NCERT Grade 11-Equilibrium-Answers

NCERT Solutions for Class 11 Chemistry

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1.      (a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.

(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.

(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

2.    The equilibrium constant (Kc) for the give reaction is:

Hence, Kc for the equilibrium is 12.239 M-1.

3.     Partial pressure of I atoms,

Partial pressure of I2 molecules,

Now, for the given reaction,

4.     

5.     The relation between Kp and Kc is given as:

Kp = Kc (RT) Δn

(i) Here

Δn = 3 – 2 = 1

R = 0.0831 bar L mol-1 K-1

T = 500 K

Kp = 1.8 × 10-2

Now,

Kp = Kc (RT) Δn

(ii) Here,

Δn = 2 – 1 = 1

R = 0.0831 bar L mol-1 K-1

T = 1073 K

Kp = 167

Now,

6.    It is given that Kc for the forward reaction is Kc = 6.3 × 1014

As we know that from the given information, the rate of backward reaction is the reciprocal of the rate of forward reaction:

Then, Kc for the reverse reaction will be

7.    For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

8.    Let the concentration of N2O at equilibrium be x.

The given reaction is:

The value of equilibrium constant i.e., Kc = 2.0 × 10-37 is very small. Therefore, the amount of N2 and O2 reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N2 and O2.

Then,

9.     The given reaction is:

Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.

Again, 2 mol of NOBr are formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr are formed from  mol of Br, or 0.0259 mol of NO. The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br2] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518

= 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br2] = 0.0437 – 0.0259

= 0.0178 mol.

10.   For the given reaction,

11.   The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:

Hence, the value of Kp for the given equilibrium is 4.0.

12.   The given reaction is:

N2(g) + 3H2(g) ↔ 2NH3(g)

The given concentrations of various species is

Now, reaction quotient Qc is:

Since Qc ≠ Kc, the reaction mixture is not at equilibrium.

Again, Qc > Kc, Hence, the reaction will proceed in the reverse direction.

13.   The balanced chemical equation corresponding to the given expression can be written as: 4NO(g) + 6H2O(g) ↔ 4NH3(g) + 502(g)

14.   The given reaction is:

Therefore, the equilibrium constant for the reaction,

We know that:

15.    It is given that equilibrium constant Kc for the reaction

H2(g) + I2(g) ↔ 3HI(g) is 54.8.

Therefore, at equilibrium, the equilibrium constant K’c for the reaction

Let the concentrations of hydrogen and iodine at equilibrium be x molL-1

[H2] = [I2] x mol L-1.

Therefore,

Hence, at equilibrium, [H2] = [I2] = 0.068 mol L-1.

16.   The given reaction is:

Hence, at equilibrium,

[H2] = [I2] = 0.167M

[Hl] = (0.78 – 2 × 0.167)M = 0.446M.

17.   Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.

Now, according to the reaction,

We can write,

18.   (i) Reaction quotient, 

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.

The given reaction is:

Therefore, equilibrium constant for the given reaction is:

(iii) Let the volume of the reaction mixture be V.

Therefore, the reaction quotient is,

Since Qc < Kc, equilibrium has not been reached.

19.   let the concentrations of both PCl3 and Cl2 at equilibrium be x mol-1. The given reaction is:

It is given that the value of equilibrium constant. Kc is 8.3 × 10-3

Now we can write the expression for equilibrium as:

Therefore, at equilibrium, [PCl3] = [Cl2] = 0.02 molL-1

20.   For the given reaction,

It is given that Kp = 0.265

Since Qp > Kp, the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO2 will increase while the pressure of  will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO2 be p.

Then, we can write,

Therefore, equilibrium partial of CO2, pco2 = 0.80 – 0.339 = 0.461 atm

And, equilibrium partial pressure of CO, pco = 1.4 + 0.339 = 1.739 atm.

21.   The given reaction is:

Since, Qc ≠ Kc, the reaction is not at equilibrium.

Since Qc < Kc, the reaction will proceed in the forward direction to reach equilibrium.

