Solved Problems. Objective. K a = the degree of hydrolysis of 0.001M solution of the salt is (a) 10 3 (b) 10 4 (c) 10 5 (d) 10 6.

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1 Solved Problems SM-IE-CH-1 Objective Problem 1: In the hydrolytic equilibrium A HO HA OH K a = the degree of hydrolysis of 0.001M solution of the salt is (a) 10 (b) 10 (c) 10 5 (d) h K K h w C Ka C (a) = Problem : For preparing a buffer solution of ph 6 by mixing sodium acetate and acetic acid, the ratio of concentration of salt and acid (K a = 10 5 ) should be (a) 1:10 (b) 10:1 (c) 100:1 (d) 1:100. [salt] ph = pka + log [acid] 10 = 5 log if [salt]/[acid] = 10:1, then ph = 6 1 (b) Problem : Let the solubiliteis of AgCl in H O, 0.01 M CaCl ; 0.01 M NaCl and 0.05M AgNO be S 1, S, S, S respectively. What is the correct relationship between these quantities. (a) S 1 > S > S > S (b) S 1 > S = S > S (c) S 1 > S > S > S (d) S > S > S > S 1. Solubility of AgCl in water = K sp = s1 In 0.01 M CaCl it is given by Ksp Ksp = s (0.01 +s) s 0.0 In 0.01 M NaCl it is given by Ksp Ksp s (0.01s) s 0.01 In 0.05 M AgNO it is given by Ksp Ksp s (0.6s) s 0.05 The solubilities are derived by neglecting s in comparison to 0.0, 0.01 and (c)

2 SM-IE-CH- Problem : The ph at which Mg(OH) begins to precipitate from a solution containing 0.10M Mg + ions [K p of Mg(OH) = ] is (a) 5 (b) 9 (c) (d) 10. When Mg(OH) starts precipitation, then [Mg + ] [OH ] = Ksp of Mg(OH) [0.1] [OH ] = [OH ] = 10 5 M poh = 5 ph = 1 poh ph = 1 5 = 9 (b) Problem 5: The concentration of hydroxyl ion in solution left after mixing 100 ml of 0.1 M MgCl and 100 ml of 0. M NaOH (K sp of Mg(OH) = ] is (a).8 10 (b).8 10 (c).8 10 (d) MgCl + NaOH Mg(OH) + NaCl mm before mm after reaction thus, 10 m mole of Mg(OH) are formed. The product of [Mg + ] [OH ] is therefore Which is more than Ksp of Mg(OH). Now solubility (S) of Mg(OH) can be derived by Ksp = S K sp S [OH ] = S =.8 10 (c) Problem 6: 1 ml of.1mhcl is added into 99 ml of water. Assume volumes are additive, what is ph of resulting solution. (a) (b) 1 (c) (d) nhcl = MV =.1M 10 L = 10 mol Vfinal = (1+99) ml = 100 ml =.1L [HCl] = n/v = 10 M = [HO + ] ph = log [HO + ] = log 10 = (a)

3 Problem 7: What is ph of water at 100 C. K w at 100 C is (a) 7 (b) 6.15 (c) 6.9 (d) 7.1 SM-IE-CH- Kw = [HO + ] [ OH] In water, [HO + ] = [ OH] Kw = [HO + ] log Kw = log [HO + ] ph = log [HO ] = log Kw = log { } = (b) Problem 8: What is ph of 0.0 M solution of ammonium chloride at 5 C? K b (NH ) = (a) 5.77 (b) 8.5 (c) 7 (d).87 For a salt of weak base and strong acid, at 5 C ph = 7 1 pkb 1 log C = 7 1 ( log ) 1 log 0.0 = 5.77 (a) Problem 9: The ph of 0.1M CH COOH is.87. What is ph of 0.1M NH OH. K a (CH COOH) = and K b (NH OH) = (a) (b).87 (c) 7 (d) 9.5 Since Ka(CHCOOH) = Kb(NH and concentration are equal and so ph (CHCOOH) = poh (NHOH) ph =.87 ph = 1 poh = 1.87 = (a) Problem 10: Liquid ammonia ionises to a slight extent. At - 50 C, its ion product is K NH = [NH ][NH ] = 10. How many amide ions, NH are present per mm of pure liquid ammonia. (a) (b) (c) (d) [NH ] KNH mol 1L 9 n - MV10 10 mol NH 6 L 10 mm Number of NH ions = ions = ion (c)

4 SM-IE-CH- Problem 11: A solution of HCl is diluted so that its ph changes by 0.. How does concentration of H + ion change? (a) 0.5 times of initial value (b) 0. times of initial value (b) 10 times increases (d) None Let H + ion concentration changes by x factor. ph = log [HO + ] ph + ph = log {x(ho + ]} = log x log [HO + ] or ph = log x = 0. x = 0.5 (a) Problem 1: The ph of 0.1M CH COOH is.87. What is ph of 0.1M NH OH. K a (CH COOH) = and K b (NH OH) = (a) (b).87 (b) 7 (d) 9.5. Since Ka(CHCOOH) = Kb(NHOH) and concentration are equal so ph (CHCOOH) = poh (NHOH) ph =.87 ph = 1 poh = 1.87 = (a) Problem 1: Liquid ammonia ionises to a slight extent. At - 50 C, its ion product is K NH = NH = How many amide ions, NH liquid ammonia. (a) (b) (b) (d) NH are present per mm of pure [NH ] K NH 15 mol 1L n MV10 = 10 9 mol NH 6 L 10 mm Number of NH ions = ions = ion (b) Problem 1: To a 50 ml of 0.1 M HCl solution, 10 ml of 0.1 M NaOH is added and the resulting solution is diluted to 100 ml. What is change in ph of the HCl solution? (a).98 (b).98 (b) 0.1M (d) None. Before adding HCl solution ph = 1 [ [HCl] = [HO] + = 10 1 M] nhcl (initially) = MV = 0.1 M 0.05 L = 5 10 mol

