in order to change its pH by one unit

Higher the buffer capacity better is buffering action

Question 7: Calculate the buffer capacity of an acidic buffer solution having a volume of 2 liters and a pH of 5. If 2 moles of HCl are added, its pH changes to 1.

Options:

(1) 0.5

(2) 0.25

(3) 0.125

(4) None of these

Ans 2

Solubility

Solubility (s) of a substance is equal to no. of moles of that substance that dissociates into its ions in 1L solution

Unit of solubility equals to moles per L

Question: The solubility in water of a sparingly soluble salt AV2 is 1 × 10⁻⁵ mol/L. Its solubility product (Ksp) will be:

Option 1: 4 × 10⁻¹⁵

Option 2: 4 × 10⁻¹⁰

Option 3: 1 × 10⁻¹⁵

Option 4: 1 × 10⁻¹⁰

Question 1: Which is the correct representation of the solubility product constant (Ksp) of Ag2CrO4?

Option 1: \[Ag+\]²\[CrO4²⁻\]

Option 2: \[Ag+\]\[CrO4²⁻\]

Option 3: \[2Ag+\]\[CrO4²⁻\]

Option 4: \[2Ag+\]²\[CrO4²⁻\]

Q. Given that the solubility product (Ksp) of A₃B₂ is 2 × 10⁻³⁰, what will be its solubility in moles per liter?

Options:

(1) (1.85 × 10⁻³²)^(1/5)

(2) (2 × 10⁻³⁰ / 108)^(1/5)

(3) (10⁻²⁸ / 5400)^(1/5)

(4) All of the above

8

Question: The Ksp of Cr(OH)₂ is 1.6 × 10⁻³⁰. The molar solubility of this compound in water is:

Options:

(1) √(1.6 × 10⁻³⁰)

(2) (1.6 × 10⁻³⁰)^(1/4)

(3) (1.6 × 10⁻³⁰ / 27)^(1/4)

(4) (1.6 × 10⁻³⁰) /27

Question: The solubility of CaF₂ in water at 20°C is 15.6 mg/dm³. What will be the solubility product (Ksp) of CaF₂?

Options:

(1) 4 × 10⁻⁴

(2) 8 × 10⁻⁸

(3) 32 × 10⁻¹²

(4) None of these

Question: The solubility product of AgCl is 2.8 × 10⁻¹⁰ at 25°C. Calculate the solubility of the salt in 0.1 M AgNO₃ solution.

Options:

(1) 2.8 × 10⁻⁹ mol/L

(2) 2.8 × 10⁻¹⁰ mol/L

(3) 3.2 × 10⁻⁹ mol/L

(4) 3.2 × 10⁻¹² mol/L

Question: The molar solubility of a sparingly soluble salt MX₄ is S. Thus, the corresponding solubility product (Ksp.s) is given in terms of S by the relation:

Options:

(1) S = (Ksp/128)^(1/4)

(2) S = (Ksp/256)^(1/5)

(3) S = (256 Ksp)^(1/5)

(4) S = (128 Ksp)^(1/4)

Question: The pH of an aqueous solution of Ba(OH)₂ is 10. If the Ksp of Ba(OH)₂ is 1 × 10⁻⁹, then the concentration of barium ions in the solution, in mol/L, is:

Options:

(1) 1 × 10⁻² mol/L

(2) 1 × 10⁻⁴ mol/L

(3) 1 × 10⁻¹ mol/L

(4) 1 × 10⁻⁵ mol/L

Question: The solubility product (Ksp) of a sparingly soluble salt MX₂ is 4 × 10⁻¹². The concentration of M²⁺ ions in the aqueous solution of the salt is:

Options:

(1) 2 × 10⁻⁶ M

(2) 1 × 10⁻⁴ M

(3) 1.6 × 10⁻⁴ M

(4) 4 × 10⁻¹⁰ M

Question: In a 1 L vessel, [Ca²⁺] = 0.01 M and [F⁻] = 0.01 M. Will CaF₂ precipitate? Given that the Ksp of CaF₂ is 4 × 10⁻⁵.

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Question: A precipitate of AgCl is formed when equal volumes of the following solutions are mixed. The Ksp of AgCl is 10⁻¹⁰. Which of the following combinations will form a precipitate?

Options:

(1) 10⁻⁴ M AgNO₃ and 10⁻⁷ M HCl

(2) 10⁻⁵ M AgNO₃ and 10⁻⁷ M HCl

(3) 10⁻⁵ M AgNO₃ and 10⁻⁴ M HCl

(4) 10⁻⁶ M AgNO₃ and 10⁻⁶ M HCl

Question: Solid BaCO₃ is gradually dissolved in a 1 × 10⁻⁴ M Na₂CO₃ solution. At what concentration of Ba²⁺ will the precipitate begin to form?

Given: Ksp of BaCO₃ = 5.1 × 10⁻⁹

Options:

(1) 4.1 × 10⁻⁵ M

(2) 5.1 × 10⁻⁵ M

(3) 8.1 × 10⁻⁶ M

(4) 8.1 × 10⁻⁷ M

Question: The Ksp of Mg(OH)₂ is 10⁻⁶. At what minimum pH will Mg(OH)₂ start precipitating from a 0.01 M MgCl₂ solution?

Options:

(1) 2 + log 2

(2) −log 2

(3) 12 + log 2

(4) 12−log 2

Question: At what pH will Cu(OH)₂ start precipitating in a 0.01 M CuSO₄ solution? Given: Ksp of Cu(OH)₂ = 10⁻¹⁹.

Options:

(1) 5

(2) 13

(3) 10

(4) None of these

Question: A 0.1 M NaOH solution is added to 2 L of 0.1 M CH₃COOH (acetic acid). Given that Ka = 10⁻⁵, calculate the pH of the mixture when half of the acid has been neutralized.

Options:

(1) 5

(2) 4

(3) 6

(4) None of these