Ionic Equilibrium Solved Problems

Page # 15 IONIC EQUILIBRIUM SOLVED PROBLEMS Ex.1 Sol. Ex.2 Sol. Ex.3 Sol. Ex.4 Sol. How many hydrogen ions are present in 1 ml of a solution of pH =13 – (A) 6.02 x 1013 (B) 6.02 x 1012 7 (C) 6.02 x 10 (D) 6.02 x 105 (Ans. C) pH = 13 –log [H+] = 13 or [H+] = 10–13 mole/litre [H+] in 1 ml = 10–16 moles 1 mole H+ contains 6.023 × 1023 H+ ions  No. of H+ ions in 10–16 moles = 6.023×1023×10–16 = 6.023 ×107 The pH of a 0.05 M solution of H2SO4 in water is nearly – (A) 0.05 (B)1 (C) –1 (D) 0 (Ans. B) pH= – log10 H+ The concentration of H+ ions is expressed in gm equivalent Molarity of H2SO4 = 0.05  Normality = 0.05 × 2 = 0.1  pH = –log 0.1 or pH = 1 A salt M2 X3 dissolves in water such that its solubility is x g. mole/litre. Its KSP is – (A) x5 (B) 6x2 (C) 108 x5 (D) 6x5 (Ans. C) Solubility of M2X3 = x gm mole/litre M2 X3 2M+3 + 3X–2 = 2x  [M+3] [X –2 ] = 3x Solubility product KSP = (2x)2.(3x)3=108 x5 The solubility of AgCl in water, in 0.02 M CaCl2, in 0.01M NaCl and in 0.05 M AgNO3 are S0, S1,S2,S3 respectively. Which of the following relationships between these quantities is correct – (A) S0 > S1 > S2 > S3 (B) S0 > S2 > S1 > S3 (C) S0 > S1= S2 > S3 (D) S0 > S2 > S3 > S1 (Ans. B) So lub ility Pr oduct (B) Solubility = Concentration of Common ion K SP = 50 KSP  S1 = 0.02 K SP S2 = = 100 KSP 0.01 S3 = K sp = 20 KSP 0.05 So, S2 > S1 > S3 Again solubility will be greatest in water. So, S0 > S2 > S1 > S3 1 Ex.5 Sol. Ex.6 The pH of a 0.01M solution of a monobasic acid is four. Which one of the following statement about the acid is incorrect – (A) When a little NaOH is added, it will form a buffer solution (B) It is a weak acid (C) It's sodium salt will be acidic (D) It's sodium salt will be basic (Ans. C) Concentration of monobasic acid = 0.01 M pH = 4 If the acid is completely ionised the pH of the acid would be pH = –log 0.01 = –log 10–2 = 2 So it is a weak acid. The sodium salt of a weak acid when dissolved the anions will be hydrol ysed gi vi ng rise to OH – i on concentration. The solution will be basic. So statement (C) is incorrect. The dissociation constant of a monobasic acid which is 3.5% dissociated in solution at 20ºC is – (A) 3.5 × 10–2 (B) 5 × 10–3 –5 (C) 6.34 × 10 (D) 6.75 × 10–2 (Ans.C) Sol. Concentration of acid = N 20 N = 0.05 N 20 Out of 100 molecules, 3.5 molecules have been dissociated  Out of 1 molecules the no. of dissociated molecules = 35 = 0.035 =  100 0.05  (0.035) 2 C 2 Ka = = = 6.34×10–5 (1   ) 1  0.035 Ex.7 Sol. Solubility product of AgCl is 2.8 × 10–10 at 25ºC. Calculate solubility of the salt in 0.1 M AgNO3 solution – (A) 2.8 × 10–9 mole/litre (B) 2.8 × 10–10 mole/litre (C) 3.2 × 10–9 mole/litre (D) 3.2 × 10–12 mole/litre (Ans. A) In 0.1 M AgNO3 AgNO3 Ag + + NO3– AgCl Ag+ + Cl¯ + KSP = [Ag ] [Cl¯] Now [Ag+] can be taken as [AgNO3] while [Cl¯] is the solubility of AgCl Page # 16 IONIC EQUILIBRIUM  Cl = KSP 2.8  1010 = [Ag  ] 01 . Sol  Solubility of AgCl = 2.8×10–9 