5 SM-IE-CH-5 nnaoh added = MV = 0.1 M 0.01 L = 1 10 mol HCl + NaOH NaCl + HO t = mol 1 10 mol 0 10 mol 0 Vfinal = 100 ml = 0.1 L n 10 mol [HCl] = = 10 M V 0.1L ph = log [HO + ] = log = log = 0.01= 0.60 = 1.98 Increase in ph = (1.98 1) = 0.98 (b) Problem 15: What amount of solid sodium acetate be added into 1 litre of the 0.1 M CH COOH solution so that the resulting solution has ph almost equal to pk a (CH COOH) =.7 (a) 1gm (b) 5 gm (b) 10 gm (d) 1.9 gm. Since the resulting solution be acidic buffer, one may use Henderson equation. [CHCOO ] ph = pka log [CH COOH] Let n mol of CHCOONa be added to do so nmol or, ph =.7 + log vl 0.1 mol vl n or, 5 =.7 + log 0.1 n or, log 0.1 = 0.6 n or, = antilog 0.6 = n = mol 0.18 mol Amount of sodium acetate = gm = 1.9 gm (d) Problem 16: To a 100 ml solution of 0.1 M CH COONa and 0.1 M CH COOH, 0. gm of solid NaOH was added. Assuming volume remains constant, calculate the change in ph value? Given that pk a (CH COOH) =.7. (a) 0.15 (b) 0.5 (b) 0.01 (d) Before NaOH addition, ph = pka =.7 [Since [CHCOO ] = [CHCOOH]] The following reaction occurs due to NaOH addition.

6 SM-IE-CH-6 HCCOOH + NaOH HCCOONa + HO t = mol mol mol mol mol ( ) mol mol n H CCOONa = 0.01 mole After reaction, n CH COOH = mol n = ( ) mol = mol H CCOO [CHCOO ] ph = pka + log [CHCOOH] =.7 + log 0.011/ V / V =.7 + log11 = change in ph = log = (d) Problem 17: A weak base (BOH) with K b = 10 5 is titrated with a strong acid, HCl. At / th of the equivalent point, ph of the solution is: (a) 5 + log (b) 5 log (c) log (d) 8.5. Let the initial equivalent of BOH be x BOH + HCl BCl + HO Initial equivalent x /x 0 0 At / th x x eqv. pt. x 0 [salt] x poh = pkb log 5 log [Base] x ph = 1 5 log = 8.5 (d) x x Problem 18: When equal volumes of the following solutions are mixed, precipitation of AgCl (K sp = ) will occur only with (a) 10 M [Ag + ] and 10 M[HCl] (b) 10 5 M[Ag + ] and 10 5 M [Cl ] (c) 10 6 M [Ag + ] and 10 6 M [Cl - ] (d) M [Ag + ] and M [Cl ]. We look for that case where the ionic product exceeds the Ksp 1 5 [Ag ][Cl ] (a)

7 SM-IE-CH-7 Problem 19: The solubility products of Al(OH) and Zn(OH) are and respectively. If NH OH is added to a solution containing Al + and Zn + ions, then substance precipitated first is: (a) Al(OH) (b) Zn(OH) (c) Both together (d) None at all. Solubility of Al(OH) is lesser than Zn(OH). (a) Problem 0: How many gram of CaC O will dissolve in one litre of saturated solution? K sp of CaC O is mol and its molecular weight is 18. (a) 0.006g (b) 0.018g (c) 0.00g (d) 0.060g. CaC O Ca C O sp S S K SS S s K sp = ( ) 1/ = mol litre 1 w = g (a) w Subjective Problem 1: The statement given above which suggests us to calculate from the expression, = K a C and than compare its value with 0.1 looks a little odd. How can we compare the value of ( which is derived after making the assumption) with 0.1 and then claim that it is correct if 0.1? Can you suggest the validity of this statement? C The actual expression for calculating the value of is, Ka =. If we make the 1 assumption that is very small compared to one and ignore it in the denominator, we get, Ka = C. By ignoring the in the denominator, we increase the value of the denominator. This will consequently increase the value of the numerator (since the ratio is a constant) consequently the value of calculated after the assumption will be greater than the actual value of. If this value of is less than or equal to 0.1 then it means that the actual value of will be much less compared to 0.1. Therefore the assumption is valid. If this value of is greater than 0.1, than the actual value of may be greater, equal to or less than 0.1. So here we cannot predict and so we do not make the assumption. Problem : Calculate the ph at the equivalence point of the titration between 0.1M CH COOH ( 5 ml) with 0.05 M NaOH. Ka (CH COOH) =

8 SM-IE-CH-8 We have already seen that even though when CHCOOH is titrated with NaOH the reaction does not go to completion but instead reaches equilibrium. We can assume that the reaction is complete and then salt gets hydrolysed because, this assumption will help us to do the problem easily and it does not effect our answer. [H + KwKa ] = C First of all we would calculate the concentration of the salt, CHCOONa. For reaching equivalence point, N1V1 = NV = 0.05 V V = 50 ml Therefore [CHCOONa] = [H ] = = ph = log = 8.6 Problem : Given the solubility product of Pb (PO ) is 1.5 x Determine the solubility in gms/litre. Solubility product of Pb (PO) = Pb (PO) Pb + + PO If x is the solubility of Pb (PO) Then Ksp = (x) (x) = 108 x Ksp x = 5 = x = moles/lit Molecular mass of Pb(PO) = 811 x = g/lit = g/lit Problem : What is ph of 1M CH COOH solution? To what volume must one litre of this solution be diluted so that the ph of resulting solution will be twice the original value. Given : K a = HCOOH + HO HCOO + HO + t = 0 1M 0 0 -xm xm xm t = teq (1-x)M x x x Ka = 1 x x 1 x = K a =. 10 = [HO + ] ph = log [HO + ] = log {. 10 } = log. =.7