22.  Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

Therefore, at equilibrium,

[BrCl] = 3.3 × 10-2 – (2 × 1.5 × 10-3)

= 3.3 × 10-3 – 3.0 × 10-3

= 0.3 × 10-3

= 3.0 × 10-4 molL-1

23.   Let the total mass of the gaseous mixture be 100 g.

Mass of CO = 90.55 g

And, mass of CO2 = (100 – 90.55) = 9.45 g

Now, number of moles of CO, 

Number of moles of CO 2,  

Partial pressure of CO,

Partial pressure of CO2,

For the given reaction,

Δn = 2 – 1 = 1

We know that,

24.   (a) For the given reaction,

ΔG° = ΔG°( Products) – ΔG°( Reactants)

ΔG° = 52.0 – {87.0 + 0}

= -35.0 kJ mol-1

(b) We know that,

ΔG° = RT log

ΔG° = 2.303 RT log Kc

Hence, the equilibrium constant for the given reaction Kc is 1.36 × 106

25.   (a) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(b) The number of moles of reaction products will decrease.

(c) The number of moles of reaction products remains the same.

26.  The reactions given in (i), (iii), (iv), (v), and (vi) will get affected by increasing the pressure.

The reaction given in (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.

The reactions given in (i), (iii), (v), and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products.

27.   Given, Kp for the reaction i.e., H2(g) + Br2(g) ↔ 2Hbr(g) is 1.6 × 105.

Therefore, for the reaction 2Hbr(g) ↔ H2(g) + Br2(g) the equilibrium constant will be,

Now, let p be the pressure of both H2 and Br2 at equilibrium.

Now, we can write,

Therefore, at equilibrium,

[H2] = [Br2] = 2.49 × 10-2 bar

[HBr] = 10 – 2 × (2.49 × 105) bar

= 9.95 bar = 10 bar (approx).

28.   (a) For the given reaction,

(b) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

29.   (a) According to Le Chatelier’s principle, on addition of H2, the equilibrium of the given reaction will shift in the forward direction.

(b) On addition of CH3OH, the equilibrium will shift in the backward direction.

(c) On removing CO, the equilibrium will shift in the backward direction.

(d) On removing CH3OH, the equilibrium will shift in the forward direction.

30.   (a) 

(b) Value of Kc for the reverse reaction at the same temperature is:

(c) (i) Kc would remain the same because in this case, the temperature remains the same.

(ii) Kc is constant at constant temperature. Thus, in this case, Kc would not change.

(iii) In an endothermic reaction, the value of  increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc  will increase if the temperature is increased.

31.   Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:

Hence, at equilibrium, the partial pressure of H2 will be 3.04 bar.

32.   If the value of Kc lies between 10-3 and 103, a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

33.   The given reaction is:

Hence, the concentration of O3 is 2.86 × 10-28 M.

34.   Let the concentration of methane at equilibrium be x.

It is given that Kc = 3.90.

Therefore,

Hence, the concentration of CH5 at equilibrium is 5.85 × 10-2 M.

35.   A conjugate acid-base pair is a pair that differs only by one proton.

The conjugate acid-base for the given species is mentioned in the table below.

36.   Lewis acids are those acids which can accept a pair of electrons. For example, BF3, H+, and are Lewis acids.

37.   The table below lists the conjugate bases for the given Bronsted acids.

38.   The table below lists the conjugate acids for the given Bronsted bases.

39.   The table below lists the conjugate acids and conjugate bases for the given species.

40.   (a) OH is a Lewis base since it can donate its lone pair of electrons.

(b) F is a Lewis base since it can donate a pair of electrons.

(c) H+ is a Lewis acid since it can accept a pair of electrons.

(d) BCl3 is a Lewis acid since it can accept a pair of electrons.

41.   Given,

[H+] = 3.8 × 10-3

∴pH value of soft drink

= log [H+]

= -log (3.8 × 10-3)

= -log 3.8 – log 10-3

= -log 3.8 + 3

= -0.58 + 3

= 2.42.

42.   Given,

pH = 3.76

It is known that,

pH = -log[H+]

⇒ log [H+] = – pH

⇒ [H+] = anti log (-pH)

= anti log (-3.76)

= 1.74 × 10-4 M.

Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 × 10-4 M.

43.   It is known that,

Given,

Ka of HF = 6.8 × 10-4

Hence, Kb of its conjugate base F

Given,

Ka of HCOOH = 1.8 × 10-4

Hence, Kb of its conjugate base HCOO

Given,

Ka of HCN = 4.8 × 10-9

Hence, Kb of its conjugate base CN

44.   Ionization of phenol:

As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.

Noe, let be the degree of ionization of phenol in the presence of 0.01 M C6H5ONa.

Also,

45.   (i) To calculate the concentration of HS ion:

Case I (in the absence of HCl):

Let the concentration of HS be x M.

