CHEMICAL EQUILIBRIUM
MCQs
Q.1 A reaction is reversible because
(a) reactants are reactive (b) products are reactive
(c) products are stable (d) reactants are stable
Q.2 A large value of Kc means that at equilibrium
(a) less reactants and more products
(b) more reactants and less product
(c) same amount
(d) none
Q.3 Extent to H2 + I2 ® 2HI can be increased by
(a) increasing pressure (b) increasing product
(c) increasing temp (d) adding a catalyst
Q.4 Strength of an acid can be determined by
(a) PKa (b) PKp
(c) POH (d) PKw
Q.5 In an exothermic reversible reaction increase in temp shifts the equilibrium to
(a) reactant side (b) product side
(c) remains unchanged (d) none
Q.6 Units of Kw are
(a) mole dm–3 (b) mole2 dm–3
(c) mole2 dm–6 (d) mole2 dm–3
Q.7 A basic Buffer solution can be prepared by mixing
(a) weak acid and its salt with strong base
(b) strong acid and its salt with weak base
(c) weak base and its salt with strong acid
(d) strong base and its salt with weak acid
Q.8 Buffer action can be explained by
(a) common ion effect (b) law of mass action
(c) Le–Chatlier’s principle (d) all above
Q.9 Ionization of weak acid is expressed in term of following constant
(a) Kw (b) Kn
(c) Ka (d) Kb
Q.10 Solubility of Ca(OH)2 is exothermic. If solubility will increase
(a) at high temp (b) at low temp
(c) temp independent (d) none
Q.11 For which system does the equilibrium constant, Kc has units of concentration
(a) N2 + 3H2 2NH3 (b) H2 + I2 2 HI
(c) 2NO2 N2O4 (d) 2HF H2 + F2
Q.12 Which statement about the following equilibrium is correct
2SO2(g) + O2(g) 2SO3(g) D H = – 188.3 kJ mol–1
(a) the value of Kp falls with a rise in temp
(b) the value of Kp falls with increasing pressure
(c) adding V2O5 catalyst increase the equilibrium yield of sulphur trioxide
(d) the value of Kp is equal to Kc
Q.13 The PH of 10–3 mole dm–3 of an aqueous solution of H2SO4 is
(a) 3.0 (b) 2.7
(c) 2.0 (d) 1.5
Q.14 The solubility product of AgCl is 2.0 x 10–10 mole2 dm–6. The max concentration of Ag+ ions in the solution is
(a) 2.0 x 10–10 mol dm–3 (b) 1.41 x 10–5 mol dm–3
(c) 1.0 x 10–10 mol dm–3 (d) 4.0 x 10–20 mol dm–3
Q.15 An excess of aqueous silver nitrate to added to aqueous barium chloride and precipitate is removed by filtration what are the main ions in the filtrate
(a) Ag+ and NO only (b) Ag+ and Ba2+ and NO3
(c) Ba2+ and NO only (d) Ba+2 and NO and Cl–
Q.16 For N2 + 3H2 2NH3
(a) Kc = Kp (b) Kp = Kc RT
(c) Kp = Kc (RT)–2 (d) Kp = Kc (RT)–1
Q.17 H2 + I2 2HI
In the above equilibrium system, if the conc. of reactants of 25oC is increased, the value of Kc will
(a) increase (b) decrease
(c) remains constant
(d) depends upon nature of reactants
Q.18 In a chemical reaction, equilibrium is said to have established when
(a) opposing reactions stops
(b) concentrations of reactants and products are equal
(c) rate constants of opposing reactions are equal
Q.19 The relation between Kc and Kp is
(a) Kc = Kp (RT) Dn (b) Kp = Kc (RT) Dn
(c) Kp = Kc (RT) Dn (d) Kp = Kc
Q.20 The precipitation occurs if the ionic concentration is
(a) less than Ksp (b) more than Ksp
(c) equal to Ksp (d) is present at any moment
Q.21 The PH of oranges is