mole/litre Ex.8 Sol. What is the maximum concentration of Ni +2 ions in water (at 25ºC) that is saturated with H2S (0.1M at 25ºC) and maintained at pH 3 with HCl . KSP of NiS is 3 x 10–21. H2S dissociates in two steps, each with an equilibrium constant – H2S H+ + HS– Ka1 = 9.1 × 10–8 HS– H+ + S–2 Ka2 = 1.1 × 10–12 (A) 3 ×10–7 M (B) 4 × 10–7 M –7 (C) 5 × 10 M (D) 6 × 10–7 M (Ans. A) The overall dissociation of H2S will be the Ex.11 The dissociation constant of weak acid HA is 4.9×10–8. After making the necessary approximations. Calculate pH in 0.1 M acid – (A) 1.155 (B) 2.155 (C) 3.155 (D) 4.155 (Ans. D) Sol. For weak acid Ka = 2 .C   [H+ ]2 [S -2 ] 10 -20 –2 [S ] = = = 10–14 M [H+ ]2 10 -6 Since NiS will precipitate, if the S.P. is exceeded, the highest value which [Ni+2] concentration can have is: [Ni+2] = Sol. = = = = = K a1 × K a2 = (9.1×10–8) × (1.1×10–12) = 1 × 10–19 Saturated H2S is approximately 0.1 M and dissociation of H2S is very slight So, [H+]2 [S–2] = 1×10–19×10 –1 = 10–20 Since, pH is 3, [H+] = 10–3 M Ex.9 = pH product of K a1 and K a2 . K The equation for this reaction is HC2O4– + Cl2+ 5OH¯ 2Cl¯ + CO3–2 + 3H2O Here, OH¯ is the limiting reagent and will be used up completely. Since, 5 moles of OH¯ produce 2 moles of Cl¯ 0.1 mol of OH¯ will produce 0.04 mole of Cl¯  Moles of Cl¯ in the final solution = 0.04 3 x10 21 = 3 × 10–7 M 10 14 Calculate the pH of solution having H+ ion concentration of 5 × 10–4 mole/litre – (A) 3.3 (B) 2.26 (C) 1.26 (D) 0.26 (Ans. A) [H+] ion concentration = 5 × 10–4 mole/litre pH = – log [5 × 10–4] = – (log 5 + log 10–4) = – 0.7 + 4 = 3.3 Ex.10 Suppose the change HC2O4–+Cl2CO3–2+Cl– is to be carried out in basic solution. Starting with 0.1 mole of OH, 0.1 mole of HC2O4– and 0.05 mole of Cl2, How many moles of Cl¯ would be expected to be in the final solution – (A) 3.04 (B) 2.04 (C) 1.04 (D) 0.04 (Ans. D) Ka 4.9 x 10 8 = C 0.1 7 × 10–4 –log H+ = –log  C –log 7×10–4 ×10–1 4.1549 = 4.155 Ex.12 Ka for cyanoacetic acid is 4 × 10–3. What is the value of degree of hydrolysis of 0.4 M sodium cyano acetate solution ? (A) 4.5 × 10–6 (B) 5.5 × 10–6 (C) 2.5 × 10–6 (D) 3.5 × 10–6 (Ans. C) Sol. Kh = Kw -14 = 10 -3 Ka 4×10 = 0.25 × 10–11 h = Kh c -11 = 0.25×10 0.4 = 2.5 × 10–6 Ex.13 Calculate the pH of solution obtained by mixing10ml of 0.1M HCl and 40ml. of 0.2M H2SO4– (A) 0.3685 (B) 0.4685 (C) 1.3685 (D) 1.4684(Ans. B) Sol. Milli equivalent of H+ from HCl = 10 × 0.1 = 1 Milli equivalent of H+ from H2SO4 = 40 × 0.2 × 2 = 16 Total Meq of H+ in solution = 1 + 16 = 17 Meq 17 = 3.4 × 10–1 [ [H+] = ] V in ml 50 pH = – log [H+] = – log [0.34] = 0.4685 [H+] = Ex.14 The dissociation constants for aniline, acetic acid and water at 25ºC are 4 × 10–10, 2 × 10–5 and 10–14 respectively. Calculate degree of hydrolysis of aniline acetate in a decinormal solution– (A) 0.025 (B) 0.015 (C) 0.035 (D) 0.045 (Ans. C) Page # 17 IONIC EQUILIBRIUM Sol. Aniline++Acetate–+H2O Aniline +Acetic acid Before hydrolysis 1 1 0 After