Case II (in the presence of HCl):

In the presence of 0.1 M of HCl, let [HS] be y M.

(ii) To calculate the concentration of [S2-]:

Case I (in the absence of 0. M HCl):

HS ↔ H+ + S2-

[HS] = 9.54 × 10-5 M (From first ionization, case I)

Let [S2-] be X.

Also, [H+] = 9.54 × 10-5 M(From first ionization, case I)

Case II (in the presence of 0.1 M HCl):

Again, let the concentration of HS be X’ M.

[HS] = 9.1 × 10-8 M (From first ionization, case I)

[HS] = 0.1M (From HCl, case II)

46.   Method 1

1) CH3COOH ↔ CH3COO + H+    Ks = 1.74 × 10-5

2) H2O + H2O ↔ H3O+ + OH        K = 1.0 × 10-14

Since Ka >> Kw;

CH3COOH ↔ CH3COO + H+ + H3O+

Degree of dissociation,

CH3COOH ↔ CH3COO + H+

Thus, concentration of CH3COO = c.a

= .05 × 1.86 × 10-2

= 0.93 × 10-2

= .00093M

Sin ce [oAc] = [H]+

[H]+ = .00093 = .093 × 10-2

pH = -log [H+]

= – log (0.93 × 10-2)

∴ pH = 3.03

Method 2: Degree of dissociation,

CH3COOH ↔ CH3COO + H+

Thus, concentration of CH3COO = c.a

= .05 × 1.86 × 10-2

= 0.93 × 10-2

= .00093M

Sin ce [oAc] = [H+]

[H+] = .00093 = .093 × 10-2

pH = -log [H+]

= – log (0.93 × 10-2)

∴ pH = 3.03

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

47.   Let the organic acid be HA.

⇒ HA ↔ H+ + H

Concentration of HA = 0.01 M

pH = 4.15

-log [H+] = 4.15

[H+] = 7.08 × 10-5

Then,

48.   (i) 0.003 MHCl:

H2O + HCl ↔ H3O+ + Cl

Since HCl is completely ionized,

[H2O+] = [HCl]

⇒ [H3O+] = 0.003

Now,

pH = -log [H3O+] = -log(.003)

Hence, the pH of the solution is 2.52.

(ii) 0.002 HBr:

HBr + H2O ↔ H3O + Br

[H3O+] = .002

∴ pH = -log [H3O+]

= -log (0.002)

= 2.69

Hence, the pH of the solution is 2.69.

(iv) 0.002 M KOH:

KOH(aq) ↔ K+(aq) + OH(aq)

[OH] = [KOH]

Now, pOH = -log [OH] = 2.69

∴ pH = 14 – 2.69

= 11.31.

Hence, the pH of the solution is 11.31.

49.   (a) For 2g of TlOH dissolved in water to give 2 L of solution:

(b) For 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution:

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:

13.6 × 1 mL = M2 × 1000 mL

(Before dilution) (After dilution)

13.6 × 10-2 = M2 × 1L

M2 = 1.36 × 10-2

[H+] = 1.36 × 10-2

pH = -log (1.36 × 10-2)

= (0.1335 + 2)

= 1.866 = 1.87.

50.   Degree of ionization, a = 0.132

Concentration, c = 0.1 M

Thus, the concentration of H3O+ = c.a

= 0.1 × 0.132

= 0.0132

pH = -log [H+]

= -log (0.0132)

= 1.879 : 1.88

Now,

Ka = Ca2

= 0.1 × (0.132)2

Ka = .00017

pKa = 2.75

51.   c = 0.005M

pH = 9.95

pOH = 4.05

pH = -log (4.105)

4.05 = -log [OH]

[OH] = 8.91 × 10-5

ca = 8.91 × 10-5

Thus, Kb = ca2

= 0.005 × (1.782)2 × 10-4

= 0.005 × 3.1755 × 10-4

= 0.0158 × 10-4

Kb = 1.58 × 10-6

Pkb = – log Kb

= -log (158 × 10-6)

= 5.80.

52.   K = 4.27 × 10

c = 0.001M

pH = ?

a = ?

kb = ca2

4.27 × 10-10 = 158 × a2

4270 × 10-10 =  a2

65.34 × 10-5 = a = 6.53 × 10-4

Then, [anion] = ca = .001 × 65.34 × 10-5 = .065 × 10-5

pOH = -log (0.65 × 10-5) = 6.187

pH = 7.813

Now,

Ka × Kb = Kw

∴ 4.27 × 10-4 × Ka = Kw

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10-5.