9 SM-IE-CH-9 Now, let 1L of 1M ACOH solution be diluted to VL to double the ph and the conc. of diluted solution be C. HCOOH + HO HCOO + HO + t = 0 C t = teq C New ph = old ph =.7 =.7 ph = log [HO + ] =.7 [HO + ] = [CHCOO ] [HO ] Ka = [CHCOOH] = C = L on dilution C M1V1 = MV 1M 1L = L V V = L Problem 5: Find the concentration of H +, HCO and carbonic acid if the ph of this is.18 Ka 1 (H CO ) = and Ka = ph = log[h + ].18 = log [H + ] [H + ] = HCO H + + HCO [H ][HCO ] Ka = or, = [HCO ] [HCO ] = again, HCO H + + CO [H ][[CO ] Ka = = = [HCO ] CO ] = [ CO, in a 0.01M solution of [HCO ] [CO ] 5 Problem 6: An aqueous solution of metal bromide MBr (0.05M) in saturated with H S. What is the minimum ph at which MS will ppt.? K sp =(MS) = Concentration of standard H S =.1 Ka 1 (H S) = Ka (H S) =

10 SM-IE-CH-10 In saturated solution of MS MS(s) M ++ + S [S 1 KSP 6 10 ] = = [M ].05 The precipitate of MS will form only if [S ] exceeds the concentration of HS H + + HS Ka1 HS H + + S -- Ka HS H + + S K = [H ] [S [ K = [H S] = [H + ] =.109 ph =.96 [H ] Problem 7: How much Ag + would remain in solution after mixing equal volumes of M AgNO and 0.08N HOCN. Given that : Ksp for AgCN = Ka(HOCN) =. 10 HOCN H + + OCN Ag + + OCN AgOCN(s) Due to two fold dilution, Let [Ag + ] = x Amount of [Ag + ] pptted = (.0 x)m Amount of CN ppted = (.0 x)n [HOCN] = x [ OCN] = xm [H + ] = 0. - x [OCN] K Then, a.10 or, [.10 n OCN} = [HOCN [H ].0 x.0 x And [Ag + ][.10 x OCN] =.0 x = Ksp = or, x = [Ag + ] = 5 10 Problem 8: The self ionization constant for pure formic acid, K = [HCOO H ][HCOO ] has been estimated as 10 6 at room temperature. What percentage of formic acid molecules in pure formic acid are converted to formate ion? The density of formic acid is 1. g/cm. Given density of formic acid = 1. g/cm Weight of formic acid in 1 litre solution = 1. 10

11 1.10 Thus, [HCOOH] = 6.5 M 6 Since in case of auto ionization [HCOO H ] [HCOO ] = 10 6 [HCOO ] = [HCOO H ] = 10. [HCOO ] Now % dissociation of HCOOH = 100 = 0.00% [HCOOH] 6.5 SM-IE-CH-11 Problem 9: Calculate the concentration of all species of significant concentrations present in 0.1 M H PO solution. K 1 = , K = , K = I step HPO H + + HP O ; K1 = II step HP O H + + HP O ; K = III step HP O H + + P O ; K = for I step : HPO H + + HP O C C C [H ][HPO ] C.C K1 [H PO ] (0.1 C) C = (0.1 C) C = 0.0 [H + ] = 0.0 M [H PO ] = 0.0 M [HPO] = = M The value of K1 is much large than K and K. Also dissociation of II and III steps occurs in presence of H + furnished in I step and thus, dissociation of II and III steps is further suppressed due to common ion effect. For II step HP O H + + HP O ; (0.0 y) (0.0 + y) y The dissociation of HP O occurs in presence of [H + ] furnished in step I. [H ][HPO ] Thus, K = or (0.0 y)y = [HPO ] (0.0 y) y is small 0.0 y 0.0 and neglecting y = 0.0y 0.0 y = or [HPO ] = K = For III step : HP O H + + P O ( x) (0.0 + x) x

12 SM-IE-CH-1 [H ][PO ] (0.0 x).x K 8 [HPO ] (6.10 x) Again neglecting x and assuming, x = x = x Problem 10: If CH COOH (Ka = 10 5 ) reacts with NaOH at 98 K, then find out the value of the maximum rate constant of the reverse reaction at 98 K at the end point of the reaction. Given that the rate constant of the forward reaction is mol 1 L sec 1 at 98 K. Also calculate Arrhenius parameter for backward reaction if H98 = kcal and E a(f) = 9 kcal. CHCOOH + NaOH CHCOONa + HO; Kf = mol 1 L sec 1 The backward reaction is of hydrolysis of sodium acetate 1 Kfor 1 K W Ka K C (K h) Kb K ac a KW 11 1 Kfor Kb K ac W 10 5 Ka 10 Given, H98 = kcal and E a f = 9 kcal H = Ea E f ab E abac /RT = 9 E a b Ea A b b e = Ab e 98 Ab = Problem 11: The ph of pure water at 5 C and 5 C are 7 and 6 respectively. Calculate the heat of formation of water from H + and OH. At 5 C; [H + ] = 10 7 KW = 10 1 At 5 C; [H + ] = 10 6 KW = 10 1 Now using K w H T T 1.0 log10 Kw R T 1 1T 1 10 H log H = cal/mol = kcal/mol Thus HO H + + OH ; H = kcal/mol H + + OH HO; H = kcal/mol