hydrolysis 1–h 1–h h Let concen. salt be C mole litre–1 Kh = Let 'a' millimoles of NH4+ are added to a solution having milli moles of NH4OH = 500× 0.2 = 100 0 h Ch . Ch h2 = C(1  h) . C(1  h) (1  h) 2 h = 1 h Kw K a. K b 10 2  10  4  10 10 6 0.1  V 2V [CH3COONa] = 0.1  V 2V 01 . = 0.05 M 2 Now calculate pH by hydrolysis of CH3COONa CH3COONa + H2O [OH–] = C.h = p[OH] pH CH3COOH + NaOH Kw . C = Ka 10 14  0.05 2  10 5 = 5 × 10–6 = 6 × 0.699 = 5.301 = [14 –5.301] = 8.699  8.7 Ex.16 Calculate the amount of (NH4)2SO4 in gm which must be added to 500 ml of 0.2 M NH3 to yield a solution of pH = 9, Kb for NH3 = 2 × 10–5– (A) 3.248 g (B) 4.248 g (C) 1.320 g (D) 6.248 g (Ans. C) Sol. = [Base] = 100 500 Given Kb for NH4OH = 2 x10–5 and pH = 9 = Ex.15 Calculate the pH at the equivalence point when a solution of 0.1M acetic acid is titrated with a solution of 0.1M NaOH, Ka for acid = 2 × 10–5 – (A) 5.7 (B) 6.7 (C) 7.7 (D) 8.7 (Ans. D) Sol. Let V ml of acid and V ml of NaOH be used concentration of both acid and NaOH are same. CH3COOH + NaOH CH3COONa + H2O Concentration before reaction 01 . V 0.1  V = 2V 2V Concentration after reaction 0 [NH4OH] a / 500 100 / 500  a = 200 mili moles = 0.2 mol moles of (NH4)2 SO4 added % hydrolysis = 54.95 % h = 0.035 0 and a 500  5 = –log 2 × 10–5 + log 14 =  [NH4+] = [salt] = [ NH 4  ] pOH = –log Kb + log [ NH 4 OH ] a = 0.1 mol 2  W(NH = 0.1 × 132 = 1.32 4)2 SO4 Ex.17 The Ka for formic acid and acetic acid are 2 × 10 –4 and 2 × 10 –5 respectively. Calculate the relative strength of acids with same molar concentration – (A) 10 (B) 7 (C) 8 (D) 5 (Ans. A) Sol. Relative strength of weak acids = FK GH K a1  a2 C1 C2 I JK  Relative strength = FK I GH K JK a1 a2 ( C1 = C2)  2  10  4   =  5  2  10   Relative strength for HCOOH to CH3COOH = 10 : 1 Ex.18 Find the pH of solution prepared by mixing 25ml of a 0.5 M solution of HCl, 10ml of a 0.5 M solution of NaOH and 15ml of water (A) 0.8239 (B) 1.0029 (C) 1.0239 (D) 1.8239 (Ans. A) Sol. We know that for HCl and NaOH, m.e. = m.e.  m.e. of HCl = 0.5 × 25 = 12.5 m.e. of NaOH = 0.5 × 10 = 5.0 m.e. of HCl in the resultant mixture = 12.5 – 5.0 = 7.5 total volume = (25 + 10 + 15) ml = 50 ml m. e. 7.50  Normally of HCl = = Vol(ml) 50 7.50 7.50  Molarity =  [H+] = [HCl] = 50 50  pH = –log 7.50 = 0.8239 50 Page # 18 IONIC EQUILIBRIUM Ex.19 The solubility procduct of chalk is 9.3 × 10–8. Calculate its solubility in gram per litre – (A) 0.3040 gram / litre (B) 0.0304 gram / litre (C) 2.0304 gram / litre (Ans. B) (D) 4.0304 gram / litre Sol. CaCO3 Ca2+ + CO32– Let the solubility of CaCO3 be s mole per litre  KSP = [Ca2+] [CO32–] = s.s  s = KSP = 9.3  108 = 0.000304 mole / litre Solubility in g/l = mole/litre × Moleqular weight of CaCO3 = 0.000304 × 100 = 0.0304 gram / litre Ex.20 Maximum Conductivity would be of – (A) K3Fe(CN)6 [0.1 M solution] (B) K2Ni(CN)4 [0.1M solution] (C) FeSO 4 .Al 2 (SO 4 ) 3 .24H 2 O [0.1 M solution] (D) Na[Ag(S2O2)3] [0.1 M solution] (Ans. C) Sol. Double