53.   c = 0.05 M

When HCl is added to the solution, the concentration of H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case I: When 0.01 M HCl is taken.

Let x be the amount of acetic acid dissociated after the addition of HCl.

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.

Case II: When 0.1 M HCl is taken.

As in the same case, let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

54.   Kb = ca2

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

Percentage of dimethyamine is equal to 0.0054 * 100=0.54%.

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

55.   The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH = -log[H+]

(a) Human muscle fluid 83:

pH = 6.83

pH = – log [H+]

∴ 6.83 = – log [H+]

[H+] = 1.48 × 10-7 M

(b) Human stomach fluid, 2:

pH =1.2

1.2 = – log [H+]

∴[H+] = 0.063

(c) Human blood, 38:

pH = 7.38 = – log [H+]

∴ [H+] = 4.17 × 10-8 M

Human saliva, 4:

pH = 6.4

6.4 = -log [H+]

[H+] = 3.98 × 10-7

56.  The hydrogen ion concentration in the given substances can be calculated by using the given relation:

pH = -log [H+]

(i) pH of milk = 6.8

Since, pH = -log [H+]

6.8 = -log [H+]

log [H+] = -6.8

[H+] = anitlog(-6.8)

= 1.5 × 10-7 M

(ii) pH of black coffee = 5.0

Since, pH = -log [H+]

5.0 = -log [H+]

log [H+] = -5.0

[H+] = anitlog(-5.0)

= 10-5 M

(iii) pH of tomato juice = 4.2

Since, pH = -log [H+]

4.2 = -log [H+]

log [H+] = -4.2

[H+] = anitlog (-4.2)

= 6.31 × 10-5 M

(iv) pH of lemon juice = 2

Since, pH = -log [H+]

2.2 = -log [H+]

log [H+] = -2.2

[H+] = anitlog(-2.2)

= 6.31 × 10-3 M

(v) pH of egg white = 8

Since, pH = -log [H+]

7.8 = -log [H+]

log [H+] = -7.8

[H+] = anitlog(-7.8)

= 6.31 × 10-3 M.

57.   

KOH(aq) ↔ K+(aq) + OH(aq)

[OH] = 0.05M = [K+]

= (10-14/0.05) = 2 × 10-13 M

pH = -log [2 × 10-13]

= 12.70.

58.   Solubility of sr (OH)2 = 19.23 g/L

Then, concentration of Sr(OH)2

59.   Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

In the presence of 0.1 m of HCl, let a ‘ be the degree of ionization.

60.   c = 0.1 M

pH = 2.34

61.   NaNo2 is the salt of a strong base (NaOH) and a weak acid (HNO2).

Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:

Therefore, degree of hydrolysis

62.   pH = 3.44

We know that,

pH = -log [H+]

63.  (i) NaCl:

Therefore, it is a neutral solution.

(ii) KBr:

Therefore, it is a neutral solution.

(iii) NaCN:

Therefore, it is a basic solution.

(iv) NH4NO3

Therefore, it is a acidic solution.

(v) NaNO2

Therefore, it is a basic solution.

(vi) KF

Therefore, it is a basic solution.

64.   It is given that Ka for ClCH2COOH is 1.35 × 10-3.

ClCH COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH.

65.   Ionic product,

Kw = [H+] [OH]

Let [H+] = x

Sin ce[H+] = [OH], Kw = x2

⇒ Kw at 310 K is 2.7 × 10-14

∴ 2.7 × 10-14 = x2

⇒ x = 1.64 × 10-7

⇒ [H+] = 1.64 × 10-7

⇒ pH = -log [H+]

= -log [1.64 × 10-7]

= 6.78

Hence, the pH of neutral water is 6.78.

66.   

Since there is neither an excess of H3O+ or OH, the solution is neutral.

Hence, pH = 7.

67.   (1) Silver chromate:

Ag2CrO4 → 2Ag+CrO42-

Then,

Kxp = [Ag+]2 [CrO42-]

let the solubility of Ag2CrO4 be s.

⇒ [Ag+] 2s and [CrO42-] = s

Then,

Ksp = (2s)2 .s = 4s3

⇒ 1.1 × 10-12 = 4s3

.275 × 10-12 = s3

s = 0.65 × 10-4 M

Molarity of Ag+ = 2s = 2 × 0.65× 10-4 = 1.30 × 10-4 M

Molarity of CrO42- = s = 0.65 × 10-4 M

(2) Barium chromate:

BaCrO4 → Ba2+ + CrO42-

Then,

Ksp = [Ba2+] [CrO42-]

Let s be the solubility of BaCrO4.

