Chapter 09: Acid Base Chemistry

Short Questions & Flashcards Study Portal

Short Questions

Common Ion Effect

Q.1

Define the following example for each: (i) Ionization constant (ii) Solubility product (iii) Common ion effect (iv) Acid-base Indicator

Answer

(i) Ionization constant It is the extent of ionization of acids and bases which is much less than 100%. It is the quantitative measure of the strength of an acid or base. 'Ka' is ionization constant of acids and 'Ko' is ionization constant of a bases. HA + H2OF H2O* + A [HO*][A] f K. = [HA] (ii) Solubility product It is the product of concentration of ions raised to the exponent equal to the coefficient of the balanced equation. It is represented by • Ksp For example: (ill) Common ion effect The suppression of ionization of a weak electrolyte by adding a common ion to it is called common ion effect. For example: When HCl gas is passed through the saturated solütion of NaCl, due to formation of common Clions is increased and NaCl is crystallized out NaCC (8) Na (aq) + Cl (aq) H HC (g) (ag) + CC (a4) (iv) Acid-base Indicator An indicator is a substance that changes colour to mark a titration's and point. Acid base indicators exhibit one colour in acid and another in base. They are mostly weak organic acids or weak organic bases. For example: Phenolphthalein, methyl orange etc.

Illustration (added) - Solubility Equilibrium & Ksp + - Undissolved Solid ⇌ Dissolved Ions

Buffer Solutions

Q.2

Differentiate between: (i) Hydrolysis and dissolution (ii) Acidic and basic buffer solutions

Answer

(i) Hydrolysis: The process in which substance dissolves in water, dissociate into ions and interact with water molecules is called hydrolysis. (i) It is a chemical process. (ii) Formation of new substances occurs. Example: Hydrolysis of ester Dissolution The process in which substance dissolves in solvent and form a homogeneous mixture. (i) It is a physical process. (ii) Formation of new substance does not occur Example: Dissolution of sugar in water. (ii) Difference of acidic and basic buffer solution. Acidic buffer: The buffer formed by mixing weak acid and salt of strong base is acidic buffer. Example: Mixture of CH:COOH and CH:COONa Basic Buffer: The buffer formed by mixing weak base and salt of strong acid is basic buffer Example: Mixture of NH4Cl and NH4OH

Concepts of Acids and Bases

Q.3

Explain the concept of conjugate acid-base pairs. How are they related in terms of proton transfer?

Answer

Conjugate acid: A specie formed when a base accepts a proton from the acid is called conjugate acid. Conjugate acid: A specie formed when an acid donates a proton to a base is called conjugate acid. In this reaction, Cl and HOt are conjugate acid- base pairs. HCC + H2O = HOt + Ce acid base Conjacid Conj.base

Illustration (added) - Conjugate Acid-Base Pairs HA Acid A⁻ Conjugate Base Lose H⁺

Q.3

Describe the Bronsted-Lowry theory of acids and bases. Provide examples of conjugate acid-pairs and explain clearly their relationship.

Answer

See Q.2 from theory.

Q.4

What is the relationship between the strength of an acid and the strength of its conjugate base?

Answer

There is an inverse relationship between the strength of an acid and strength of its conjugate base. The strong acid produces a weak conjugate base. Similarly, the strong base produces a weak conjugate acid. Example: HCl strong acid and its conjugate base 'Cl" is a weak base. 05. For the following three reactions, identify the reactants that are Arrhenius bases, Bronsted-Lowry bases, or Lewis bases. State which type of bases each reactant is. Explain your answers. (i) NaOHs) =Na (ag) + OH (ag) → F(ag) + H, O(a) Ans. (i) NaOHs) Na(aq) + OH (ag) In this reaction, NaOH dissociates in aqueous solution to produce OH ion. Hence, NaOH is an Arrhenius Base. (i) HF(a4) + H,0(0) Faq) + HO (ag) In this reaction, H2O behaves as Bronsted- Lowry base because it accepts a proton (H*) from HF. (i) H(ag) + NH 3(a4) =H4(aq) In this reaction, NH3 behaves as Lewis base because it donates the pair of electrons to Ht can

Lewis Concept

Q.4

Define the Lewis theory of acids and bases. How does this theory differ from the Bronsted-Lowry theory? Give examples of Lewis acids and bases that do not involve proton transfer.