salt on ionization gives more ions. One molecule of the salt gives Fe+2, 2Al +3, 4SO4–2 ions. Hence its conductance would be highest. Ex.21 What amount of sodium propanoate should be added to one litre of an aqueous solution containing 0.02 mole of propanoic acid (Ka = 1.0 × 10–5 at 25ºC) to obtain a buffer solution of pH 6 – (A) 0.1 M (B) 0.2M (C) 0.3 M (D) 1.3 M (Ans. B) Sol. Using the expression pH = pKa + log [Salt] [ Acid] we get,6 = –log(1.0 ×10–5)+ log Which gives or 6 = 5 + log [Salt] [0.02 M] [Salt] [0.02 M] [Salt] = 10 or [Salt] = 0.2 M [0.02 M] Ex.22 What will be the pH of the solution, if 0.01 mole of HCl is dissolved in a buffer solution containing 0.03 mole of propanoic acid (Ka = 1.0 × 10–5) and 0.02 moles of salt, at 25ºC (A) 3.699 (B) 4.699 (C) 5.11 (D) 6.11 (Ans. B) Sol. pH = pKa + log [Salt] [ Acid] = –log(1.0 × 10–5) + log (0.02  0.01) (0.03  0.01)  1 = 5 + log   = 5 – 0.6 = 4.4 4 Ex.23 20 ml of 0.2 M NaOH is added to 50 ml, of 0.2 M CH 3COOH to give 70ml, of the solution. What is the pH of the solution ? The ionization constant of acetic acid is 2 × 10–5 – (A) 4.522 (B) 5.568 (C) 6.522 (D) 7.568 (Ans. A) Sol. The addition of NaOH converts equivalent amount of acetic acid into sodium acetate. Hence, Concen. of acetic acid after the addition of 30 NaOH = × 0.2 M 70 Concen. of CH2COONa after the addition of 20 NaOH = × 0.2 M 70 Hense, Using the expression [Salt] pH = pKa + log [ Acid] 20 = –log (2 × 10–5) + log 30 = 4.699 – 0.177 = 4.522 FG IJ H K Ex.24 The concentration of H+ ion in a 0.2 M solution of HCOOH is 6.4 × 10–3 mole –1 . To this solution HCOONa is added so as to adjust the concentration of HCOONa to one mole per litre. What will be the pH of this solution? Ka for HCOOH is 2.4 × 10–4 and the degree of dissociation of HCOONa is 0.75 – (A) 3.19 (B) 4.19 (C) 5.19 (D) 6.19 (Ans. B) Sol. Assuming that the addition of HCOONa suppresses the ionization of HCOOH, we can use the expression [Salt] [ Acid] to compute pH of the solution, since salt is 75% dissociated we will get, 0.75 Ph = –log(2.4 × 10–4) + log 0.2 = 3.62 + 0.57 = 4.19 pH = pKa + log Page # 19 IONIC EQUILIBRIUM Ex.25 What amount of HCl will be required to prepare one litre of a buffer solution (containing NaCN and HCN) of pH 10.4 using 0.01 mole of NaCN. Given Kion (HCN) = 4.1 × 10–10 – (A) 8.55 × 10–3 mole (B) 8.65 × 10–3 mole (C) 8.75 × 10–3 mole (D) 9.9 × 10–4 mole (Ans. D) Sol. The addition of HCl converts NaCN into HCN. Let x be the amount of HCl added. We will have. [NaCN] = (0.01 – x) [HCN] = x Substituting these values along with pH and Ka in the expression. [Salt] pH = – log Ka + log [ Acid] We get 10.4 = –log[4 × 10–10] + log or 10.4 = 9.4 + log 0.01  x x 0.01  x x 0.01  x = 1 x 0.01  x or = 10  11x = 10–2 x or x = 9.9 × 10–4 M or log Ex.26 Calculate the pH of aqueous solution of 1.0 M HCOONH 4 assuming complete dissociation (pKa of HCOOH = 3.8 and pKb = of NH3 = 4.8) – (A) 3.5 (B) 4.5 (C) 5.5 (D) 6.5 (Ans. D) Sol. Ammonium