Thus, [Ba2+] = s and [CrO42-] = s

⇒ Ksp = s2

⇒ 1.2 × 10-10 = 42

⇒ s = 1.09 × 10-5 M

Molarity of Ba2+ = Molarity of CrO42- = s = 1.09 × 10-5 M

(3) Ferric hydroxise:

Fe(OH)3 → Fe2+ + 3OH

Then,

Ksp = [Fe2+] [OH]3

Let s be the solubility of Fe(OH)3.

Thus, [Fe2+] = s and [OH] = 3s

⇒ Ksp = s.(3s)2

= s.27s3

Ksp = s.27s4

1.0 × 10-38 = 27s4

0.37 × 10-38 = s4

.00037 × 10-36 = s4 ⇒ 1.39 × 10-10 M = S

Molarity of Fe2+ = s = 1.39 × 10-10 M

Molarity of OH = 3s = 4.17 × 10-10 M

(4) Lead chloride:

PbCl2 → Pb2+ + 2Cl

Ksp = [Pb2+] [Cl]2

Let Ksp be the solubility of PbCl2.

[Pb2+] = s and [Cl] = 2s

Thus, Ksp = s.(2s)= 4s3

⇒ 1.6 × 10-5 = 4s3

⇒ 0.4 × 10-5 = s3

4 × 10-6 = s3 ⇒ 1.58 × 10-2 M = S.1

Molarity of PB2+ = s = 1.58 × 10-2 M

Molarity of chloride = 2s = 3.16 × 10-2 M

(5) Mercurous iodide:

Hg2I2 → Hg2+ + 2I

Ksp = [Hg2+] [I]2

Let s be the solubility of Hg2I2.

⇒ [Hg2+] = s and [I] = 2s

Thus, Ksp = s.(2s)⇒ Ksp = 4s3

4.5 × 10-29 = 4s3

1.125 × 10-29 = s3

⇒ s = 2.24 × 10-10 M

Molarity of Hg2+ = s = 2.24 × 10-10 M

Molarity of I = 2s = 4.48 × 10-10 M

68.   Let s be solubility of Ag2CrO4.

Then, Ag2CrO↔ Ag2+ + 2CrO4

Ksp = (2s)2.s = 4s3

1.1 × 10-12 = 4s3

s = 6.5 × 10-5 M

Let s ‘ be the solubility of AgBr.

AgBr(s) ↔ Ag+ + BrO

Ksp = s’= 5.0 × 10-13

∴ s ‘ = 7.07 × 10-7 M

Therefore, the ratio of the molarities of their saturated solution is

69.   When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.

Then,

Now, the solubility equilibrium for copper iodate can be written as:

Cu(103)2 → Cu2+(aq)+ 210(aq)

Ionic product of copper iodate:

= [Cu2+] [103]2

= (0.001) (0.001)2

= 1 × 10-9

Since the ionic product (1 × 10-9) is less than Ksp (7.4 × 10-8), precipitation will not occur.

70.   Since pH = 3.19,

Let the solubility of C6H5COOAg be x mol/L.

Then,

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10-6 mol/L. Now, let the solubility of C6H5COO Ag be x’ mol/L.

Hence, C6H5COO Ag is approximately 3.317 times more soluble in a low pH solution.

71.   Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., 

If the concentrations of both solutions are equal to or less than 5.02 × 10-9 M, then there will be no precipitation of iron sulphide.

72.   CaSO4(s) → Ca2+(aq) + SO42-(aq)

Ksp = [Ca2+] [SO42]

Let the solubility of CaSO4(s) be s.

Then, Ksp = s2

9.1 × 10-6 = s2

9.1 × 10-5 = s2

s = 3.02 × 10-3 mol/M

Molarity of CaSO4 = 136 g/mol

Molarity of CaSO4 in gram/L = 3.02 × 10-3 × 136 = 0.41 g/L

This means that we need 1L of water to dissolve 0.41g of CaSO4.

Therefore, to dissolve 1g of CaSO4 we require  of water.

73.   For precipitation to take place, it is required that the calculated ionic product exceeds the Ksp value.

Before mixing:

This ionic product exceeds the Ksp of Zns and CdS. Therefore, precipitation will occur in and CdCl2 and ZnCl2 solutions.

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