Answer

See Q.3 from theory.

Q.5

Discuss applications and implications of the common ion effect in various fields. NUMERICAL PROBLEMS

Answer

pH and pOH

Q.6

An amphoteric substance behave as either an acid or a base. Identity whether water behaves as an acid or a base in each of the following reactions. (i) NH, + H,0=-NH; + OH- (i) HNO, + H,0 H2O* + NO; (iv) CH, COOH+ H,0--CH, COO-+ H,0+

Answer

(i) H2O+ HC=→H,0*+CE In this reaction, H20 behaves as a base because it accepts a proton (HT) from HCl. (ii) NH, + H,0 = ENH; + OH In this reaction, H2O behaves as an acid because it donates a proton (HT) to NH3. (ii) HNO, + H2O--H,0* + NO; In this reaction, H2O behave as a base because it accepts a proton (HT) from HNO3. (iv) CH,COOH + H,0==CH,COO- + H,0+ In this reaction, H2O behave as a base because it accepts a proton (Ht) from CH COOH

Solubility Product

Q.6

What is the solubility product for sparingly soluble salts. Give its two applications.

Answer

See Q.12 from theory.

Q.7

Which salt would you expect to dissolve more readily in acidic solution: Barium carbonate or Copper sulfide? Explain. (sp(aco,)) = 1.1×10-10, K sp(cus) = 8x10-34

Answer

Barium carbonate (BaCO3) dissolves more readily than CuS in acidic solution. When BaCO3 dissociates. BaCO, Ba* + CO3- Its conjugate base CO? reacts with Ht ions and form a weak acid BaCO3. CO3 + 2H* ÷H,CO, Hence, due to formation of H,CO3, solubility of BaCO3 increase. While in CuS. CuS -Cu2+ + S2- Its conjugate base does not react with Ht ions. Hence, its solubility is less than BaCO3. More over the Ksp of BaCO3 is more than the Kop of CuS. So, BaCO3 is more soluble than CuS.

Q.7

Describe the general shape of a titration curve for a strong acid titrated with a strong base. How can you identify the equivalence point on this titration curve?

Answer

See Q.15 from theory. A 5 = 4.8+ log [HA] = 5-4.8 log [HA] A = 0.2 log [HA] Taking antilog on both side [A] = antilog0.2 [HA] LA] =1.5848 [HA] Answer: The ratio of [A] to [HA] is 1.5848.

Illustration (added) - Acid-Base Burette Setup Burette (Acid) Conical Flask (Base + Indicator)

Q.8

Why does common ion effect decrease solubility of a less soluble salt? The addition of a common ion to the

Answer

solution of a less soluble electrolyte suppresses (decreases) its ionization and the concentration of unionized species increase, which may come out as precipitate. That's why solubility of salt decrease due to common ion effect. State the basic principle of 09. Mention factors solubility product. affecting solubility product. Ans. The solubility product is the product of the concentrations of ions raised to an exponent equal to the co-efficient of the balanced equation. Ksp is the measure of dissociation of sparingly soluble salt. It is the product of molar solubilities of two ions at equilibrium stage. Factors affecting Ksp: Ksp is temperature (1) Temperature dependent value is usually very small at room temperature. (ii) Common ion effect: The solubility of slightly soluble salts decreases due to common ions. SLO BASED SHORT QUESTION ANSWERS Bronsted-Lowry concept are

Q.8

A buffer solution has a pH of 5.0. It is made from a weak acid HA with a pKa of 4.8. What is the ratio of the concentration of the conjugate base [A] to the concentration of the weak acid [HA] in this buffer?

Answer

Given:
pH = 5.0

pKa = 4.8

To find: Ratio [A-] / [HA]



Solution:

According to the Henderson-Hasselbalch equation:


pH = pKa + log([A-] / [HA])


Substitute the values:
5.0 = 4.8 + log([A-] / [HA])

5.0 - 4.8 = log([A-] / [HA])

0.2 = log([A-] / [HA])


Take antilog on both sides:
[A-] / [HA] = 10^0.2 ≈ 1.58


So, the ratio of the concentration of conjugate base to weak acid is 1.58 : 1.

Q.9
Calculate the solubility of sparingly soluble salt lead (II) iodide (PbI2) in water. It has Ksp = 1.4 × 10-8.