(a) 3.5 (b) 3.1
(c) 4.6 (d) 4.2
Q.22 Which one of following solution have zero PH
(a) 1M HCl (b) 0.5 MH2SO4
(c) 0.1 M HNO3 (d) 1M CH3COOH
Q.23 The solubility product expression for BaF2 can be written as
(a) [Ba2+] [F–] (b) [Ba2+] [2F]
(c) [Ba2+] [F–]2 (d) [Ba+] [F–]2
Q.24 To prepare a buffer with PH close to 9.0, you could use a mixture of
(a) NH4OH and NH4Cl
(b) CH3COOH and CH3COONa
(c) HNO2 + NaNO2
(d) NaHCO3 + H2CO3
Q.25 For which reaction the numerical value of Kc and Kp are same
(a) N2 + 3H2 2NH3 (b) 2SO2 + O2 2SO3
(c) H2 + Cl2 2HCl (d) N2O4 2NO2
Q.26 For which system does the equilibrium constant Kc have units (mole dm–3)–1
(a) H2 + I2 2HI (b) N2 + 3H2 2NH3
(c) 2NO2 N2O4
(d) CH3COOH + C2H5OH CH3COOC2H5 + H2O
Q.27 What can affect the magnitude of equilibrium constant Kp of a reversible gaseous reaction
(a) temperature (b) pressure
(c) catalyst (d) none of above
Q.28 Which gas can change the PH towards acidic
(a) argon (b) carbon dioxide
(c) nitrogen (d) oxygen
Q.29 The solution having zero PH will be
(a) basic (b) high basic
(c) neutral (d) highly acidic
Q.30 A solution have H+ ions concentration 1 x 10–7 its PH will be
(a) acid (b) basic
(c) neutral (d) zero
Q.31 Which one of the following has highest PH
(a) 0.1 M HCl (b) 1.0 M HCl
(c) gastric juice (d) lemons
Q.32 Which PH is considered as basic
(a) 1 (b) 7
(c) 2 (d) 11
Q.33 The sum of PH and POH is
(a) 2 (b) 7
(c) 14 (d) 13.5
Q.34 A buffer solution can be prepared by mixing
(a) a strong acid and weak base
(b) a weak acid and weak base
(c) a strong acid and its salt
(d) a weak base and its salt with strong acid
Q.35 Law of mass action was presented by
(a) Henderson
(c) Guldberg and Waage (d) Arrehenius
Q.36 The unit of Kc for reaction
N2 + O2 2NO
(a) mol dm–3 (b) mol–1 dm3
(c) mol–2 dm6 (d) no units
Q.37 PH of pure water is
(a) 3.2 (b) 4.2
(c) 7.0 (d) 0
Q.38 Which of following change will favour the formation of more SO3 at equilibrium
2SO2 + O2 2SO3 + heat
(a) by adding SO3 at equilibrium
(b) by increasing temp
(c) by decreasing temp
(d) by decreasing pressure
Q.39 When pressure is applied to the given equilibrium
ice water which of the following will happen
ice water which of the following will happen
(a) more ice will be formed
(b) more water will be formed
(c) equilibrium will not be disturbed
(d) water will formed
Q.40 Which of following change will favour the formation of more HI in the given reaction
H2 + I2 2HI
(a) increasing pressure
(b) decreasing pressure
(c) by adding more HI
(d) by adding more H2 and I2ANSWERS
| Questions | 1 | 2 | 3 | 4 | 5 |
| Answers | b | a | c | a | A |
| Questions | 6 | 7 | 8 | 9 | 10 |
| Answers | a | a | d | c | b |
| Questions | 11 | 12 | 13 | 14 | 15 |
| Answers | a | a | b | b | c |
| Questions | 16 | 17 | 18 | 19 | 20 |
| Answers | c | a | c | b | b |
| Questions | 21 | 22 | 23 | 24 | 25 |
| Answers | a | a | c | a | c |
| Questions | 26 | 27 | 28 | 29 | 30 |
| Answers | b | b | b | d | c |
| Questions | 31 | 32 | 33 | 34 | 35 |
| Answers | d | d | c | d | c |
| Questions | 36 | 37 | 38 | 39 | 40 |
| Answers | d | c | c | d | d |
SHORT QUESTION WITH ANSWERS
Q.1 What is weak electrolyte?
