Ionic Equilibrium-III Exercises

Ionic Equilibrium (Advanced) IONIC EQUILIBRIUM-III  Marked questions are recommended for Revision. PART - I : SUBJECTIVE QUESTIONS Section (A) : Solubility, Solubility product and Simple solubility calculations Commit to memory : Solubility (s; in mol/L) of a sparingly soluble salt AxBy : Ksp = xx.yy.(s)x+y A-1. Although it is a violent poison if swallowed, Mercury(II) cyanide Hg(CN) 2, has been used as a typical skin antiseptic. (Ksp of Hg(CN)2 = 1.35 × 10–23, Atomic mass of Hg = 200 u). Assume no ion to be undergoing hydrolysis. (a) What is the molar solubility of this salt in pure water ? (b) How many milligrams of Hg(CN)2 dissolve per litre of pure water ? (c) How many milliliters of water are required to dissolve 1.134 g of the salt ? A-2. How many milligrams of gold(III) ion are there per litre of solution in a saturated aqueous solution of gold (III) iodide ? AuI3(s) Au3+(aq) + 3I–(aq). Ksp of AuI3 = 2.7 × 10–47, atomic mass of Au = 197 u. 3+ Assume no hydrolysis of Au ion. Section (B) : Condition for precipitation, Common ion effect on solubility Commit to memory : Condition for precipitation of AxBy : Ionic product or KIP > Ksp ; [Ay+]x [Bx–]y > Ksp of AxBy Common ion effect on solubility : Solubility (s') of AxBy (sparingly soluble ; Ksp) in a solution of 'c' M ApDq/ErBp (both readily soluble) : (pc)x(ys')y = Ksp ; s' = ..... B-1. What can be the maximum concentration of equimolar solutions of ferrous sulphate (FeSO 4) and sodium sulphide (Na2S), so that when mixed in equal volumes, there is no precipitation of iron sulphide (FeS) ? (For iron sulphide, Ksp = 6.25 × 10–18) Assume no ion to be undergoing hydrolysis. B-2. Calculate the solubility of silver phosphate (Ag3PO4) in mg/mL (a) in pure water and (b) in a solution that is 0.03 M in AgNO3. [Ksp(Ag3PO4) = 2.7 × 10–23]. Assume no ion to be undergoing hydrolysis. Atomic mass of Ag = 108 u. Section (C) : Simultaneous Solubility, Complex Formation Commit to memory : Simultaneous solubility (s1 & s2) of AxBy & AxCy (both sparingly soluble) : y s  (ys1 )y [B x – ]y = = =  1 x– y y (ys2 ) K sp (A x Cy ) [C ]  s2  x y Further, (xs1 + xs2) (ys1) = Ksp(AxBy) & (xs2 + xs1)x (ys2)y = Ksp(AxCy). Put s1 in terms of s2 from eq.1 in eq.2 or eq.3 to obtain s2 & then obtain s1. Complex formation : Upon mixing Mn+ & L and analyzing the reaction Mn+ + xL [MLx]n+ through chemical equilibrium (assuming Mn+ to be completely consumed), concentrations of L and [MLx]n+ can be obtained. When they are put in expression of Kc, [Mn+]eq can be obtained. K sp (A xBy ) C-1. Calculate the simultaneous solubility of AgCl and AgI. Assume no hydrolysis of Ag + ion. Ksp (AgCl) = 1 × 10–10, Ksp (AgI) = 8.5 × 10–17 C-2. On passing 1 mole NH3 gas through 1 L of 0.2 M CuSO4 solution, determine [Cu2+] in the resulting solution. Given : Kf of [Cu(NH3)4]2+ = 2 × 1013. Assume no hydrolysis of Cu2+ ion. Ionic Equilibrium (Advanced) Section (D) : Selective Precipitation, Solubility exchange Commit to memory : Selective precipitation: On gradually adding ApBq(s ; readily soluble) to a solution containing C 1M Xw– & C2M Yz– [Aq+] req. for AwXq pptn = w K sp (A w Xq ) ( C1 ) q ; [Aq+] req. for AzYq pptn = z K sp (A z Yq ) ( C2 )q Minimum value of [Aq+] req. will tell which salt will precipitate first. Solubility exchange: Upon treating a solution of CpDq / CpDq(s) (readily soluble) with AxBy(s) / saturated solution of AxBy (sparingly soluble) such that some AxBy gets converted to ApDy(sparingly soluble), then : x K sp (A xBy ) [Bx − ]fy = p K sp (ApDy ) [Dp− ]fy D-1. A solution which is 0.1 M in NaI and also 0.1 M in Na2SO4 is treated with solid Pb(NO3)2. Which compound, PbI2 or PbSO4, will precipitate first ? What is the concentration of anion of the least soluble compound when the more soluble one starts precipitating ? Ksp(PbI2) = 9 × 10–9, Ksp(PbSO4) = 1.8 × 10–8. Assume no hydrolysis of Pb2+ ion. D-2. 