Answer

Ksp of PbI2 = 1.4 × 10-8

Solubility (S) = ?


Solution:
Consider the dissociation of PbI2 in water:
PbI2(s) <=> Pb2+(aq) + 2I-(aq)

Equilibrium: S 2S



Ksp = [Pb2+][I-]^2

1.4 × 10-8 = (S)(2S)^2 = 4S^3

S^3 = (1.4 × 10-8) / 4 = 3.5 × 10-9


Taking cube root on both sides:
S = (3.5 × 10-9)^(1/3) ≈ 1.52 × 10-3 mol dm-3


So, the solubility of PbI2 in water is 1.52 × 10-3 mol dm-3.

Q.10

Prove by equations what happens when NaCrOs is added to saturated solution of PbCrO4. When the solution of NazCrO4 and

Answer

PbCr04 are mixed, the concentration of is increased according to Le- CrO 3(ag) Chatelier's principle. The solubility of PbCrO4 decreases as it precipitate out. This is due to common ion effect of CrO 4(ag) ion. Na, CrO 4(ag) 2Na* + CO*-

Illustration (added) - Le Chatelier's Equilibrium Balance Shift R P Stress Applied → Shift Counteracts

Q.10

The molar solubility of silver chromate (AgzCrO4) in pure water at 298 K is 6.5 × 10-5 moldm3. Calculate the Ksp of silver chromate at this temperature.

Answer

Solubility of Ag2CrO4 = S = 6.5 × 10-5 mol dm-3

Ksp = ?


Solution:
Consider the dissociation of Ag2CrO4 in water:
Ag2CrO4(s) <=> 2Ag+(aq) + CrO4^2-(aq)

Equilibrium: 2S S



Ksp = [Ag+]^2 [CrO4^2-] = (2S)^2 * S = 4S^3

Ksp = 4 * (6.5 × 10-5)^3

Ksp = 4 * (2.746 × 10-13)

Ksp = 1.1 × 10-12


So, the Ksp of silver chromate is 1.1 × 10-12.

Q.11

According to the Lewis acid-base concept, boron trifluoride (BF3) can act as an acid. Is this statement correct?

Answer

Yes, according to the Lewis acid-base concept, Boron trifluoride (BF3) can act as an acid. Because in BF3, central atom 'B' has incomplete octet or deficit of 2 electrons and it has ability to accept an electron pair

Q.12

If the concentration of hydrogen ions in a solution is 1 × 10-5 M, what is the pH of the solution?

Answer

Given:
[H+] = 1 × 10-5 M

pH = ?


Solution:
pH = -log[H+]

pH = -log(1 × 10-5)

pH = 5

Q.13

What are the microscopic characteristics of acids and bases?

Answer

Acids:
•They have sour taste.

•They change blue litmus to red.

•They react with bases to neutralize them.


Bases:
•They have bitter taste.

•They change red litmus to blue.

•They feel slippery/soapy to the touch.

Q.14

What is the Bronsted-Lowry concept of acids and bases?

Answer

In Bronsted-Lowry theory, an acid is a proton (HT) donor, and a base is a proton acceptor This concept is broader than Arrhenius' Example: NH3 + H2O → NH4* + OH-

Q.15

What is a conjugate acid-base pair?

Answer

A conjugate acid-base pair ditters by one proton. When an acid donates a proton, it forms its conjugate base, and vice versa. Example: HCl / Cl and NH4* / NH3 are conjugate pairs. 01. What i meant by amphoteric substances? Ans. Amphoteric substances can act as both acids and bases depending on the environment. Example: Water (H2O) can accept or donate a proton.

Q.17

What is meant by monoprotic and polyprotic acids? Monoprotic acids donate one proton

Answer

per molecule (e.g., Hcl), while polyprotic acids donate more than one (e.g., H2SO4, H3PO4).

Q.18

What happens when Hcl reacts with NH?

Answer

HCl donates Ht to NH3: Hcl + NH3 → NH4Cl It's a classic Bronsted acid-base reaction.

Q.19

What is meant by conjugate acid and conjugate base?

Answer

When a base accepts a proton, it becomes a conjugate acid. When an acid donates a proton, it forms a conjugate base. Example: NH3 → NH4; HгCO3 → HСO3 Why are acids sour and bases

Q.20

bitter?

