Explore topic-wise InterviewSolutions in .

(a) Lewis acid

Lewis acid are electrophile because they accept electron pair

O^2- is the strong base

(b)  NO^-₃  is the very weak base.

(b)  2 4 1 3

(a)  3 1 4 2

(a)  2 3 4 1

(a) Both A and R are correct and R explains A

(d)  3 4 1 2

(a)  2 1 4 3

(a)  3 4 2 1

(c)  3 1 4 2

(a)  4 3 1 2

1. increases 

2. H₂CO₃ and NaHCO₃

3. Vanslyke, buffer capacity

4. Neutralization

5. CH₃COO^-

6. Calcium oxalate

7. greater 

8. strong base

9. Lewis acids

10. Lewis bases

(a) (ii) only

1. Acid tastes sour, turns the blue litmus to red and reacts with metals such as zinc and produces hydrogen gas. 

2. Base tastes bitter, turns the red litmus to blue and soapy to touch.

1. Arrhenius theory does not explain the behaviour of acids and base in non-aqueous solvents such as acetone, tetrahydro furan. 

2. This theory does not account for the basicity of the substances like ammonia which do not possess hydroxyl group.

(a) CH₃COOH + CH₃COONa

(c) 1 M NaOH

NaOH has maximum [OH^-] and minimum of [H⁺] and so maximum pH value.

(c) Basic

If pH = 7 solution is neutral 

pH < 7 solution is acidic 

pH > 7 solution is basic

(a) Neutral

pH = 5 means [H⁺] = 10^-5

After dilution [H⁺] = 10^-5 / 100 = 10^-7 M

[H⁺] from H₂O cannot be neglected. 

Total [H⁺] = 10^-7 + 10^-7 = 2 x 10^-7 

pH = 7 – 0.3010 = 6.6990 = 7 

pH = 7 (Neutral)

pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

pH = – log₁₀ [H₃O] (or) pH = – log₁₀ [H⁺]

Ionisation eqn. of Ca(Oh)2: Ca(OH)2 = Ca2+ + 2OH– 

ksp of Ca(OH)2=[Ca 2+] [OH-]2=s*(2s)²=4s³ 

=>s=cube-root(ksp/4)=sqrt(4.42*10^-5 /4)=2.23*10^-2 M Ca 2+ ions in 500 ml  soln.=(2.23*10^-2)/2 =0.011 500 ml soln+500 ml NaOH=1000ml,molarity of OH- in new solution=0.2M 

[Ca]2+ =ksp/[OH-]² = (4.42*10^-5)/(0.2*0.2)=0.0011M no. of moles of Ca2+ ppt=0.011-0.0011=0.0102M. Mass of Ca(OH)2= 0.0102*74.

1. The product of concentration of H⁺ and OH^- ions in water at a particular temperature is known as ionic product.

2. The ionic product of water at room temperature (25°C) is,

K_w = [H⁺] [OH⁺] (or)

K_w = [H₃O⁺] [OH⁺] 

K_w =(1 x 10^-7) (1 x 10^-7)

K_w = 1 x 10^-14 mol² dm^-6

(a) pH will increase

KCN + H₂O ⇌ KOH + HCN.

KOH is a strong base and HCN is a weak acid. So pH will increase.

(b) an increase in alkalinity

Adding Na₂CO₃ to water makes the solution basic and hence pH increases from 7.

In the hydration of ion, each of six water molecules donates a pair of electron to Cr³⁺ to form the hydrated cation, hexaaquachromium (III) ion, thus, the lewis acid is Cr and the Lewis base H₂O

(a) Aqueous NaCl solution

F^- is the weak base

(d) (ii) & (iii)

[H₃O⁺] = 4 x M

pH = – log₁₀ [H₃O⁺] 

pH = – 1og₁₀ [4 x 10⁵] 

pH = – log₁₀ [4] – log₁₀ [10^-5]

pH = – 0.6020 – ( – 5) = – 0.6020 + 5

pH = 4.398 

Therefore, the solution is acidic.

(d) mol² L^-2

(d) ClO₄ is the conjugate base of HCIO₄

(a) Both A and R are correct and R is the correct explanation of A

1. Strong acid: HCIO₄ , H₂SO₄ 

2. Strong base: NH₂^- , O^2-

(a) 4 x 10^-8 

pH = – log [H⁺] 

7.4 = – log [H⁺] 

7.4 = log 1 – log [H⁺] 

log [H⁺] = log 1 – 7.4 

log [H⁺] = 8.6 

[H⁺] Antilog of 8.6 

= 4 x 10^-8

(b) CH₃COO^-

Those substances which give a proton is called Bronsted acid, while CH₃COO^- doesn’t have a proton. So it is not a Bronsted acid.

1. Water

2. 1 x 10^-14

3. increases

4. endothermic

5. acidic, basic

6. 1 x 10^-17

7.  zero

8. 14

9. degree of dissociation

10. 10 times

The pH of the battery acid is 0

The pH of drain cleaner is 14

The pH value of water is 7

If the pH of a solution is more than 7, it is called Basic solution.

it is called Acidic solution.

(a) will decrease

(a) Both A and R are correct and R is the correct explanation of A

CO₂ is acidic oxide which on dissolution in water develops acidic nature.

(b) A is wrong but R is correct

(c) 4

For a monobasic acid [H⁺] = c.α

= \(\frac{1}{2}\) x 0.001 = 10^-4 

pH = – log₁₀[10^-4] = 4

(b) 5.09

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

= 4.57 + log \(\frac{0.10}{0.03}\) = 5.09

From Henderson equation of acidic buffer solution we know, 

pH = pKa +log [A^-]/[salt]

degree of ionization (∞) = √Kc/C = 3.4e-3

 So,[CH₃COO^-] = C∞  = 3.4e-3 × 0.09 = 0.306e-3

 So, pH = 4.7 + log [ 0.306e-3]/1.5 = 4.7 + log 0.0204   (log 0.0204  = -1.69)

= 4.7 - 1.69 

= 3.01

NH₃ , NH₄ and HCOOH