0.1 mole AgCl(s) is added to 1 litre H2O. Next, crystals of NaBr are added until 75% of the AgCl is converted to AgBr(s), the less soluble silver halide. What is [Br–] in the resulting solution ? Ksp of AgCl is 1.75 × 10–10 and Ksp of AgBr is 5.25 × 10–13. Assume no hydrolysis of Ag+ ion. PART - II : ONLY ONE OPTION CORRECT TYPE Section (A) : Condition for precipitation, Common ion effect on solubility Commit to memory : Condition for precipitation of AxBy : Ionic product or KIP > Ksp ; [Ay+]x [Bx–]y > Ksp of AxBy Common ion effect on solubility : Solubility (s') of AxBy (sparingly soluble; Ksp) in a solution of 'c' M ApDq/ErBp (both readily soluble) : (pc)x(ys')y = Ksp; s' = ..... A-1. What is the minimum pH at which Mg(OH)2 starts precipitating from a solution containing 0.02 M MgCl2? Ksp of Mg(OH)2 = 1.8 × 10–11. (A) 4.52 (B) 9.04 (C) 9.48 (D) 8.48 Section (B) : Simultaneous solubility and Complex formation Commit to memory : Simultaneous solubility (s1 & s2) of AxBy & AxCy (both sparingly soluble) : s  (ys1 )y [B x – ]y = = =  1 x– y y (ys2 ) K sp (A x Cy ) [C ]  s2  K sp (A xBy ) y Further, (xs1 + xs2)x (ys1)y = Ksp(AxBy) & (xs2 + xs1)x (ys2)y = Ksp(AxCy). Put s1 in terms of s2 from eq.1 in eq.2 or eq.3 to obtain s2 & then obtain s1. Complex formation : Upon mixing Mn+ & L and analyzing the reaction Mn+ + xL [MLx]n+ through chemical equilibrium (assuming Mn+ to be completely consumed), concentrations of L and [MLx]n+ can be obtained. When they are put in expression of Kc, [Mn+]eq can be obtained. Ionic Equilibrium (Advanced) B-1. Simultaneous solubility of two sparingly soluble salts A & B having some common ion : (A) is greater than their respective solubilities in separate solutions. (B) lies in between their solubilities in separate solutions. (C) is less than its solubility in separate solutions for more soluble salt, but is greater than its solubility in separate solution for less soluble salt. (D) is less than their respective solubilities in separate solutions. B-2. The solubility product values of AgCl and AgCNS are 1.8 × 10–10 and 1.6 × 10–11 respectively. When a solution is saturated with both solids, calculate the ratio [Cl –]/[CNS–] and also [Ag+] in the solution : Assume no reaction of cation/anion. (A) 1.125, 4 × 10–6 M (B) 11.25, 1.4 × 10–5 M (C) 1.25, 4 ×10–5 M (D) 1.25, 4 × 10–6 M B-3. Which of the following options is true for a solution saturated with AgCI and AgBr, if their solubilities in mole per litre in separate water samples are x and y respectively ? Given : x > y. Assume no hydrolysis of Ag+ ion. (A) [Ag+] < [Br–] + [CI–] (B) x > [CI–] > [Br–] (C) [Br–] > y (D) [Ag+] = x + y B-4. Upon mixing equal volumes of 0.02 M AgNO3 & 0.16 M KCN solutions : (Assume no hydrolysis of any ion). Given : Ag+ (aq.) + 2CN– (aq) [Ag(CN)2]– (aq) ; KC = 1018 B-5. (A) [NO3– ] = 0.01 M ; [K+] = 0.08 M (B) [CN–] = 0.06 M (C) [Ag(CN)2– ] = 0.01 M (D) All of these In the above question (B-4), [Ag+] is about : (A) 2.78 ×10–18 M (C) No free Ag+ ions will be left in the solution (B) 1.67 × 10–19 M (D) Cannot be determined Section (C) : Selective Precipitation, Solubility exchange Commit to memory : Selective Precipitation : On gradually adding ApBq(s ; readily soluble) to a solution containing C1 M Xw– & C2 M Yz– [Aq+] req. for AwXq pptn = w K sp (A w Xq ) ( C1 ) q ; [Aq+] req. for AzYq pptn = z K sp (A z Yq ) ( C2 )q Minimum value