Question 1

What is the pH of 1 molar HCl solution? (assume complete dissociation)

Accepted Answer

HCl is a strong acid and dissociates completely in water. HCl H^+ + Cl^- For a 1 M solution of HCl, the concentration of hydrogen ions is: [H^+] = 1 M pH = - _ 10 [H^+] pH = - _ 10 (1) = 0 Answer: Zero

Question 2

Calculate the pH of buffer solution containing 0.1M CH _3 COOH and 0.1 M CH _3 COONa [K_a = 1.8 10^ -5 ]

Accepted Answer

Using the Henderson-Hasselbalch equation for an acidic buffer: pH = p K_a + ( [ Salt ] [ Acid ] ) Given [ Salt ] = 0.1 M and [ Acid ] = 0.1 M pH = - (1.8 10^ -5 ) + ( 0.1 0.1 ) pH = 5 - (1.8) + (1) pH = 5 - 0.255 + 0 = 4.745 Answer: 4.745

Question 3

There are two different solutions, A and B. The pH of solution A is 4. Find the pH of solution B having [ H ^+] three times that of solution A.

Accepted Answer

The pH of solution A is given as 4. [ H ^+]_A = 10^ -4 M For solution B, the hydrogen ion concentration is three times that of solution A. [ H ^+]_B = 3 [ H ^+]_A = 3 10^ -4 M pH of solution B = - [ H ^+]_B pH = - (3 10^ -4 ) pH = 4 - 3 Using 3 = 0.4771: pH = 4 - 0.4771 = 3.5229 Answer: 3.5229

Question 4

A weak monobasic acid is 4\% dissociated in 0.05 M solution. What is percent dissociation in 0.1 M solution ?

Accepted Answer

For a weak monobasic acid, the dissociation constant K_a is given by Ostwald's dilution law: K_a = C ^2 where C is the concentration and is the degree of dissociation. Given C_1 = 0.05 M and _1 = 4\% = 0.04: K_a = 0.05 (0.04)^2 = 0.05 0.0016 = 8 10^ -5 For the new concentration C_2 = 0.1 M, let the degree of dissociation be _2: K_a = C_2 _2^2 8 10^ -5 = 0.1 _2^2 _2^2 = 8 10^ -5 0.1 = 8 10^ -4 _2 = 8 10^ -4 = 2.828 10^ -2 The percent dissociation in 0.1 M solution is 2.828\%. Answer: 2.828\%

Question 5

Calculate the pOH of 0.01 M monobasic acid that is completely dissociated, at 298 K.

Accepted Answer

For a completely dissociated monobasic acid, the concentration of hydrogen ions is equal to the concentration of the acid. [H^+] = 0.01 M = 10^ -2 M The pH of the solution is calculated as: pH = - [H^+] = - (10^ -2 ) = 2 At 298 K, the relationship between pH and pOH is: pH + pOH = 14 Substituting the value of pH: pOH = 14 - 2 = 12 Answer: 12

Question 6

What is the normal pH range of human blood?

Accepted Answer

The normal pH of human blood is slightly basic, typically maintained in a narrow range between 7.35 and 7.45. Among the given options, the range 7.36 to 7.42 correctly represents this physiological pH range. This pH is tightly regulated by various buffer systems in the body, primarily the bicarbonate buffer system. Answer: 7.36 to 7.42

Question 7

Which of the following equimolar solutions is the best conductor of electricity at same temperature ?

Accepted Answer

Electrical conductivity of an aqueous solution depends on the number of ions present and their ionic mobility. HCl is a strong electrolyte and dissociates completely in water to give H ^+ and Cl ^- ions. CH _3 COOH and NH _4 OH are weak electrolytes and dissociate only partially, producing fewer ions. C _6 H _ 12 O _6 (glucose) is a non-electrolyte and does not dissociate into ions. Furthermore, the H ^+ ion has exceptionally high ionic mobility in water. Thus, the equimolar solution of HCl has

Question 8

Calculate the molar concentration of weak monobasic acid if K_a = 1.8 10^ -5 and degree of dissociation is 0.01 ?