Answer

Due to their chemical nature: acids release Ht which activate sour taste receptors, while bases release OH which give a slippery, bitter feel

Lewis concept

Q.21

Define Lewis acid and Lewis base.

Answer

A Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor Example: NH3 donates an electron pair to BF3, which accepts it. ionic product of water

Q.22

What happens to water's ionization with temperature?

Answer

Ionization of water increases with temperature, so Kw and [H+], [OH] increase, though pH may decrease slightly. Define ionization constant of water

Ionic Product of Water

Q.23

(Kw).

Answer

at 25°C. It reflects self-ionization of water. pH and pOH is pH and how is it

Q.24

What calculated?

Answer

pH is a measure of hydrogen ion concentration. It is calculated as:

Q.25

What is the pH of a neutral solution at 25°C?

Answer

A neutral solution has [H+] = 1x10-7 M, so pH = 7. Water is neutral at 25°C

Q.26

What is pOH and how is it related to pH?

Answer

pOH = -log[OH]. It is related to pH by the equation: pH + pOH. = 14 at 25°C.

Q.27

What is the pH of a strong acid like 0.01 M Hcl?

Answer

Since HCl is a strong acid, it ionizes completely: [Ht] = 0.01 M pH = -log(0.01) =2 ionization Constant of Acids (ka)

Q.28

How is strength of acid measured?

Answer

By acid dissociation constant Ka. Stronger base has higher Ka and produces more Ht ions. The Ka is the ratio of concentrațion of acid ionized to the concentration of acid initially added.

Q.29

How is pKa related to acid strength?

Answer

pKa = -log Ka Lower pKa means stronger acid. It's a logarithmic measure of acid strength

Common ion effect

Q.30

What is common ion effect? Give example.

Answer

Common Ion Effect Definition: The suppression of ionization of a weak electrolyte by adding a common ion to it is called common ion effect. Purification of sodium chloride: NaCl is purified by passing hydrogen chloride gas through brine (saturated solution of Nacl). Sodium chloride is fully ionized in the solution. Equilibrium constant expression for this process can be written as follows: NaC1(6) Na(ag) + Claq)

Buffer solution

Q.31

What is meant by a buffer solution?

Answer

A buffer resists pH changes when small amounts of acid or base are added. It usually consists of a weak acid and its salt or weak base and its salt.

Q.32

Give an example of an acidic buffer.

Answer

A mixture of CHCOOH and CH COONa is an acidic buffer that maintains pH around 4.75.

Q.33

What are the applications of buffers?

Answer

Buffers are used in blood, and fermentation, pharmaceuticals, laboratory titrations to maintain stable pH

Solubility product

Q.34

What is solubility product? Give example.

Answer

Solubility Product Definition: The solubility product is the product of the concentrations of ions raised to an exponent equal to the co-efficient of the balanced equation. Example: The Ksp of PbSO4 is 1.6 × 10-16

Salt hydrolysis

Q.35

Why NaCl cannot hydrolyzed water? In salts of strong acids and strong

Answer

bases like sodium chloride (NaCl). The conjugate base of a strong acid (Cl from HCl) is very weak and does not significantly react with water. The conjugate acid of a strong base (Nat from not NaOH) is also very weak and does significantly react with water. NaCl (s) = NO → Na"., + CC Nat is the conjugate acid of NaOH (a strong base) and does not affect the pH Cl is the conjugate base of Hcl (a strong acid and does not affect the pH.

Salt Hydrolysis

Q.36

What is salt hydrolysis?

Answer

Salt hydrolysis Definition: The breakdown of a salt (made up of weak acid or base) by reacting with water, resulting in the bond breaking in that salt which changes the pH of water is called hydrolysis. Acid-base indicators

Q.37

What is meant by titration?

Answer

Titration is a technique used to determine the concentration of an unknown acid or base using a solution of known concentration.

Q.38

What is equivalence point in titration?

Answer

It is the point where moles of acid equal moles of base in titration. The pH at this point depends on the nature of acid/base.

Q.39

What is the role of indicators in titration?

Answer

Indicators change color near the equivalence point to help detect end of titration. Example: Phenolphthalein turns pink in base. DESCRIPTIVE QUESTIONS

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