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

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.
  • \(HCl+NaOH\rightarrow NaCl+H_{2}O\)  
• 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.

Correct answer: 3)

Concept:

• The acidic or basic strength of any solution can be described on the basis of pH value.
• pH can be expressed as the decimal logarithm of the reciprocal of the hydrogen ion activity, a_H⁺, in a solution.
• pH can be expressed as,

          pH=-log[aH+]

• pH scale typically ranges from 0-14 at room temperature. 
• pH 7 is neutral, pH below 7 represents an acidic solution, and pH above 7 represents a basic solution.
• Since, the solution is NaOH, which is basic in nature and thus it should have a value of pH more than 7 and it should be between 7 to 14.

Explanation:

Now, \({{\rm{N}} \over {{\rm{10}}}}\) NaOH solution

= 0.1 M NaOH

= 0.1 M

=10^-1 M

Hence, [OH^-]=[NaOH]= 10^-1 M

Now, the ionic product of water is given by,

​[H+][OH-]=K_w

[H⁺][OH^-]=10^-14 (Kw=10-14)

[H⁺]= \(\frac{[10^{-14}]]}{[10^{-1}]}\)

[H+]= 10^-13

pH= -log [H⁺]

pH = -log[10^-13]

pH= 13

Conclusion:

Hence, pH value of \({{\rm{N}} \over {{\rm{10}}}}\) NaOH is 13.

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 \(\rightleftarrows \) H⁺ + F^-) 
• NH₃ is a Bronsted Lowry base. (H₂O +NH3 \(\rightleftarrows \)  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: \(\text{AB} (s) \rightleftharpoons \text{A}^+ (aq) + \text{B}^- (aq)\)

The Ksp is given by: \(K_{sp} = [\text{A}^+][\text{B}^-]\)

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⁺] = \(\frac{(1.5 × 10^{–5} moles)}{(0.050 L)}\) = 3.0 × 10^–4 M
• Concentration of SO₄^2– after mixing:
  • [SO₄^2–] = \(\frac{(1.2 × 10^{–5} moles)}{(0.050 L)}\) = 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:

\(H_2SO_4 → H^+ + HSO_4^-\) (complete dissociation)

Second dissociation:

\(HSO_4^- ⇌ H^+ + SO_4^{2-}\) (partial dissociation)

Explanation:

Given a 0.10 M H₂SO₄ solution:

First dissociation: \(0.10 M\ H_2SO_4 → 0.10 M\ H^+ + 0.10 M\ HSO_4^-\)

Second dissociation: some of the\( HSO_4^-\) will dissociate further:

\(HSO_4^- ⇌ H^+ + SO_4^{2-}\)

This additional dissociation will increase the concentration of H⁺ and produce some \(SO_4^{2-}.\)

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

\([H^+] > [HSO_4^-]\) 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: \([H^+] > [HSO_4^-]\)

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.

\(HCl(aq)\;+\;NaOH(aq)\;\rightarrow \;NaCl(aq)\;+\;H_{2}O(aq)\)

\(Normality\;=\;no\;of\; moles\;\times \;Molarity\)

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

\(pH\;=\;-log[H^{+}]\)

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.

\(HCl(aq)\;\rightleftharpoons \;H^{+}(aq)\;+\;Cl^{-}(aq)\)

The relationship between normality and molarity is:

\(Normality\;=\;no\;of\; moles\;\times \;Molarity \)

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

So, 0.1 N = 0.1 M

Now, \(Molarity\;=\;\dfrac{No\;of\;moles}{Volume}\)

\(No\;of\;moles\;=\;0.1\;M\;\times \;1\;ml\)

\(No\;of\;moles\;=\;0.1\)

Step 2: Calculating the final concentration of \([H^{+}]\) present in the solution when HCl reacts with NaOH and form NaCl:

\(HCl(aq)\;+\;NaOH(aq)\;\rightarrow \;NaCl(aq)\;+\;H_{2}O(aq)\)

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

 = 1 ml + 999 ml

= 1000 ml

\(Molarity\;=\;\dfrac{No\;of\;moles}{Volume}\)

\([H^{+}]\;=\;\dfrac{0.1}{1000}\)

\([H^{+}]\;=\;10^{-4}\;M\)

Step 3: Calculate the pH of the resultant solution:

\(pH\;=\;-log[H^{+}]\)

\(pH\;=\;-log[10^{-4}]\)

\(pH\;=\;4\)

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.

\(pH\;=\;-log[H^{+}]\)

• 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 \([H^{+}]_{acid}\;=\;10^{-8}\;M\)

• When HCl being an acid is diluted with water,  the concentration of \([H^{+}]\) ions becomes very less. 
• In this case, the concentration of \([H^{+}]\) present in water cannot be neglected. In these types of cases, total \([H^{+}]\) concentration can be calculated by using:

\([H^{+}]_{total}\;=\;[H^{+}]_{acid}\;+\;[H^{+}]_{water}\)    .......(1)

Also, the concentration of \([H^{+}]_{water}\) at standard temperature i.e; \(25^{o}C\) is \(10^{-7}\;M\).

Substitute the \([H^{+}]_{water}\) and \([H^{+}]_{acid}\) value in equation (1)

\([H^{+}]_{total}\;=\;10^{-8}\;M\;+\;10^{-7}\;M\)

\([H^{+}]_{total}\;=\;10^{-7}\;(0.1+1)\)

\([H^{+}]_{total}\;=\;1.1\times 10^{-7}\)

The formula used to calculate pH is:

\(pH\;=\;-log[H^{+}]\)

\(pH\;=\;-log[1.1\times 10^{-7}]\)

\(pH\;=\;6.9589\)

This value is slightly less than 7.

Conclusion:

The pH value of \(\mathbf{1\times 10^{-8}\;M}\) 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:
  \([Zn^{2+}] [S^{2-}] \leq K_{sp}\)
  \([S^{2-}] = \frac{K_{sp}}{[Zn^{2+}]} = \frac{1.25 \times 10^{-22}}{0.05} = 2.5 \times 10^{-21}\)
• H₂S dissociates as:
  \(H_2S \leftrightarrow 2H^+ + S^{2-} \\ [S^{2-}] = \frac{K_{NET} \times [H_2S]}{[H^+]^2}\)
• Substituting values:
  \(2.5 \times 10^{-21} = \frac{1 \times 10^{-21} \times 0.1}{[H^+]^2} \\ [H^+]^2 = \frac{1 \times 10^{-22}}{2.5 \times 10^{-21}} = 0.04 \\ [H^+] = \sqrt{0.04} = 0.2 \, M\)

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