Accepted Answer

For a weak monobasic acid, the dissociation constant is given by: K_a = C ^2 1- Since = 0.01, which is very small compared to 1, we can approximate 1- 1. K_a = C ^2 Rearranging for C: C = K_a ^2 Substituting the given values: C = 1.8 10^ -5 (0.01)^2 C = 1.8 10^ -5 10^ -4 C = 1.8 10^ -1 M Answer: 1.8 10^ -1

Question 9

If [ H _3 O ^+] of a solution is 1 10^ -4 . What is the value of pOH at 298\, K ?

Accepted Answer

Given [ H _3 O ^+] = 1 10^ -4 M. The pH of the solution is given by: pH = - [ H _3 O ^+] pH = - (1 10^ -4 ) = 4 At 298\, K , the relationship between pH and pOH is: pH + pOH = 14 Substituting the value of pH: 4 + pOH = 14 pOH = 10 Answer: 10

Question 10

Calculate the [ H _3 O ^+] if pH of the solution is 4.25

Accepted Answer

pH = - [ H _3 O ^+] [ H _3 O ^+] = 10^ - pH [ H _3 O ^+] = 10^ -4.25 [ H _3 O ^+] = 10^ 0.75 10^ -5 [ H _3 O ^+] = 5.623 10^ -5 M Answer: 5.623 10^ -5 M

Question 11

What is the pH of solution containing 5 10^ -4 \ M \ H ^+ ion?

Accepted Answer

The pH of a solution is given by the negative logarithm of the hydrogen ion concentration: pH = - [ H ^+] Given [ H ^+] = 5 10^ -4 \ M pH = - (5 10^ -4 ) pH = -( 5 + 10^ -4 ) pH = -( 5 - 4) pH = 4 - 5 Using the value 5 0.699: pH = 4 - 0.699 = 3.301 Answer: 3.301

Question 12

A weak base is 1.3\% dissociated in its aqueous solution. If K_b for weak base is 1.69 10^ -5 at 298 K. Find the concentration of aqueous solution of weak base.

Accepted Answer

For a weak base, the dissociation constant K_b is given by Ostwald's dilution law: K_b = C ^2 1 - Since the degree of dissociation = 1.3\% = 0.013 is very small ( 1), we can approximate 1 - 1. The expression simplifies to: K_b = C ^2 Rearranging for concentration C: C = K_b ^2 Substituting the given values K_b = 1.69 10^ -5 and = 0.013: C = 1.69 10^ -5 (0.013)^2 C = 1.69 10^ -5 1.69 10^ -4 C = 10^ -1 = 0.1 M Answer: 0.1 M

Question 13

Which of the following aqueous salt solutions has the least pH value ?

Accepted Answer

Na _2 CO _3 is a salt of a strong base ( NaOH ) and a weak acid ( H _2 CO _3), so its aqueous solution is basic with pH > 7. CuCl _2 is a salt of a weak base ( Cu(OH) _2) and a strong acid ( HCl ), so its aqueous solution undergoes cationic hydrolysis to give an acidic solution with pH CH _3 COONa is a salt of a strong base ( NaOH ) and a weak acid ( CH _3 COOH ), so its aqueous solution is basic with pH > 7. NaNO _3 is a salt of a strong base ( NaOH ) and a strong acid ( HNO _3), so its aqueous

Question 14

An organic monobasic acid has dissociation constant 1.96 10^ -8 . What is its percentage dissociation in 0.01 M solution ?

Accepted Answer

For a weak monobasic acid, the dissociation constant is given by K_a = C ^2, where is the degree of dissociation and C is the concentration. = K_a C Substituting the given values K_a = 1.96 10^ -8 and C = 0.01 M: = 1.96 10^ -8 0.01 = 1.96 10^ -6 = 1.4 10^ -3 Percentage dissociation = 100 Percentage dissociation = 1.4 10^ -3 100 = 0.14\,\% Answer: 0.14\,\%

Question 15

The solubility of salt AX _2 is 6 10^ -12 \ mol dm ^ -3 . What is solubility product of the salt ?