Ans.
A compound which is only partially ionized in aqueous solution is called as weak electrolyte.e.g CH3COOH(Acetic aci)
Q.2 What is meant by state of chemical equilibrium?
Ans.
The state of reversible reaction, in which forward and reverse rates are equal.(Rf=Rr)
Q.3 What are reversible reaction?
Ans.
Those reactions in which the reactants products are converted into each other under same set of conditions.
Q.4 Define Le–Chatlier’s principle.
Ans.
If a system in equilibrium is disturbed at equilibrium it will move in that direction where change is minimized.
Q.5 What are conjugate acid and bases?
Ans.
When an acid is dissolved in water it will form H3O+ ions and an anion the anion of an acid is called conjugate base and H3O+ is called conjugate acid e.g.
CH3OOH + H2O H3O+ + CH3COO–
Acid Base Conjugate acid Conjugate base
Q.6 Why we need buffer solution in daily life?
Ans.
Buffers can resist the charge of its PH value therefore they are required in chemical analysis, pharmaceuticals, electroplating, photography, beverage industry, microbiology molecular biology, soil science and in quantitative analysis.Our blood is best example of buffer solution having PH 7.35. If it decreases upto 7 or goes upto 8 death may occur.
Q.7 Discuss the factors on which buffer PH depends.
Ans.
PH of buffer depends upon two factors.
(i) Pka or ka value of an acid
(ii) Concentration of salt and acid taken for the buffer.
Q.8 What is buffer capacity?
Ans.
The amount of acid or base which a buffer can absorb without significant charge in PH is called buffer capacity. Buffer capacity is the ability of buffer to resist the change in its PH value.
Q.9 How can we calculate the PH of buffer?
Ans.
We can calculate the PH of a buffer the PH of Henderson equation which can be derived as
HA ® H+ + A–
Kc =
H+ =
Take log on both sides
Log [H+] = log
Log [H+] = log ka +
Multiply equation by (– 1) on both sides
– log [H+] = – log ka – log
PH = Pka – log
or
PH = Pka + log
PH = Pka + log
In similar way
POH = Pkb + log
Q.10 How Kc can be applied to calculate direction of chemical reaction?
Ans.
(i) If < Kc
Then the reaction will proceed in the forward direction until the equilibrium is established.
(ii) If > Kc
Then the reaction will proceed in the backward direction until the equilibrium is established.
If = kc
Then the reaction is already in equilibrium.
Q.11 How value of Kc can help to predict the extent to which a chemical reaction can take place?
Ans.
The extent of reaction depends upon the magnitude of Kc. i.e.
(i) Very small value of Kc:
when Kc is very small the forward reaction will not occur to an appreciable extent and the reverse reaction will go almost to complete.
(ii) Kc is very large:
When Kc is very large the reverse reaction will not occur to an appreciable extent and the forward reaction is almost complete.
(iii) Kc has an intermediate value:
When the value of Kc is neither very large nor very small, the equilibrium mixture contains appreciable amounts of both reactant and products.
Q.12 What are optimum conditions for the formation of NH3 by Haber’s process?
N2(g) + 3H2(g) 2NH3(g) DH = -92.46 kJ/m
Ans.
(i) As the reaction is exothermic max yield is obtained at low temp.
(ii) By applying pressure, more product is obtained. According four moles of reactants produce two moles of product so by applying pressure more yield of NH3 is gained.
(iii) Continuous withdrawl of NH3 will also increase the yield.
(vi) Rate of reaction can also be increase with the help of catalyst like iron or iron oxide.
Q.13 A catalyst does not change the position of equilibrium but this equilibrium position approach earlier. Why?
Ans.
