Ionic Equilibrium MCQ Quiz in मल्याळम - Objective Question with Answer for Ionic Equilibrium - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is option 4 i.e., Na.

• Ethanoic acid liberates hydrogen gas when it reacts with Na (Sodium). 
• Ethanoic acid is commonly known as Acetic acid.
• Chemical formula - CH₃COOH.
• The Chemical reaction
  • Acid + Metal → Salt + Hydrogen gas
    • 2CH₃​COOH + 2Na → 2CH₃​COONa + H_2​ (gas)
• Please note that Zinc can also replace Sodium and hydrogen gas would still be liberated in the reaction.
• CH₃COONa is nothing but a salt called as Sodium Acetate which is used as additives in food, industry, concrete manufacture, etc.

The correct answer is More than one of the above  

Concept:

• In chemistry, organic and inorganic substances can be classified as acids, bases, and salts.
• This classification is grounded on certain properties unique to each kind.

Explanation:

• According to the Arrhenius definition, acid is something that donates a proton.
• According to the Bronsted-Lowry definition, acid is something that gives H⁺ ions in water.
• According to the Lewis definition, acid is something that accepts a pair of electrons.
• Therefore, all of these represent the properties of acids.

Additional Information

• Salt is a product of the reaction between an acid and a base.
• For example, the production of NaCl is a result of the reaction between HCl and NaOH.
  •   
• The pH of an acid is always less than 7 while that of a base is always greater than 7.
• The acid in the litmus test turns the blue litmus red while a base turns the red litmus blue.

Concept:

According to acid-base theory,

• Lewis acid is a species that can accept a pair of electrons.
• Lewis base is a species that can donate a pair of electrons to other species.
• For example, NH₃ is a lewis base because it can donate a pair of electrons and BF3 is lewis acid because it can accept a pair of electrons.   

Explanation:

Lewis structure of BF₃ - 

• The central atom 'B' has electronic configuration B(5) = 1s2 2s2 2p1.
• It has a total of 3 valence electrons in its valence shell i.e. 2s2 2p1only.
• So, it can form a maximum of 3 covalent bonds only when one of the 2s electrons is excited to 2p.
• Each F atom has 7 valence electrons and needs one electron to complete its octet.
• So, three F atom makes 3 covalent bonds with B.
• Boron now has a total of 6 electrons (3 of B and 3 from 3F's).
• Thus, the octet of B is not complete, it has a vacant p orbital in its structure which can accommodate 2e^-.

Structure -

→ Lewis acid is a species that can accept a pair of electrons.

In BF3,  the octet of B is not complete, it has a vacant p orbital in its structure which can accommodate 2e- or accept a pair of electrons thus act as Lewis acid. 

Conclusion:

Therefore, the acidity of BF3 can be explained on the basis of Lewis's concept of acid and bases.

Hence, the correct answer is option 3.

Additional Information

Arrhenius concept -

• ​An acid is a substance that increases the concentration of H⁺ ions in an aqueous solution.
• A base is a substance that increases the concentration of OH^- ions in the solution.
• For example,
  • HCl is acid as it dissociates in an aqueous solution into H⁺ and Cl^- ions and increases the concentration H⁺ ions in the solution.
  • NaOH is a base as it increases the OH^- ion concentration in the solution.

Bronsted Lowry concept - 

• Bronsted Lowry acid is a substance that donates H⁺ ion or a proton to other species and forms its conjugate base.
• Bronsted Lowry base is a substance that accepts an H⁺ ion or a proton and forms a conjugate acid.
• HF is a Bronsted Lowry acid. (HF  H⁺ + F^-) 
• NH₃ is a Bronsted Lowry base. (H₂O +NH3   OH^- + NH₄⁺ )

Concept:

Solubility product (K_sp)

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble salt. It represents the maximum amount of the salt that can dissolve in water.

For a salt AB that dissociates as:

The Ksp is given by:

Ion Product ( Q )

The ion product ( Q ) is the product of the concentrations of the ions in a solution at any given time. It helps in predicting whether a precipitate will form.

For the same salt: ]

Comparison

• If Q

• If Q = Ksp : The solution is at equilibrium. No net change will occur.

• If Q > Ksp : The solution is supersaturated. A precipitate will form.

Explanation:

• The dissociation reaction for PbSO₄ is:
  • PbSO₄ ⇌ Pb²⁺ + SO₄^2–
• The K_sp expression is:
  • K_sp = [Pb²⁺][SO₄^2–]
• Given K_sp for PbSO₄ = 1.6 × 10^–8
• Calculation of Ion Concentrations
  • First, find the final concentrations of Pb²⁺ and SO₄^2– after mixing the solutions.
• Initial moles of Pb(NO₃)₂:
  • (10.0 mL) × (1.5 × 10^–3 M) = 1.5 × 10^–5 moles
• Initial moles of Na₂SO₄:
  • (40.0 mL) × (3.0 × 10^–4 M) = 1.2 × 10^–5 moles
• Total volume after mixing:
  • 10.0 mL + 40.0 mL = 50.0 mL = 0.050 L
• Concentration of Pb²⁺ after mixing:
  • [Pb²⁺] = = 3.0 × 10^–4 M
• Concentration of SO₄^2– after mixing:
  • [SO₄^2–] = = 2.4 × 10^–4 M
• Calculation of Ion Product (Q)
  • Ion product Q is given by:
    • Q = [Pb²⁺][SO₄^2–]
      • Q = (3.0 × 10^–4 M) × (2.4 × 10^–4 M)
        • Q = 7.2 × 10^–8
• Comparison with Ksp
  • Given K_sp (PbSO₄) = 1.6 × 10^–8
  • Since Q (7.2 × 10^–8) > Ksp (1.6 × 10^–8), the ion product exceeds the solubility product, indicating that a precipitate will form.

