Ionic Equilibria MCQ Quiz in తెలుగు - Objective Question with Answer for Ionic Equilibria - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

According to Henderson – Hasselbalch equation:

pH = pK_a + log \(\rm \frac{[Salt]}{[acid]}\)

pKa = 4.6

Therefore, 

pH = 4.6 + log \(\frac{0.01}{0.1}\)

= 4.6 - 1 = 3.6

ph ∼ 3.8

:The basic colour of the indicator is visible when the concentration of In_B is equal to In_A. 

We represent general indicator by the formula HIn

Its ionization in solution will be, HIN ⇋ H⁺ + In^-

we define equilibrium constant as,  

\(\rm K_{ai}=\frac{[H^+][In^-]}{[HIn]}\)

upon rearranging and taking logarithms we get,

-log K_ai = -log [H⁺] - \(\rm \log\frac{[In^-]}{[HIn]}\)

pK_ai = pH - \(\rm \log\frac{[In^-]}{[HIn]}\)

Thus, when [In^-] = [HIn], we have pK_ai = pH

That means, when the two forms have equal concentration, the color change is most noticeable. When [H⁺] is greater than 10 K'ai In^-  color dominates, whereas color due to HIn dominates if [H⁺ ] < K_ai/10 ^- 

Note, here In- and HIn represents In_B and In_A respectively.

According to Henderson – Hasselbalch equation:

pOH = pK_b + log \(\rm \left(\frac{[salt]}{[base]}\right)\)

pK_b = -log K_b = − log 1.81 × 10^-5 = 4.7423

Therefore,

pOH = 4.7423 + log \(\left(\frac{0.25}{0.2}\right)\)

= 4.7423 + 0.0969 = 4.821

ph = 14 - pOH = 14 – 4.821 = 9.17

Hence the pH of a buffer solution is 9.17

pH = −log[H⁺]

(a) value of pH = 0.1M HNO₃ = −log [0.1] =1

(b) Value of pH =0.05 H₂SO₄= −log [2 × 0.05] =1

(c) Value of pH = 0.1M CH₃COOH ≠ 1 as it is a weak acid so, will not undergo complete ionization.

(d) Milli-equivalent of acid = 50 × 0.4 = 20

Milli-equivalent of base = 50 × 0.2 = 10

Remaining acid = 20 −10 = 10 milli-eq.

[H+ ] = 1050 + 50 = 10100 = 0.1

pH = −log[H⁺] =1.

Since HCl is a strong acid, it will dissociate completely into it's ions at equilibrium. The dissociation of HCl in aqueous medium is given by: 

Given C = 0.03M

pH = -log[H⁺] = -log[H₃O⁺]

= -log[0.03]

= -log[3 × 10^-2]

= 2 – 0.48

= 1.53

We Know, pH + pOH = 14

pOH = 14 – 1.53

pOH = 12.48

Since HCl is a strong acid, so its pH will be less than 7.

The concentration of HCl = 10⁻⁸ M

Total [H⁺ ] =[H⁺ ] obtained from HCl +[H⁺ ] obtained from H₂O[H⁺ ]

HCl being a strong acid, it completely ionizes.

[H⁺ ] HCl = 1.0 × 10⁻⁸

The concentration of H⁺ from ionization is equal to that of OH⁻ from water, so let us consider it to be ‘x’

.[H⁺] from H₂O = [OH⁻ ] from H₂O = x

Total [H⁺] = x + 1.0× 1 0^-8  …(1)

But we know

[H⁺] [OH^-] = 1.0× 10^-14

(1.0 × 10^-8x + x) (x) = 1.0 × 10^-14

x² + 10^-8 x - 10^-14 = 0

Solving x, we get

x = 9.5 × 10^-8

Substituting x value in equation (1)

Total, [H⁺] = 9.5 × 10^-8 + 1.0 × 10^-8

Total, [H+] = 1.05 × 10^-7

Consider an equation

⇒ pH = − log [H⁺ ]

⇒ pH = − log [1.05×10⁻⁷ ]

pH = 6.98

H₂PO₄⁻ → HPO₄⁻² + H⁺ , conjugate base of H₂PO₄⁻ is HPO₄⁻²

H₂SO₄ ionizes completely in water and thus, is a strong acid.

