Question 1

Given is a concentrated solution of a weak electrolyte A_xB_y of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :

Accepted Answer

The dissociation reaction of the weak electrolyte A_xB_y is: A_xB_y xA^ y+ + yB^ x- Initial concentrations: c 0 0 Equilibrium concentrations: c(1- ) cx cy The dissociation constant K is given by: K = [A^ y+ ]^x [B^ x- ]^y [A_xB_y] K = (cx )^x (cy )^y c(1- ) K = c^ x+y x^x y^y ^ x+y c(1- ) K = c^ x+y-1 x^x y^y ^ x+y 1- For a weak electrolyte, the degree of dissociation is very small, so 1- 1. K = c^ x+y-1 x^x y^y ^ x+y Solving for : ^ x+y = K c^ x+y-1 x^x y^y = ( K c^ x+y-1 x^x y^y )^ 1 x+y Answe

Question 2

M_3 A_2 is a sparingly soluble salt of molar mass y g mol^ -1 and solubility x g L^ -1 . The ratio of the molar concentration of the anion (A^ 3- ) to the solubility product of the salt is

Accepted Answer

The molar solubility S of the salt is given by S = x y mol L^ -1 . The dissociation of the sparingly soluble salt M_3A_2 is: M_3A_2 3M^ 2+ + 2A^ 3- The molar concentration of the anion A^ 3- is: [A^ 3- ] = 2S The solubility product K_ sp of the salt is: K_ sp = [M^ 2+ ]^3 [A^ 3- ]^2 = (3S)^3 (2S)^2 = 27S^3 4S^2 = 108S^5 The ratio of the molar concentration of the anion to the solubility product is: [A^ 3- ] K_ sp = 2S 108S^5 = 1 54S^4 Substituting S = x y into the expression, we get: Ratio = 1 5

Question 3

Arrange the following resultant mixtures in increasing order of their pH values A. 10 mL 0.2 M Ca(OH)_2 + 25 mL 0.1 M HCl B. 10 mL 0.01 M H_2SO_4 + 10 mL 0.01 M Ca(OH)_2 C. 10 mL 0.1 M H_2SO_4 + 10 mL

Accepted Answer

For mixture A: Millimoles of OH^ - = 10 0.2 2 = 4 Millimoles of H^ + = 25 0.1 1 = 2.5 Remaining OH^ - = 4 - 2.5 = 1.5 mmol The solution is basic, so pH > 7. For mixture B: Millimoles of H^ + = 10 0.01 2 = 0.2 Millimoles of OH^ - = 10 0.01 2 = 0.2 The solution is completely neutralized, so pH = 7. For mixture C: Millimoles of H^ + = 10 0.1 2 = 2 Millimoles of OH^ - = 10 0.1 1 = 1 Remaining H^ + = 2 - 1 = 1 mmol The solution is acidic, so pH Therefore, the increasing order of pH values is C Answer

Question 4

20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What

Accepted Answer

For complete neutralization of 20 mL of acetic acid, the millimoles of NaOH required is: 28.4 0.1 = 2.84 mmol Thus, 20 mL of the acetic acid solution contains 2.84 mmol of CH_3COOH. For the preparation of solution (X), the millimoles of NaOH added is: 14.2 0.1 = 1.42 mmol The reaction between acetic acid and sodium hydroxide is: CH_3COOH + NaOH CH_3COONa + H_2O Millimoles of CH_3COOH remaining after the reaction = 2.84 - 1.42 = 1.42 mmol Millimoles of CH_3COONa formed = 1.42 mmol Since the solut

Question 5

The pH of a solution obtained by mixing 5 mL of 0.1 M NH_4OH solution with 250 mL of 0.1 M NH_4Cl solution is _____ 10^ -2 . (Nearest integer) Given: pK_b(NH_4OH) = 4.74 2 = 0.30 3 = 0.48 5 = 0.70