Conclusion:

A precipitate will form when 10.0 mL of 1.5 × 10^–3 M Pb(NO₃)₂ is added to 40.0 mL of 3.0 × 10^–4 M Na₂SO₄: Yes because ion product > Ksp

Concept:

Sulfuric acid (H₂SO₄) is a strong acid that dissociates in two steps in an aqueous solution:

First dissociation:

(complete dissociation)

Second dissociation:

(partial dissociation)

Explanation:

Given a 0.10 M H₂SO₄ solution:

First dissociation:

Second dissociation: some of the will dissociate further:

This additional dissociation will increase the concentration of H⁺ and produce some

Therefore, the concentration of H⁺ will be greater than just 0.10 M due to the second dissociation:

[HSO_4^-]\) due to the additional H⁺ ions produced in the second dissociation.

Conclusion:

Thus, the true statement with regard to a 0.10 M H₂SO₄ solution is:  [HSO_4^-]\)

Concept:

Neutralisation Reaction:

In the reaction in which equal moles of a strong acid such as HCl reacts with equal moles of a base like NaOH, a neutral salt NaCl is formed.

pH: It is expressed as the negative logarithm of the concentration of hydrogen ions present in the solution.

Explanation:

Given Data:

Normality of HCl = 0.1 N

Volume of HCl = 1ml

Volume of NaCl = 999 ml

Step 1: Calculating the number of moles of HCl

HCl being a strong acid gets completely dissociated to give one mole of hydrogen and chloride ions.

The relationship between normality and molarity is:

When the number of moles is 1, then normality = molarity

So, 0.1 N = 0.1 M

Now, 

Step 2: Calculating the final concentration of  present in the solution when HCl reacts with NaOH and form NaCl:

The total volume of the solution = Volume of 0.1 N HCl + Volume of NaCl solution 

 = 1 ml + 999 ml

= 1000 ml

Step 3: Calculate the pH of the resultant solution:

Conclusion:

The pH of the resultant solution is 4

Concept:

pH: The negative logarithm of the hydrogen ion concentration is used to calculate the pH of a solution.

• A solution is acidic when its pH is lower than 7.
• The solution is regarded as a base when the pH is higher than 7.
• The solution is neutral when the pH value is 7.

Explanation: 

Given data:

Concentration of HCl 

• When HCl being an acid is diluted with water,  the concentration of  ions becomes very less. 
• In this case, the concentration of  present in water cannot be neglected. In these types of cases, total  concentration can be calculated by using:

    .......(1)

Also, the concentration of  at standard temperature i.e;  is .

Substitute the  and  value in equation (1)

The formula used to calculate pH is:

This value is slightly less than 7.

Conclusion:

The pH value of  HCl solution in water is 6.9589, which is slightly less than 7.

Explanation:-

Solubility product (K_sp):

The dissociation of zirconium phosphate in water can be written as follows:

(Zr⁴⁺)₃ (PO₄)₄ ⇌ 3 Zr⁴⁺ + 4 PO₄³⁻

Let's denote the molar solubility of zirconium phosphate by 's'. At equilibrium, we have:

[Zr⁴⁺] = 3s

[PO₄³⁻] = 4s

The solubility product constant (K_sp) expression for zirconium phosphate is given by:

K_sp = [Zr⁴⁺]³ [PO₄³⁻]⁴

Substituting the concentrations in terms of 's', we get:

K_sp = (3s)³ (4s)⁴

K_sp = 27s³ × 256s⁴

K_sp = 6912s⁷

To find the molar solubility (s):

s = (K_sp / 6912)^1/7

Therefore, the correct answer is (K_sp / 6912)^1/7

CONCEPT:

Acid Strength and Anion Stability

• The strength of an acid is also influenced by the stability of the conjugate base (anion) it forms after donating a proton (H⁺).
• The more stable the conjugate base, the stronger the acid.
• Stability of the conjugate base depends on factors such as resonance, electronegativity, and the ability to disperse negative charge.

EXPLANATION:

• HCl forms Cl⁻, a highly stable ion due to its large size and electronegativity making it a very strong acid pka of -6.3.
• H₂SO₄ forms HSO₄⁻, which is stabilized by resonance, making H₂SO₄ a  strong acid, pka of -3.
• HNO₃ forms NO₃⁻, which is also resonance-stabilized, but slightly less stable than HSO₄⁻.
• CH₃COOH forms CH₃COO⁻, which has limited resonance and less stability, making it the weakest acid here.

The correct answer is: HCl

CONCEPT:

Precipitation of Metal Sulfides in Acidic Solutions

• The solubility product constant ( K_sp ) of a salt is the product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced equation.
• For sulfides, K_sp determines the concentration of S²⁻ ions in solution that will lead to precipitation of the metal sulfide, such as ZnS.
• The concentration of S²⁻ in acidic solution is affected by the dissociation of H₂S, which depends on the concentration of H⁺ ions. The overall dissociation constant ( K_NET} ) gives the relationship between H⁺ concentration and S²⁻ concentration.

CALCULATION:

• The solubility product expression for ZnS is given by:
  
  
• H₂S dissociates as:
  
• Substituting values:
  

The correct minimum molar concentration of H⁺ required is 0.2 M, corresponding to option 3.