13 SM-IE-CH-1 Problem 1: Calculate the ph of solution obtained by mixing 10 ml of 0.1 M HCl and 0 ml of 0. M H SO. Milli-equivalent of H + from HCl = = 1 Milli-equvalent of H + from HSO = 0 0. = 16 Total meq. of H + in solution = = 17 [H ] =. 10 [H ] 50 ph = log [H + ] = log 0. ph = Meq. V in ml Problem 1: Calculate the ph of a solution which contains 100 ml of 0.1 M HCl and 9.9 ml of 1.0 M NaOH. HCl + NaOH NaCl + HO t = t = t = [H ] left from HCl = M ph = log H + = log ph =.009 Problem 1: A 0.1 M solution of weak acid HA is 1% dissociated at 5 C. What is its K a? If this solution is with respect to NaA 0. M, what will be the new degree of dissociation of HA and ph? 1 For weak acid HA : HA = 0.01, [HA] = 0.1 M Ka C 0.1 (0.01) 10 Now 0. M NaA, a salt of HA,is added to it resulting a buffer solution of [HA] = 0.1 M and [NaA] = 0. M. ph = log log ph = Also HA H + + A (1 ) [A ] is provided by NaA since dissociation of HA in presence of NaA is suppressed due to a common ion effect. [H ][A ] (C ) 0. 5 Ka 10 [HA] C(1 )

14 SM-IE-CH-1 Problem 15: Calculate the amount of (NH ) SO in g which must be added to 500 ml of 0. M NH to yield a solution of ph = 9.5. K b for NH = poh = log Kb + log [Salt] [Base] or poh = log Kb + log [NH ] [NHOH] [NH ] is obtained from salt (NH)SO. ph = 9.5 poh = =.65 Millimole of NHOH in solution = = 100 Let millimole of NH added in solution = a a 100 [NH ] ; [NHOH] = log a/500 + log 100 / 500 a.65 = log a = Millimole of (NH)SO added = a w w(nh ) SO 5.8g

15 Assignments (New Pattern) SECTION I Single Choice Questions SM-IE-CH A solution of HCl is diluted so that its ph changes by 0.. How does concentration of H + ion change? (a) 0.5 times of initial value (b) 0. times of initial value (b) 10 times increases (d) None The ph of 0.1M CHCOOH is.87. What is ph of 0.1M NHOH. Ka(CHCOOH) = and Kb(NHOH) = (a) (b).87 (b) 7 (d) Liquid ammonia ionises to a slight extent. At - 50 C, its ion product is 0 K [NH ][NH ] 10. How many amide ions, NH are present per mm of pure NH liquid ammonia. (a) (b) (b) (d) To a 50 ml of 0.1 M HCl solution, 10 ml of 0.1 M NaOH is added and the resulting solution is diluted to 100 ml. What is change in ph of the HCl solution? (a).98 (b).98 (b) 0.1M (d) None. 5. What amount of solid sodium acetate be added into 1 litre of the 0.1 M CHCOOH solution so that the resulting solution has ph almost equal to pka (CHCOOH) =.7 (a) 1gm (b) 5 gm (b) 10 gm (d) 1.9 gm. 6. To a 100 ml solution of 0.1 M CHCOONa and 0.1 M CHCOOH, 0. gm of solid NaOH was added. Assuming volume remains constant, calculate the change in ph value? Given that pka (CHCOOH) =.7. (a).15 (b).5 (b).01 (d) A weak base (BOH) with Kb = 10 5 is titrated with a strong acid, HCl. At / th of the equivalent point, ph of the solution is: (a) 5 + log (b) 5 log (c) log (d) When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = ) will occur only with (a) 10 M [Ag + ] and 10 M[HCl] (b) 10 5 M[Ag + ] and 10 5 M [Cl ] (c) 10 6 M [Ag + ] and 10 6 M [Cl - ] (d) M [Ag + ] and M [Cl ].

16 SM-IE-CH The solubility products of Al(OH) and Zn(OH) are and respectively. If NHOH is added to a solution containing Al + and Zn + ions, then substance precipitated first is: (a) Al(OH) (b) Zn(OH) (c) Both together (d) None at all. 10. How many gram of CaCO will dissolve in one litre of saturated solution? Ksp of CaCO is mol /lt and its molecular weight is 18. (a) 0.006g (b) 0.018g (c) 0.00g (d) 0.060g. 11. In the hydrolytic equilibrium A HO HA OH Ka = the degree of hydrolysis of 0.001M solution of the salt is (a) 10 (b) 10 (c) 10 5 (d) For preparing a buffer solution of ph 6 by mixing sodium acetate and acetic acid, the ratio of concentration of salt and acid (Ka = 10 5 ) should be (a) 1:10 (b) 10:1 (c) 100:1 (d) 1: Let the solubilities of AgCl in HO, 0.01 M CaCl; 0.01 M NaCl and 0.05M AgNO be S1, S, S, S respectively. What is the correct relationship between these quantities. (a) S1 > S > S > S (b) S1 > S = S > S (c) S1 > S > S > S (d) S > S > S > S1. 1. The ph at which Mg(OH) begins to precipitate from a solution containing 0.10M Mg + ions [Kp of Mg(OH) = ] is (a) 5 (b) 9 (c) (d) The concentration of hydroxyl ion in solution left after mixing 100 ml of 0.1 M MgCl and 100 ml of 0. M NaOH (Ksp of Mg(OH) = ] is (a).8 10 (b).8 10 (c).8 10 (d) The solubility of Fe(OH) is x mol L 1. Its Ksp would be (a) 9x (d) x (c) 7x (d) 9x. 17. The solubility of Fe(OH) would be maximum in (a) 0.1 M NaOH (b) 0.1 M HCl (c) 0.1 M KOH (d) 0.1 M HSO. 18. pka values of three acids A, B and C are.5,.5 and 6.5 respectively. Which of the following represents the correct order of acid strength? (a) A > B > C (b) C > A > B (c) B > A > C (d) C > B > A.