Accepted Answer

The given mixture forms a basic buffer solution. Millimoles of base NH_4OH = 5 0.1 = 0.5 mmol Millimoles of salt NH_4Cl = 250 0.1 = 25 mmol Using the Henderson-Hasselbalch equation for a basic buffer: pOH = pK_b + ( [ Salt ] [ Base ] ) pOH = 4.74 + ( 25 0.5 ) pOH = 4.74 + (50) We know that (50) = (5 10) = 5 + 10 = 0.70 + 1 = 1.70 pOH = 4.74 + 1.70 = 6.44 The pH of the solution is given by: pH = 14 - pOH pH = 14 - 6.44 = 7.56 pH = 756 10^ -2 Answer: 756

Question 6

The first and second ionization constants of a weak dibasic acid H_2 A are 8.1 10^ -8 and 1.0 10^ -13 respectively. 0.1 mol of H_2 A was dissolved in 1 L of 0.1 M HCl solution. The concentration of HA

Accepted Answer

Given: Weak dibasic acid H_2A with K_ a1 = 8.1 10^ -8 and K_ a2 = 1.0 10^ -13 Initial concentration: [H_2A]_0 = 0.1 M The solution also contains 0.1 M HCl , which dissociates completely to give [H^+]_0 = 0.1 M . The main source of HA^- is the first dissociation: H_2A H^+ + HA^- Let x be the amount of H_2A that dissociates. Then at equilibrium: [H_2A] = 0.1 - x [H^+] = 0.1 + x [HA^-] = x Applying the equilibrium expression: K_ a1 = [H^+][HA^-] [H_2A] 8.1 10^ -8 = (0.1 + x)(x) 0.1 - x Since K_ a1

Question 7

At 25°C, 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is

Accepted Answer

For part (a), at the start of the titration, the solution contains only the weak acid HX. The concentration of HX is C = 0.2 M. Using the formula for the hydrogen ion concentration of a weak acid with 1: [H^+] = K_a C = 5 10^ -4 0.2 = 10^ -4 = 10^ -2 M pH = - (10^ -2 ) = 2.0 For part (b), when 10 mL of 0.2 M NaOH is added to 20 mL of 0.2 M HX: Initial millimoles of HX = 20 0.2 = 4 mmol Millimoles of NaOH added = 10 0.2 = 2 mmol The added NaOH neutralizes half of the HX to form NaX, creating an a

Question 8

The solubility product constants of Ag_2CrO_4 and AgBr are 32x and 4y respectively at 298 K. The value of ( molarity of Ag_2CrO_4 molarity of AgBr ) can be expressed as :

Accepted Answer

Let the solubility of Ag_2CrO_4 be S_1. For Ag_2CrO_4 2Ag^+ + CrO_4^ 2- K_ sp = (2S_1)^2(S_1) = 4S_1^3 4S_1^3 = 32x S_1^3 = 8x S_1 = 2 [3] x Let the solubility of AgBr be S_2. For AgBr Ag^+ + Br^- K_ sp = S_2^2 S_2^2 = 4y S_2 = 2 y The ratio of their molarities is: S_1 S_2 = 2 [3] x 2 y = [3] x y Answer: [3] x y

Question 9

Consider a weak base ' B ' of pK _ b =5.699. ' x ' mL of 0.02 M HCl and ' y ' mL of 0.02 M weak base ' B ' are mixed to make 100 mL of a buffer of pH 9 at 25^ C . The values of ' x ' and ' y ' respect

Accepted Answer

pOH = pK_b + [ Salt Base ] 5 = 5.699 + [ Salt Base ] x y - x = 1 5 6x = y 7x = 100 x = 100 7 ml y = 600 7 ml

Question 10

Consider the dissociation equilibrium of the following weak acid HA H ^ + ( aq )+ A ^ - ( aq ) If the pKa of the acid is 4, then the pH of 10 mMHA solution is \_\_\_\_.(Nearest integer) [Given: The de

Accepted Answer

For the weak acid dissociation HA H^+ + A^- with pKa = 4 We have Ka = 10⁻⁴. For a 10 mM (0.01 M) solution where the degree of dissociation is negligible with respect to unity, the concentration of HA remains approximately 0.01 M. The equilibrium expression gives Ka = [H^+]^2 [HA] Substituting: 10^ -4 = [H^+]^2 0.01 , which yields [H^+]^2 = 10^ -6 , so [H^+] = 10^ -3 M. Therefore, pH = - (10^ -3 ) = 3.