Accepted Answer

The dissociation of the salt AX _2 is given by: AX _2 A ^ 2+ + 2 X ^- Let the solubility of the salt be s. [ A ^ 2+ ] = s [ X ^-] = 2s The solubility product K_ sp is: K_ sp = [ A ^ 2+ ][ X ^-]^2 = (s)(2s)^2 = 4s^3 Given s = 6 10^ -12 \ mol dm ^ -3 , substituting this value: K_ sp = 4 (6 10^ -12 )^3 K_ sp = 4 216 10^ -36 K_ sp = 864 10^ -36 = 8.64 10^ -34 Answer: 8.64 10^ -34

Question 16

Identify the conjugate bases of H _3 PO _3 and H _2 SO _4 respectively from following.

Accepted Answer

The conjugate base of a Bronsted acid is formed by the removal of one proton (H^+) from it. For H_3PO_3, the removal of one H^+ yields its conjugate base: H_3PO_3 H^+ + H_2PO_3^- For H_2SO_4, the removal of one H^+ yields its conjugate base: H_2SO_4 H^+ + HSO_4^- Therefore, the conjugate bases of H_3PO_3 and H_2SO_4 are H_2PO_3^- and HSO_4^- respectively.

Question 17

What is the molarity of H _2 SO _4 solution having pH = 4 ?

Accepted Answer

Given pH = 4 [ H ^+] = 10^ -4 \ M H _2 SO _4 is a strong diprotic acid and dissociates completely in dilute solutions: H _2 SO _4 2 H ^+ + SO _4^ 2- Therefore, [ H ^+] = 2 [ H _2 SO _4] [ H _2 SO _4] = [ H ^+] 2 = 10^ -4 2 = 5 10^ -5 \ M Answer: 5 10^ -5 \ M

Question 18

The solubility of a sparingly soluble a salt AB _2 is 1 10^ -6 \ mol/dm ^3. Calculate its solubility product?

Accepted Answer

The dissociation of the salt AB_2 is given by: AB_2 A^ 2+ + 2B^- Let the solubility of the salt be S. The equilibrium concentrations are: [A^ 2+ ] = S [B^-] = 2S The solubility product K_ sp is: K_ sp = [A^ 2+ ][B^-]^2 K_ sp = (S)(2S)^2 = 4S^3 Given S = 1 10^ -6 \ mol/dm ^3, substituting this value: K_ sp = 4(1 10^ -6 )^3 K_ sp = 4 10^ -18 Answer: 4 10^ -18

Question 19

Which of the following is Lewis acid?

Accepted Answer

A Lewis acid is a chemical species that can accept a pair of electrons. In BeCl _2, the central beryllium atom forms two single covalent bonds with two chlorine atoms. The valence shell of beryllium in this molecule contains only 4 electrons, which means its octet is incomplete. Due to this electron deficiency, BeCl _2 has a strong tendency to accept electron pairs from other species (Lewis bases) to complete its octet. For example, in the solid state, it forms a polymeric chain structure throug

Question 20

Which of the following is correct relationship between solubility and solubility product for silver oxalate ?

Accepted Answer

The chemical formula of silver oxalate is Ag_2C_2O_4. The dissociation reaction is: Ag_2C_2O_4(s) 2Ag^+(aq) + C_2O_4^ 2- (aq) Let the solubility of silver oxalate be S. Then, [Ag^+] = 2S and [C_2O_4^ 2- ] = S. The solubility product K_ sp is given by: K_ sp = [Ag^+]^2 [C_2O_4^ 2- ] Substituting the concentrations: K_ sp = (2S)^2 (S) = 4S^3 S^3 = K_ sp 4 S = [3] K_ sp 4 Answer: S = [3] K_ sp 4

Ionic Equilibrium — Previous Year Questions & Solutions