A catalyst decrease the energy of activation required by reacting substance so the position of equilibrium reaches earlier.
Q.14 How the change in pressure at equilibrium position effect the following reaction?
PCl5 PCl3 + Cl2
Ans.
According to Le–chatlier’s principle, by increasing pressure the reaction will move toward less number of moles. In the above reaction, by increasing pressure the reaction will move in backward direction.
Q.15 What is the relationship b/w kc, kp kc & kn?
Ans.
These four parameters are related as.
Kp = Kc (RT)Dn = Kx (P)Dn = Kn
When mole of products are reactant are equal i.e. Dn = 0 their
Kp = kc = dx = kn
Q.16 Prove that Kw is ionic product of water and its value is 1 x 10–14 at 25o.
Ans.
Water undergoes self ionization as
2H2O ® H3O+ + OH–
or H2O ® H+ + OH–
Kc [H2O] = [H+] [OH–
Kw = [H+] [OH–
At 25oC the value of Kw ionic product have been measured. Its value is 1 x 10–14.
Q.17 How can you purify NaCl by common ion effect?
Ans.
The impurities present in the common salt are CaCl2 MgCl2 and Na2SO4 in it. These impurities can be removed by common Ion effect. For this purpose a saturated solution of NaCl is prepared. Then HCl gas is passed through it. HCl is stronger electrolyte than NaCl Ionization of NaCl is suppressed by passing HCl gas and it is precipitated out.
Q.18 A buffer consists of a weak acid and is salt.
Ans.
(i) CH3COOH + CH3COONa
(ii) C6H5COOH + C6H5COONa
(iii) H2CO3 + NaHCO3
(iv) H3PO3 + NaH2PO4
Q.19 How can you co–relate Ka with Kc?
Ans.
Ka is dissociation constant of an acid while Kc is the equilibrium constant suppose an acid HA Ionize as
HA + H2O ® H3O+ + A–
Kc =
concentration of [H2O] almost remains constant so we can write.
Kc =
Kc [H2O] = ka
Ka =
Q.20 How can you calculate % Ionization of an acid?
Ans.
Percentage Ionization of an acid can be determined by the following formula.
% Ionization = x 100
initially available.
Q.21 N2(g) + 3H2(g) 2NH3(g) DH = -92.46 KJ/mol
This reaction is favourable at low temperature then why its production is favourable at 4500C ?
Ans:
This is industrial condition favourable at the industrial level because at low temperature the reaction becomes very slow because both are gases.
Q.22 What is common ion effect?
Ans:
The decrease in ionization of a weak electrolyte by adding another strong electrolyte having common ion.e.g ionization of NaCl is decreased by passing HCl through NaCl solution.
Q.23 What is solubility product?
Ans:
The product of molar solubilities of the ions of weak electrolyte at equilibrium stage is called solubility product. It is represented by Ksp
TEXT BOOK EXERCISE
Q1. Multiple Choice questions
(i) For which system does the equilibrium constant, Kc has units of (concentration)-1
(a) N2 + 3H2
2NH3
2NH3(b) H2 + I2 2HI
2HI(c) 2NO2 N2 O4
N2 O4 (d) 2HF H2 + F2
H2 + F2(ii) Which statement about the following equilibrium is correct
2SO2 + O2 2SO3
2SO3(a) The value of Kp falls with a rise in temperature
(b) The value of KP falls with increasing Pressure
(c) Adding V2 O5 catalyst increases the equilibrium yield of SO3
(d) The value of Kp is equal to Kc
(iii) The pH of 10-3 mol/dm3 of an equous solution of H2SO4 is
(a) 3 (b) 2.7 (c) 2 (d) 1.5
(iv) The solubility product of AgCI is 2 x 1010 mol2 dm-6 . The maximum concentration of Ag+ ions in the solution is
(a) 2 x 10-10 mol dm-3
(b) 1.41 x 10-10 mol dm-3
(c) 1 x 10-10 mol dm-3
(d) 4 x 10-20 mol dm-3
(v) An ecxcess of aqueous silver nitrate is added to aqueous barium chloride and precipitate is removed by filtration. What are the main ions in the filtrate?