Question 11

Consider two Group IV metal ions X ^ 2+ and Y ^ 2+ . A solution containing 0.01 M X ^ 2+ and 0.01 M Y ^ 2+ is saturated with H _ 2 ~S . The pH at which the metal sulphide YS will form as a precipitate

Accepted Answer

XS(s) X^ +2 (aq.) + S^ 2- (aq.) For precipitation of XS(s) [X^ +2 ][S^ 2- ] K_ sp (XS) [S^ 2- ] 1 10^ -22 0.01 = 10^ -20 YS(s) Y^ +2 (aq) + S^ 2- (aq) For precipitation of YS(s) [Y^ +2 ][S^ 2- ] K_ sp (YS) [S^ 2- ] 4 10^ -16 10^ -2 = 4 10^ -14 Now, H_2S(aq) 2H^+(aq) + S^ 2- (aq) [S^ 2- ][H^+]^2 H_2S = K_ a_1 K_ a_2 = 1 10^ -21 [S^ 2- ] = 1 10^ -21 [H_2S] [H^+]^2 4 10^ -14 [H^+]^2 1 4 10^ -7 10^ -1 [H^+] 1 2 10^ -4 pH 4.3 Nearest integer pH = 4.

Question 12

x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M Ba ( OH )_ 2 solution is used when the HCl solution was titrated against it. The numerical value of x is \_\_\_\_ 10^ -1 . (Near

Accepted Answer

The reaction between HCl and Ba(OH)₂ is: 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O Moles of Ba(OH)₂ used: n = 0.1 M × 0.025 L = 0.0025 mol From stoichiometry: moles of HCl = 2 × 0.0025 = 0.005 mol Mass of HCl = 0.005 mol × 36.5 g/mol = 0.1825 g = 182.5 mg Expressing as x = ___ × 10⁻¹: 182.5 = 1825 × 10⁻¹ The numerical value is 1825 (or 1.825 × 10² mg when expressed with standard scientific notation)

Question 13

Which of the following mixture gives a buffer solution with pH =9.25 ? Given : pK _ b ( NH _ 4 OH )=4.75

Accepted Answer

pOH = pK_b + Salt Base 4.75 = 4.75 + Salt Base Milimoles of [Salt] = milimoles of [Base] NH_4OH + HCl NH_4Cl + H_2O 0.2M, 0.5L 0.1M, 0.5L 100 mmole 50 mmole 50 mmole - 50 mmole Milimoles of NH_4OH = milimoles of NH_4Cl

Question 14

The first and second ionization constants of H _ 2 X are 2.5 10^ -8 and 1.0 10^ -13 respectively. The concentration of X ^ 2- in 0.1 M H _ 2 X solution is \_\_\_\_ 10^ -15 M . (Nearest Integer)

Accepted Answer

For a weak diprotic acid H_2X, the ionization steps are: H_2X H^+ + HX^- with K_ a1 = 2.5 10^ -8 HX^- H^+ + X^ 2- with K_ a2 = 1.0 10^ -13 Since K_ a1 K_ a2 , the concentration of H^+ is primarily determined by the first ionization. [H^+] K_ a1 C = 2.5 10^ -8 0.1 = 2.5 10^ -9 = 5 10^ -5 M Also, from the first ionization, [HX^-] [H^+] = 5 10^ -5 M For the second ionization: K_ a2 = [H^+][X^ 2- ] [HX^-] Substituting the values: 1.0 10^ -13 = (5 10^ -5 )[X^ 2- ] 5 10^ -5 Thus, [X^ 2- ] = K_ a2 = 1.

Ionic Equilibrium — Previous Year Questions & Solutions