of [Aq+] req. will tell which salt will precipitate first. Solubility exchange : Upon treating a solution of CpDq / CpDq(s) (readily soluble) with AxBy(s) / saturated solution of AxBy (sparingly soluble) such that some AxBy gets converted to ApDy(sparingly soluble), then : x C-1. K sp (A xBy ) [Bx − ]fy = p K sp (ApDy ) [Dp− ]fy In a solution containing equal concentrations of X –, Y– & Z–, MCl(s) (a readily soluble salt) is gradually added. If Ksp (MX) > Ksp (MY) > Ksp (MZ), then : (Assume no hydrolysis of any ion). (A) MX will precipitate first. (B) MY will precipitate first. (C) MZ will precipitate first. (D) Nothing can be said about which salt will precipitate first, as numerical data is not given. C-2. In above question, If [X–]i : [Y–]i : [Z–]i = 100 : 10 : 1 & Ksp[MX] : Ksp[MY] : Ksp[MZ] = 10 : 0.1 : 1, order of precipitation will be : (A) MZ, then MX & MY at last (B) MY, then MX & MZ at last (C) MX, then MY & MZ at last (D) MY, then MZ & MX at last Ionic Equilibrium (Advanced)  Marked questions are recommended for Revision. PART - I : ONLY ONE OPTION CORRECT TYPE 1. A solution contains [NaCl] = [Na2CO3] = C M. Upon gradually adding AgNO 3(s) : (Assume no hydrolysis of any ion). (A) Out of AgCl & Ag2CO3, one having lesser Ksp will start precipitating first. (B) Out of AgCl & Ag2CO3, one having greater Ksp will start precipitating first. (C) If, (D) If, K sp (Ag2CO3 ) C K sp (Ag2CO3 ) C   K sp (AgCl) C K sp (AgCl) C then Ag2CO3 will start precipitating first. then AgCl will start precipitating first. 2. For the reaction : [Ag(CN)2]–(aq) Ag+(aq) + 2CN–(aq), KC at 25ºC is 4 × 10–19. Calculate [Ag+] in solution which was originally 0.1 M in KCN and 0.03 M in AgNO3. (A) 1.5 × 10–17 M (B) 7.5 × 10–18 M –19 (C) 4.6875 × 10 M (D) Value cannot be determined due to mathematical complications. 3. A solution of Na2CO3 is added drop by drop to one litre of a solution containing 10 –4 mole of Ba2+ and 10–5 mole of Ag+. If Ksp for BaCO3 is 8.1 × 10–9 and Ksp for Ag2CO3 is 6.9 × 10–12, then which is not true? Assume no hydrolysis of CO32– ion. (A) No precipitate of BaCO3 will appear until [ CO32− ] reaches 8.1 × 10–5 mol per litre. (B) A precipitate of Ag2CO3 will appear when [ CO32− ] reaches 6.9 × 10–5 mol per litre. (C) No precipitate of Ag2CO3 will appear until [ CO32− ] reaches 6.9 × 10–2 mole per litre. (D) BaCO3 will be precipitated first. PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE 1. When pure water is saturated with both CaCO3 and CaC2O4, the concentration of calcium ion in the solution under equilibrium is 8 × 10–5 M. If the ratio of the solubility product of CaCO 3 to that of CaC2O4 is 3, what is the solubility product of CaCO3 in pure water ? Report your answer after multiplying by 10 9 and rounding it off to nearest whole number. 2. On mixing equal volumes of 0.1 M AgNO3 & 0.8 M KCN solutions, what % of original Ag+ is in the form of free Ag+ ions ? Given : Kd of [Ag(CN)2]– = 10–18. Report your answer after multiplying it by 4.5 × 1016. (Assume no hydrolysis of any ion) PART - III : COMPREHENSION Comprehension # 1 0.2 mole of MgCl2(s) are added in 1 litre volume of a solution (S), already containing 0.2 mole of NaOH(s). Now answer the following questions : 1. Calculate pH of obtained solution. Ksp of Mg(OH)2 is 1.6 × 10–12. (A) 8.6 (B) 8.5 (C) 8.65 (D) 8.56 2. What would be the new pH when 0.04 mole of NaOH(s) are added in the solution (S) ? (A) 8.6 (B) 8.5 (C) 8.65 (D) 8.56 3. Find the pH of the solution obtained after adding 0.04 mole of HCl(g) to the solution . (A) 8.6 (B) 8.5 (C) 8.65 (D) 8.56 Ionic Equilibrium (Advanced) Comprehension # 2 Q.4, Q.5 and Q.6 by appropriately