formate undergoes hydrolysis as NH4+ + HCOO–+H2O Kh  NH4OH + HCOOH KW Ka . Kb Moreover in the solution we have [NH4OH] = [HCOOH] [HCOOH] 2 Hence Kh = [HCOO  ] 2 or KW Ka . Kb or [H+]2 or 2pH or pH [H ] 2 = [K ] 2 a KW . Ka Kb = pKW + pKa – pKb 1 = [pKW + pKa – pKb] 2 1 = [14 + 3.8 – 4.8] = 6.5 2 = Ex.27 Calculate the pH of a solution which contains 10 ml of 1M HCl and 10 ml of 2 M NaOH (A) 11.7 (B) 12.7 (C) 13.7 (D) 10.7 (Ans. C) Sol. HCl + NaOH  NaCl + H2O Meq. before 10 × 1 10 × 2 Reaction = 10 = 20 0 0 Meq. After 0 10 10 10 Reaction 10 [OH–] left from NaOH = = 0.5 M 20 – pOH = – log OH = – log 0.5 pOH = 0.3 pH = 13.7 Ex.28 Calculate pH of a solution of given mixture (0.1 mole CH3COOH + 0.2mole CH3COONa) in 100 ml of mixture. Ka = 2×10–5. (A) 4.6 (B) 5.6 (C) 6.6 (D) 7.6 (Ans. A) Sol. We have [Salt] pH = – log Ka + log [ Acid] 3  1000 M [Salt] = = 0.366 82  100 [Acid] = 2  1000 M = 0.333 60  100 = – log 2 × 10–5 0.2  1000 100 + log = 4.6 0.1 1000 1000 Ex.29 Calculate the pH of a buffer solution prepared by dissolving 10.6 of Na2CO3 in 500 ml of an aqueous solution containing 80 ml of 1M HCl. Ka for HCO3– = 6 × 10–11 – (A) 8.6 (B) 9.6 (C) 11.6 (D) 12.6 (Ans. B) Sol. Na2CO3 + HClNaCl+NaHCO3 10.6 ×1000 80×1 106 Reaction = 100 80 0 0 Meq. After 20 0 80 80 Reaction The solution Na2CO3 and HCO3– and thus acts as buffer Meq. before 2 pH = – log Ka + [CO 3 ] log  [HCO 3 ] = – log 6 × 10–11 + log 20 = 80 9.6 Page # 20 IONIC EQUILIBRIUM Ex.30 What volume of 0.1 M HCOONa solution should be added to 50ml of 0.05 M formic acid to produce a buffer solution of pH = 4.0, pKa of formic acid = 3.7 (A) 50 ml (B) 40 ml (C) 30 ml (D) 60 ml (Ans. A) Sol. Let V ml of 0.1 M HCOONa be mixed to 50ml of 0.05 M HCOOH in mixture [HCOONa] = 0.1  V ( V  50) [HCOOH] 50  0.05 ( V  50) = V = 50 ml Ex.31 Saccharin (Ka = 2 × 10–12) is a weak acid respectively by formula HSaC. A 4 ×10–4 mole amount of Saccharin is dissolved in 200cc water of pH 3. Assuming no change in volume. Calculate the concentration of Sac. ions in the resulting solution at equilibrium (A) 2 × 10–12 M (B) 3 × 10–12 M –12 (C) 4 × 10 M (D) 5 × 10–12 M (Ans. C) 4  10 4 Sol. [HSaC] = = 2 × 10–3 M 200 / 1000 The dissociation of HSaC takes places in the presence of [H+] = 10–3 HSaC H+ + SaC– –3 conc. before 2×10 10–3 0 dissociation In presence of H+ the dissociation of HSaC is almost negligible because of common ion effect. Thus at equilibrium [HSaC] = 2 × 10–3, H+ = 10–3 [H ] [SaC  ] [HSaC] [10 3 ] [SaC  ] [2  10 3 ] BA or HA + B+ + OH– HA– A–  K = + + H2 O B+ + A– + H2O or H2 O [A ] [HA] [OH ] Also for weak acid HA HA H+ + A– Ka 10 [H ] [ A  ] K a  K W or K = = 14 = 1010 K [HA ] K 10 W Ex.33 Calculate pH of a solution whose 100 ml contains 0.2 gm NaOH dissolved in it (A) 10.699 (B) 11.699 (C) 12.699 (D) 13.699 (Ans. C) Sol. 10 0ml solution of NaOH contai ns = 2.0gm NaOH  1000 ml solution of NaOH contains = 2 gm NaOH 2 Normality of solution = = 0.05 N 40 10 14  [H+] = 0.05 10 14  – log[H+] = –log 