17 19. The value of ionic product of water at 9 K is (a) less than (b) greater than (c) equal to (d) equal to SM-IE-CH The poh of 10 8 M HCl is (a) 8 (b) 6 (b) Between 6 and 7 (d) Between 7 and Given that the dissociation constant for HO, Kw = mol litre, what is the ph of a M KOH solution? (a) (b) 10 (c) (d) 11.. An acid solution of ph 6 is diluted thousand times. The ph of solution becomes approx. (a) 6.69 (b) 6 (c) (d) 9. The ph of an aqueous solution of a 0.1 M solution of a weak monoprotic acid which is 1% ionized is (a) 1 (b) (c) (d) 11. Which of the following mixtures of solutions can function as a buffer solution? (a) 50 ml of 0. M NH + 50 ml of 0.1 M HCl (b) 50 ml of 0. M NH + 50 ml of 0. M HCl (c) 50 ml of 0. M HCl + 50 ml of 0. M NaOH (d) 50 ml of 0.1 M CH COOH + 50 ml of 0.1 M NaOH. 5. In a mixture of weak of acid and its salt with strong acid, the ratio of concentration of salt to acid is increased ten times the ph of the solution (a) Increases by 10 (b) Decreases by 10 (c) Decreases by 1 (d) Increases by 1. SECTION II May be more than one choice 1. An indicator is a weak acid and ph range of its colour is to 5. If the neutral points of the indicator lies in the centre of the hydrogen ion concentration corresponding to given ph range then ph at the equivalence points is : (a). (b).0 (c) 7.0 (d) A buffer solution contains 100 ml of 0.01 M CHCOOH and 00 ml of 0.0 M CHCOONa. 700 ml of water is added. ph before and after dilution are : (pka =.7) (a) 5.0, 5.0 (b) 5.0, 0.50 (c) 5.0, 1.5 (d) 5., 5..

18 SM-IE-CH-18. HO + HPO HO + + H PO, pk1 =.15 HO + H PO HO + + H PO, pk = 7.0 Hence, ph of 0.01 M NaHPO is : (a) 9.5 (b).675 (c).675 (d) ph of a mixture of 1 M benzoic acid (pka =.0) and 1 M sodium benzoate is.5. In 00 ml buffer, benzoic acid is : (a) 00 ml (b) 150 ml (c) 100 ml (d) 50 ml. 5. ph of mixture of HA and A buffer is 5. Kb of A = Hence [HA]/[A ] will be : (a) 1 (b) 10 (c) 0.1 (d) O COCH COOH (Aspirin) is a pain reliever with pka =. Two tablets each containing 0.09 g of aspirin are dissolved in 100 ml solution. ph will be : (a) 0.5 (b) 1.0 (c) 0.0 (d) pka (CHCOOH) is.7. x mol of lead acetate and 0.1 mol of acetic acid in one L solution make a solution of ph = 5.0. Hence, x is : (a) 0. (b) 0.05 (c) 0.1 (d) ph of a saturated solution of Ba(OH) is 1. Hence, Ksp of Ba(OH) is : (a) M (b) 5 10 M (c) M (d) 10 6 M. 9. A solution is a mixture of 0.05 M NaCl and 0.05 M Nal. The concentration of iodide ion in the solution when AgCl just starts precipitating is equal to : (Ksp AgCl = M ; Ksp AgI = M ) (a) 10 6 M (b) 10 8 M (c) 10 7 M (d) M. 10. At what ph will a M solution of an indicator with Kb = change colour : (a).0 (b).0 (c) 10.0 (d) An acidic buffer can be prepared by mixing the solutions of (a) CHCOONH + CHCOOH (b) NHCl + NHOH (c) HSO + NaSO (d) NaCl + NaOH

19 1. The strongest base among the following is (a) ClO (b) ClO (c) ClO (d) ClO SM-IE-CH The ppt. of CaF (Ksp = ) is obtained when equal volumes of following are mixed (a) 10 MCa MF (b) 10 MCa MF (c) 10 5 MCa MF (d) 10 MCa MF 1. A certain buffer solution contains equal conc. of X and HX. The Kp for X is The ph of the buffer is (a) (b) 7 (c) 10 (1) A certain weak acid has dissociation const of The equilibrium const for its reaction with a strong base is (a) (b) (c) (d) The compound insoluble in acetic acid is (a) CaO (c) CaCO (b) CaCO (d) Ca(OH) 17. The pka of acetyl salicylic acid (aspirin) is.5 The ph of gastric juice in human stomach is about - and the ph in the small intestine is about 8. Aspirin will be (a) un ionized in the small intestine and in the stomach (b) completely ionized in the small intestine and in the stomach (c) ionized in the stomach and almost un ionised in the small intestine (d) ionized in the small intestine and almost un ionised in the stomach 18. Which one is more acidic in aq. Solution (a) NiCl (c) AlCl (b) FeCl (d) BeCl 19. If pkb for fluoride ion at 5 C is 10.8, the ionization const. of HF in water at the temperature is (a) (b).5 10 (c) (d) A weak acid HX has the dissociation const it forms a salt NaX on reaction with alkali. The percentage hydrolysis of 0.1 M solution of NaX is (a) (b) 0.01 (c) 0.1 (d) M solution of NaSO is isotonic with 0.010M solution of glucose at the same temperature. The percentage dissociation of NaSO is (a) 5% (b) 50% (c) 75% (d) 85%