matching the information given in the three columns of the following table. Following 3 columns contain informations regarding water solubility of some sparingly soluble salts and its relation with their solubility product constants (Ksp). Take temperature as 25ºC. Column-1 Column-2 Column-3 Solubility increases because of Solubility increases in acidic (I) Solubility > K sp (i) presence of common anion in (P) solution as pH decreases below 3 solution 1/ 4 Solubility decreases because of Solubility increases in basic  K sp    (II) Solubility <  (ii) presence of common cation in (Q) solution as pH increases above   27  solution 11 1/ 4 Solubility decreases because of K  Solubility is unaffected by pH of (III) Solubility =  sp  (iii) presence of common anion in (R) solution  27  solution Solubility increases because of Anionic hydrolysis will make (IV) Solubility = K sp (iv) presence of common cation in (S) solubility in water greater than solution expected 4. 5. 6. For AgCN, the only correct combination is : (A) (IV) (ii) (R) (B) (I) (iii) (S) (C) (I) (i) (P) (D) (I) (iii) (P) For Al(OH)3, the only correct combination is : (A) (II) (iii) (S) (B) (II) (iv) (Q) (C) (III) (i) (P) (D) (II) (ii) (Q) For PbS, the only correct combination is : (A) (I) (iii) (P) (B) (IV) (iii) (S) (C) (I) (ii) (R) (D) (IV) (ii) (P) * Marked questions may have more than one correct option. PART - I : (ADVANCED) / PROBLEMS (PREVIOUS YEARS) 1. For a sparingly soluble salt ApBq, the relationship of its solubility product (Ls) with its solubility (S) is : (A) LS = Sp+q . pp . qq (B) LS = Sp+q . pq . qp (C) LS = Spq . pp . qq (D) LS = Spq . (pq)p+q 2. A solution which is 10–3 M each in Mn2+, Fe2+, Zn2+ and Hg2+ is treated with 10–16 M sulphide ion. If Ksp values of MnS, FeS, ZnS and HgS are 10–15, 10–23, 10–20 and 10–54 respectively, which one will precipitate first ? (A) FeS (B) MgS (C) HgS (D) ZnS 3. Solubility product constant (Ksp) of salts of types MX, MX2 and M3X at temperature T are 4.0 × 10–8, 3.2 × 10–14 and 2.7 × 10–15, respectively. Solubilities (in mol dm –3) of the salts at temperature T are in the order : (A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2 4. In 1 L saturated solution of AgCl [Ksp(AgCl) = 1.6 × 10–10], 0.1 mole of CuCl [Ksp(CuCl) = 1.0 × 10–6] is added. The resultant concentation of Ag+ in the solution is 1.6 × 10–x. The value of "x" is : 5. The solubility of a salt of weak acid (AB) at pH 3 is Y×10–3 mol L−1. The value of Y is ____. (Given that the value of solubility product of AB (Ksp) = 2×10–10 and the value of ionization constant of HB (Ka) = 1×10–8) Ionic Equilibrium (Advanced) PART - II : (MAIN) (PREVIOUS YEARS) (MAIN) 1. ONLINE PROBLEMS Zirconium phosphate [Zr3(PO4)4] dissociates into three zirconium cations of charge +4 and four phosphate anions of charge –3. If molar solubility of ziroconium phosphate is denoted by S and its solubility product by Ksp then which of the following relationship between S and Ksp is correct ? (1) S = {Ksp/(6912)1/7} (3) S = {Ksp/6912}1/7 (2) S = {Ksp/144}1/7 (4) S = {Ksp/6912}7 EXERCISE - 1 PART - I A-1. (a) 1.5 × 10–8 M (b) 3.78 × 10–3 mg B-1. 5 × 10–9 M C-1. 10–5 M AgCl, 8.5 × 10–12 M Ag. D-1. PbSO4, 0.02 M B-2. (c) 3 × 108 mL A-2. (a) 4.19 × 10–4 mg/mL (b) 4.19 × 10–16 mg/mL C-2. D-2. 1.97 × 10–7 mg 6.25 × 10–12 M 2.25 × 10–4 M PART - II A-1. (C) B-1. (D) B-2. (B) B-5. (A) C-1. (C) C-2. (B). B-3. (B) B-4. (D) 4. (C) 5. (D) 4. 7 5. 4.47 EXERCISE - 2 PART - I 1. (C) 2. (B) 3. (B) PART – II 1. 5 (actual answer = 4.8 × 10–9) 1. (A) (C) 6. (A) 2. 3. 50 PART - III 2. (D) EXERCISE - 3 PART - I 1. (A) 2. (C) 3. (D) PART - II (MAIN) ONLINE PROBLEMS 1. (2)