0.05 = – [–14 + 2 – 0.6990] = 12.699 LM N  C =1.0×10–4. Ex.32 A certain weak acid has Ka Calculate the equilibrium constant for its reaction with a strong base (A) 109 (B) 1010 (C) 1011 (D) 1012 (Ans. B) OP Q Ex.34 Calculate the ratio of pH of a solution containing 1 mole of CH3COONa + 1mole of HCl per litre and of other solution containing 1 mole CH3COONa + 1mole of acetic acid per litre. (A) 1 : 1 (B) 2 : 1 (C) 1 : 2 (D) 2 : 3 (Ans. C) Sol. Case I - pH when 1mole CH3COONa and 1 mole HCl are present CH3COONa+HClCH3COOH+NaCl Before 1 1 0 0 reaction After 0 0 1 1 reaction [CH3COOH] = 1M  [H+] C = C.  [SaC–] = 4 × 10–12 M 1 BOH strong 4 [Salt] [ Acid] (0.1  V ) / ( V  50) 4.0 = 3.7 + log 2.5 / ( V  50 )  2 × 10–12 = HA + weak Ka =  pH = – log Ka + log Ka = Sol. = Ka C = C .K a = Ka 1 1 log Ka 2 Case II - pH when 1 mole CH3COONa and 1 mole of CH3COOH, a buffer solution [Salt]  pH2 = – log Ka + log = – log Ka [Acid]  pH1 = – Page # 21 IONIC EQUILIBRIUM  [Salt] [Acid]  or = pH1 1 = pH2 2 = 1M pH1 : pH2 = 1 : 2 Ex.35 Calculate pH of 0.002 N NH4OH having 2% dissociation (A) 7.6 (B) 8.6 (C) 9.6 (D) 10.6 (Ans. C) Sol. NH 4 OH is a weak base and partial ly dissociated NH4OH Concentration 1 before dissociation Concentration 1 –  after dissociation  [OH–] pOH pH NH4+ + OH– 0 0  = C = 2 × 10–3 ×  2 100 = 4 × 10–5 M = – log[OH–] = – log 4 × 10–5 = = 14 – 4.4 = 9.6 Ex.38 Calculate the pOH – pKb for the buffer, 0.20 M NH3, 0.40M NH4Cl, Kb for ammonia = 10 (A) 0.50 (B) 0.60 (C) 0.30 (D) 0.75 (Ans. C) [Salt] Sol. pOH = pKb+ log ; [Base] 0.4 pOH – pKb = log = 0.30 0.2 Ex.39 pH of 0.01 M HS– will be pK a log C (A) pH = 7 + + 2 2 pK a log C (B) pH = 7 – – 2 2 pK 1  pK 2 (C) pH = 2 pK a  pK b (D) pH = 7 + 2 FG H 4.4 Sol. M M HCl + 100ml NaOH 10 10 M M (B) 55 ml HCl + 44 ml NaOH 10 10 M M (C) 10 ml HCl + 90 ml NaOH 10 10 M M (5) 75 ml HCl + 25ml NaOH 5 5 (Ans. D) M M 25 ml NaOH will be neutralise 25 ml 5 5 M HCl. Hence, 50 ml HCl get rest and to 5 the mixing of both solution will give total 100 ml volume. N1 V 1 = N2 V 2 50 × N = N2 5 N2 = 50  N = .1 5  100 × 100 Hence, pH = 1 H2 O H 2S = Ch = Kw + (Ans. A) OH– K w .C Ka = or pH = 1 [pKw + pKa + log C] 2 K w .C Ka = K w .K a C  [H+] Ex.40 Percentage ionisation of weak acid can be calculated using the formula - (A) 100 ml Sol. +  [OH–] Ex.36 The present species in the resultant solution which is form by the neutralisation of HCl from NaOH is (A) Na+, Cl–, H3O+ (B) Na+, Cl–, H2O (C) Na+, Cl–, OH– (D) H+, H3O+ (Ans. B) Sol. Na+ OH– + H+ Cl – Na+ Cl– + H2O Ex.37 Which of the following has pH is equal to near about one - HS– IJ K 100 Ka (B) C 1  10(pK a pH) (C) Both (A) and (B)(D) None (Ans. C) For weak acid dissociation equilibria, degree of dissociation  is given as - (A) 100 Sol. Ka C Ka C [H ] [A  ] [H ]. C [H ].  Also, Ka = = = [HA] C(1  ) (1   )  log Ka = log H+ + log (1 ) (1 ) or pKa = pH + log  (1 ) pKa – pH = log  1  pK a  pH  = 10   = or 1 =  or  =  %  = 100 10 pK a pH + 1 1 (1  10pK a pH )