20 SM-IE-CH mole of CHNH (Kb = 5 10 ) is mixed with 0.08 mole of HCl and diluted to 1 lt. What will be the H + ion conc in the solution (a) 8 10 M (b) M (c) M (d) M. A given weak acid (0.01M) has pka = 6 the ph of the solution is (a) (b) (c) 5 (d) 6. An acidic solution (0.1M) of a salt is saturated with HS (0.1 M) the conc of HS is, given k1k = 10 1 (a) 10 (b) (c) 10 0 (d) An acetic acid and sodium acetate buffer has ph = 5.6 the ratio of concentration of [OAc]/HOAc] is (pka of acetic acid =.76) (a) 6 : 1 (b) : (c) 1 : 1 (d) : 1 SECTION III Comprehension Type Questions Write-up I Acidity or alkalinity of a solution depend upon the concentration of hydrogen ion relative to that of hydroxyl ions. The product of hydrogen ion & hydroxyl ion concentration is given by K w = [H + ] [OH ] the value of which depends only on the temperature & not on the individual ionic concentration. If the concentration of hydrogen ions exceeds that of the hydroxyl ions, the solution is said to be acidic; whereas, if concentration of hydroxyl ion exceeds that of the hydrogen ions, the solution is said to be alkaline. The ph corresponding to the acidic and alkaline solutions at 5ºC will be less than and greater than seven, respectively. To confirm the above facts 0.5 M CH COOH is taken for the experiments. [Given : K a of acetic acid = ] Answer the following questions - 1. Degree of dissociation of acetic acid is - (a) (b) 6 10 (c) 10 (d) ph of the solution will be - (a).5 (b). (c) 5 (d).9. If ph of the solution is doubled, what will be the concentration of acetic acid - (a) M (b) 1.0 M (c) M (d) M

21 SM-IE-CH-1. To what volume at 5º C must 1 dm of this solution be diluted in order to double the ph- (a).7 10 dm (b). 10 dm (c) dm (d) dm 5. Now to increase the hydrogen ion concentration 100 dm of 0.1M HCl solution is added to100 dm of 0.5 M acetic acid solution, then what will be the ph of the final solution - (a) 6 (b) 1. (c) (d) 1 Write-up II In qualitative analysis, cations of group II as well as group IV both precipitated in the form of sulphides due to low value of Ksp of group II sulphides, group reagent is HS in presence of dil. HCl and due to high value of Ksp of group IV sulphides, group reagent is HS in presence of NHOH and NHCl. In a 0.1M HS solution, Sn +, Cd + and Ni + ions are present in equimolar concentration (0.1 M). Given: K a 1 Ka(HS) = 10-7, Ksp (NiS) = 10 1 K a (HS) = 10 1, K sp (SnS) = Ksp (CdS) = 10 8, 6. If HCl solution is passed slowly then which sulphide will precipitate first - (a) SnS (b) CdS (c) NiS (d) none of these 7. At what ph precipitate of NiS will form - (a) 1.76 (b) 7 (c) 1. (d) 8. Which of the following sulphide is more soluble in pure water- (a) CdS (b) NiS (c) SnS (d) all have equal solubility 9. If 0.1 M HCl is mixed in the solution containing only 0.1 M Cd + and saturated with HS then concentration of Cd + remains in the solution after CdS has precipitated - (a) (b) (c) (d) Write-up III Like poly protic acids polyhydroxy bases also dissociate or ionise in steps. However, the first step is always stronger than the subsequent ones. Under the conditions if k 1 >> k >> k ; the H O + /OH - concentration in the solution can be computed by considering only the first dissociation or ionisation. For example the base M(OH) dissociates in two steps e.g:- M(OH) MOH + + OH - and MOH + M + +OH -

22 SM-IE-CH- These are characterised by k b 1 [MOH ][OH ] and [M(OH) ] k b [M ][OH ] [MOH ] 10. When, k k then [OH ] can be computed through b1 b b1 0 b 0 (a) [OH ] k [M(OH) ] [OH ] (b) [OH ] k b [M(OH) ] 0 [OH ] (c) [OH] k [M(OH)] [OH] (d) [OH ] k b [M(OH) ] 0 [OH ] 11. If Kb 1 itself is small [M(OH) ] 0[OH ] [M(OH) ] 0and then [OH - ] can be computed as (a) [OH ] k [M(OH) ] (b) [MOH ] [OH ] b 0 (c) [OH ] k b [M(OH) 1 ] 0 (d) [OH ] k b 1. As kb k 1 b, the second dissociation would not affect the concentration of MOH + significantly. Hence (a) [MOH ] [OH ] (b) [M ] k b (c) [MOH ] [OH ] (d) [M ] k b 1. Estimate the PH of a 0.0 M Ba(OH) solution (a) 1.6 (b). (c) 1.6 (d) 10.6 Write-up IV At 5 C the specific conductance of a saturated solution of AgCl is Sm 1 and that of water with which the solution was made is Sm 1. If molar conductance at infinite dilution of AgNO, HNO and HCl are respectively 1 10, 1 10 and 6 10 Sm mol 1. Weak electrolytes molar conductance is calculated by using Kohlrausch s law. At saturated solution m of AgCl = 0 of AgCl 1. Molar conductance of AgCl at infinite dilution (a) Sm mol 1 (b) sm mol 1 (c) sm mol 1 (d) sm mol Concentration of AgCl solution (a) moles m (b) moles dm (c) mole dm (d) moles m 16. Solubility of AgCl is (a) g dm (b) g m (c).8 10 g dm (d).8 10 g m 17. Degree of dissociation of AgCl at saturated solution (a) 1 (b) 0.1 (c) 1 (d) 0.5 1