The equation pH + pOH = 14 is true at 25ºC for aqueous solutions, but it is not true at other temperatures.

Number of m.moles of Acetic Acid in the mixture = (500 × 0.2)= 100

Number of m.moles of CH₃-COONa in the mixture= (500 × 0.3)= 150

CH₃COOH + NaOH ⟶ CH₃COONa + H₂O

One mole of CH₃COOH reacts with One mole of NaOH

Thus, the excess CH₃COOH = (150-100) moles = 150 moles

Therefore,

pH= pKa + log (Salt/Acid)

pH = -log (1.75×10^-5 ) + log (150/100)

pH= 4.757 + log (1.5)

pH= 4.75 + 0.176

pH = 4.926 = -log [H⁺ ]

[H⁺ ] = 1.17 × 10^-5 mol.dm^–3

Number of m.moles of KOH in the mixture = (10 × 0.2)= 2

Number of m.moles of CH₃-COOH in the mixture= (30 × 0.1)= 3

CH₃COOH + KOH ⟶ CH₃COOK + H₂O

One mole of CH₃COOH reacts with One mole of KOH

Thus, the excess CH₃COOH = (0.003 – 0.002) moles = 0.001 moles

Therefore

pH = pKa + log salt/acid

pH = 4.75 + log 2/1

pH = 4.75+ log2

pH = 4.75 + 0.30

pH = 5.05

CONCEPT:

Buffer Solution

• A buffer solution is a solution that resists changes in pH upon the addition of small amounts of acid or base.
• Buffers are typically made by mixing a weak acid with its conjugate base, or a weak base with its conjugate acid.
• The presence of both the weak acid (or base) and its conjugate form allows the buffer to neutralize added H⁺ or OH⁻ ions, stabilizing the pH.

EXPLANATION:

• Option 1:

  CH₃COOH + CH₃COONa

  This combination forms a buffer. CH₃COOH is a weak acid, and CH₃COONa provides the conjugate base (CH₃COO⁻) in solution. Together, they maintain a stable pH when small amounts of acid or base are added.

• Option 2:

  CH₃COOH + NaCl

  This does not form a buffer. NaCl is a neutral salt and does not provide a conjugate base to pair with CH₃COOH, so it does not resist pH changes.

• Option 3:

  NaOH + NaCl

  This combination does not form a buffer. NaOH is a strong base and does not pair with NaCl to provide a weak acid-base conjugate pair.

• Option 4:

  NaOH + CH₃COONa

  This does not form a buffer, as there is no weak acid in the mixture to pair with the acetate ion (CH₃COO⁻).

Therefore, the correct option is: 1) CH₃COOH + CH₃COONa.

Concept:-

• Buffer Solution is a solution that consists of a mixture containing a weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base.
• A acidic buffer solution is a solution, containing a weak acid (A) and its salt (MA) with strong base.
• The pH of a acidic buffer solution can be determined by using Henderson's equation which is given by,

​pH = pKa + log\(\frac{[Salt]}{[Acid]}\)​

where, [salt] = Conc of the salt produced in the solution,

and [Acid] = Conc of the remaining acid present in the equilibrium.

Explanation:-

• The mixture of acetic acid and sodium acetate is an acidic buffer.
• Hence,

​pH = pKa + log\(\frac{[Salt]}{[Acid]}\)........(1)

• Now, when [Salt] = [Acid]
• Thus, from equation (1) we got

pH = pK_a

• When, [Salt] = 10 × [Acid]
• Thus, from equation (1) we got,

pH = pKa + log\(\frac{10\times[Acid]}{[Acid]}\)

or, pH = pK_a + log10

or, pH = pK_a + 1

Conclusion:-

• Hence, the pH of the buffer solution is pKa + 1