23 SECTION IV Subjective Questions SM-IE-CH- LEVEL I 1. Determine the solubility of lead iodide (Ksp = 8. x 10 9 ) in 0.0 M KI solution.. Calculate the solubility product of silver chromate at 5C if it requires 0.05 gm of AgCrO to form 1 litre saturated solution.. Sodium salt of a weak acid HA is hydrolysed to the extent of % in its 0.1M solution 5 C. Calculate Ka for the weak acid.. Calculate percentage hydrolysis in 0.00 M aqueous solution of KOCN. Ka for HOCN =.5 10 and Kw = g of Potassium alum is dissolved in enough water to make 100 ml solution. Calculate its [H + ] ion concentration. Al + + HO Al(OH) ++ + H + Kh = The ionization constant of propionic acid is Calculate the degree of ionization of the acid in its 0.05 M solution and also its ph. What will be its degree of ionization if the solution is 0.01 in HCl also? 7. Calculate ph of the following mixtures. Given that Ka and Kb = a) 50 ml of 0.10 M NaOH + 50 ml of 0.05 M CHCOOH. b) 50 ml of 0.05 M NaOH + 50 ml of 0.10 M CHCOOH. c) 50 ml of 0.10 M NaOH + 50 ml of 0.10 M CHCOOH. d) 50 ml of 0.10 M NHOH + 50 ml of 0.05 M HCl. e) 50 ml of 0.05 M NHOH + 50 ml of 0.01 M HCl. f) 50 ml of 0.10 M NHOH + 50 ml of 01.0 M HCl. g) 50 ml of 0.05 M NHOH + 50 ml of 0.05 M CHCOOH. 8. The solubility product of SrF in water is Calculate its solubility in 0.1 M NaF aqueous solution. 9. What is the ph of a 0.5 M aqueous NaCN solution? pkb of CN = What is the ph of 1 M solution of acetic acid. To what volume one litre of this solution be diluted so that ph of the resulting solution will be twice of the original value? 5 K a LEVEL II 1. The Ksp of Mg(OH) is at 5 C. If the ph of solution is adjusted to 9.0. How much Mg + ion will be precipitated as Mg(OH) from a 0.1M MgCl solution at 5 C? Assume that MgCl is completely dissociated.. A certain volume of a monobasic weak acid was titrated against 0.1M NaOH solution. The end point reached upon addition of 0 ml of alkali. The ph of the solution upon the addition of 0 ml of alkali was 6.. Find Ka for the weak acid.

24 SM-IE-CH-. An aqueous solution contains 10% ammonia by mass and has density of 0.99 gm cm. Calculate hydroxyl and hydrogen ion concentration in this solution. Ka for NH = M.. The ph of pure water at 5 C and 5 C are 7 and 6 respectively. Calculate heat of formation of water from H + and OH. 5. A solution of M CdCl contians (M) NH. What conc. of + NH ion from NHCl is necessary to prevent the precipitation of Cd(OH)? (Ksp = 10 1 and Kb = ) 6. An aqueous solution of aniline of conc. 0. M is prepared. What concentration of sodium hydroxide is needed in this solution, so that anilium ion conc. remains at M. Ka C 6H5 NH = Calculate the degree of hydrolysis of a mixture of equal volume of aniline and acetic acid each of them being 0.01 M. Ka of acetic acid = and Kb of aniline = Also calculate ph of the mixture. 8. A weak acid HA after treatment with 1 ml of 0.1 M strong base BOH has a ph of 5. At the end point, the volume of same base required is 6.6 ml. Calculate Ka of acid. 9. Two buffers, (X) and (Y) of ph.0 and 6.0 respectively are prepared from acid HA and the salt NaA. both the buffers are 0.50 M in HA. What would be the ph of the solution obtained by mixing equal volumes of the two buffers? (KHA = ). 10. A certain buffer solution contains equal concentration of X and HXKb for X is Calculate ph of buffer. LEVEL III (Judge yourself at JEE level) 1. What (HO ) must be maintained in a saturated HS solution to precipitate Pb, but not Zn from a solution in which each ion is present at a concentration of 0.01 M? 1 (K H S ; K ZnS ). sp sp. Calculate simultaneous solubility of AgCNS and AgBr in a solution of water. Ksp of AgBr = and Ksp of AgCNS = The KSP of Ca(OH) is at 5 C. A 500 ml of saturated solution of Ca(OH) is mixed with equal volume of 0. M NaOH. How much Ca(OH) in mg is precipitated?. Ksp of AgCl is ml of saturated AgCl solution is titrated with M NHSCN. Calculate volume of M NHSCN required to precipitate all Ag + from saturated AgCl solution as AgSCN.

25 SM-IE-CH mol of AgNO is added to 1 litre of a solution which is 0.1 M in Na CrO and M in NaIO. Calculate the mol of precipitate formed at equilibrium and the concentrations of and Ag, IO 1 10 respectively). and CrO. ( K sp values of Ag CrO and AgIO are M CHCOOH solution is titrated against 0.05 M NaOH solution. Calculate ph at 1/ th and / th stages of neutralization of acid. The ph for 0.1 M CHCOOH is. 7. The average concentration of SO in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 98 K. Given that the solubility of SO in water at 98 K is 1.65 mol litre 1 and the pka of HSO is 1.9, estimate the ph of rain on that day. 8. A solution contains 0.1 M HS and 0. M HCl. Calculate the conc. of S and HS ions in solution. Given K and K for HS are 10 7 and respectively. a 1 a 9. A solution contains a mixture of Ag + (0.10 M) and H g (0.10 M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitate almost completely. What % of that metal ion is precipitated. KSP of AgI = and KSP of HgI = Freshly precipitated Al and Mg hydroxides are stirred vigorously in a buffer solution containing 0.5 M of NHCl and 0.05 M of NHOH. Calcualte [Al + ] and [Mg + ] in solution. Kb for NHOH = KSP of Al(OH) = 6 10 and KSP of Mg(OH) = The KSP of AgCO at 5 C is mol L. A solution of KCO containing 0.15 mole in 500 ml water is shaken at 5 C with excess of AgCO till the equilibrium is reached. AgCO + KCO AgCO + KCO At equilibrium the solution contains mole of KCO. Assuming degree of dissociation of KCO and KCO to be same, calculate KSP of AgCO. 1. Given : Ag( NH ) Ag + + NH KC = and KSP of AgCl = at 98K. If NH is added to a water solution containing AgCl(s) only. Calculate the concentration of the complex in 1.0 M aqueous ammonia. 1. Determine the number of mole of AgI which may be dissolved in 1.0 litre of 1.0 M CN solution. KSP for AgI and KC for Ag(C N) are M and M respectively. 8 10

26 SM-IE-CH-6 1. Determine the concentration of NH solution whose one litre can dissolve 0.10 mole AgCl. KSP of AgCl and Kf of Ag(NH ) are M and M respectively. 15. An acid type indicator, HIn differs in colour from its conjugate base (In ). The human eye is resistive to colour difference only when the ratio [In ]/[HIn] is greater than 10 or smaller than 0.1. What should be the minimum change in ph of the solution to observe a 5 complete colour change? (K ). SECTION V 1. Matching the following Column A a Miscellaneous Questions Column B (a) Water 1. CHCOOH + CHCOONa (b) Buffer (in blood). Lewis acids (c) Acidic buffer. Amphiprotic (d) Ag +, Cu +. HCO + HCO (e) Blood 5. Basic The following questions ( to 5) consists of two statements, one labelled as ASSERTION (A) and REASON (R). Use the following key to chose the correct appropriate answer. (a) If both (A) and (R) are correct, and (R) is the correct explanation of (A). (b) If both (A) and (R) are correct, but (R) is not the correct explanation of (A). (c) If (A) is correct, but (R) is incorrect. (d) If (A) is incorrect, but (R) is correct. ASSERTION (A) REASON (R). Water is specially effective in screening the electrostatic interactions between the dissolved ions.. ph of HCl solution is less than that of acetic acid of the same concentration.. HCO ion acts as a strong acid as well as strong base. 5. Sb (III) is not precipitated as sulphide when in its alkaline solution, HS is passed. 6. Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate AgBr rather than AgCl. The force of ionic interaction depends upon the dielectric constant of the solvent. In equimolar solutions, the number of titrable protons present in HCl is less than that present in acetic acid. CO ions act only as weak base The concentration of S ion in alkaline medium is inadequate for precipitation. Ksp of AgCl Ksp of AgBr.

27 Answers to Assignments SM-IE-CH-7 SECTION - I 1. (a). (a). (b). (b) 5. (d) 6. (d) 7. (d) 8. (a) 9. (a) 10. (a) 11. (a) 1. (b) 1. (c) 1. (b) 15. (c) 16. (c) 17. (d) 18. (c) 19. (b) 0. (d) 1. (d). (a). (c). (a) 5. (d) 1. (a). (d). (b). (c) 5. (c) 6. (d) 7. (c) 8. (a) 9. (c) 10. (a) SECTION - II 11. (a) 1. (a) 1. (b) 1. (a) 15. (c) 16. (c) 17. (d) 18. (c) 19. (c) 0. (b) 1. (c). (b). (b). (c) 5. (d) SECTION - III 1. (b). (a). (c). (a) 5. (b) 6. (a) 7. (c) 8. (b) 9. (a) 10. (a) 11. (b) 1. (a) 1. (c) 1. (b) 15. (a) 16. (a) 17. (a) SECTION - IV LEVEL I moles/litre M. Ka = % 5. [H + ] = M 6. = ; ph =.09; in presence of 0.01 M HCl, = (a) 1.979, (b).77 (c) (d) 9.55 (e) (f) 5.78, (g) M ph =.7,.7710 litre LEVEL II mole/lit. Ka = [H + ] = M and [OH ] = M

28 SM-IE-CH-59. H = 8.55 Kcal/mole 5. ( NH ) = M 6. [NaOH] = 10 M 7. ph =.70 and h = 5.% ph = 9.55, i) ph1 = 9.079; ii) ph = 9.1 LEVEL III M mol litre 1, 10 7 mol lit mg. 1 ml [Ag ] left.10 M, (IO ] left.110 M, [CrO ] left M 6. ph =.58, ph = [HS ] = M, [S ] = M M, 99.8% 10. [Al + ] = M, [Mg + ] = M mol litre mole 1..7 M 15. SECTION - V 1. (a), (b), (c) 1, (d), (e) 5. (a). (c). (b) 5. (c) 6. (c)