(a) Ag+ and NO3-1 only
(b) Ag+ and Ba2+ and NO3-1
(c) Ba2+ and NO3-1 only
(d) Ba2+ and NO3-1 and CI-1
Ans. i)c ii)a iii)b iv)b v)b
Q2. Fill in the blanks
(i) Law of mass action states that the rate at which a reaction proceeds is directly proportional to the product of the active masses of the ______.
(ii) In an exothermic reversible reaction, _________in temperature will shift the equilibrium towards the forward direction.
(iii) The equilibrium constant for the reaction 2O3 3O2 is 1055 at 25oC, it tells that ozone is _______at room temperature.
3O2 is 1055 at 25oC, it tells that ozone is _______at room temperature.(iv) In a gas phase reaction, if the number of moles of reactants are equal to the number of moles of the products, KC of the reaction is ______to the Kp.
(i) Buffer solution is prepared by by mixing together a weak base and its salt with _______ or a weak acid and its salt with ________.
Ans. i)reactants ii)decrease iii)unstable iv)equal v)strong acid, strong base
Q3. Label the sentences True or False
(i) When a reversible reaction attains equilibrium both reactants and products are present in a reaction mixture.
(ii) The Kc of the reaction
A+B C+D is given by
C+D is given by Kc =
Therefore it is assumed that [A] = [B]=[C]=[D]
(iii) A catalyst is a compound, which increases the speed of the reaction and consequently increases the yield of the product.
(iv) Ionic product Kw of pure water at 25oC is 10-14 dm-6 and is represented by an expression
Kw =[H+][OH-
(v) AgCI is a sparingly soluble ionic solid in water. Its solution produces excess of Ag+ and CI- ions.
Ans. i)True ii)False iii)False iv)True v)False
Q4. (a) Explain the terms “reversible reaction” and “state of equilibrium”
See Section 8.1 and 8.1.2
(b) Define and explain the law of mass action and drive the expression for the equilibrium constant KC.
See Section 8.1.3
(c) Write KC for the following reactions
(i) Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)
Sn4+(aq) + 2Fe2+(aq)(ii) Ag+(aq) +Fe2+(aq) Ag(s) + Fe3+(aq)
Ag(s) + Fe3+(aq) (iii) N2(g)+ O2(g) 2NO(g)
2NO(g)(iv) 4NH3(g) 5O2(g) 4NO(g) +6H2O(g)
4NO(g) +6H2O(g)(v) PCI5(g) PCI3(g) +CI2(g)
PCI3(g) +CI2(g)| (i) Kc=  | (ii) Kc=  | (iii) Kc=  |
| (iv) Kc=  | (v) Kc=  | |
Q5. (a) Reversible reactions attain the position of equilibrium, which is dynamic in nature and not static. Explain it.
See Section 8.1.1
(b) Why do the rates of forward reactions slow down when a reversible in nature and not static. Explain it.
See Section 8.1.1
Q6. When a graph is plotted between time on X-axis and the concentration of reactants and products on Y-axis for a reversible reaction, the curve becomes parallel to time axis at a certain stage.
(a) At what stage the curves become parallel?
(b) Before the curves become parallel, the steepness of curves falls? Give reasons.
(c) The rate of decrease of concentrations of the reactants and rate of increase of concentrations of any of products may or may not be equal for various types of reactions before the equilibrium time. Explain it.
See section 8.1.1
Q7. (a) Write down the relationship of different types of equilibrium constants. i.e. Kc and Kp for the following general reactions.
aA + bB cC +dD
cC +dD See section 8.1.2
(b) Decide the comparative magnitudes of Kc, Kp for the following reactions.
Synthesis of NH3
N2(g) +3H2(g) 2NH3(g)
2NH3(g) KC is given by
Kc =
Kp is given by
Kp =
For this reaction, change in number of moles is given by
n =number of mole of product-number of moles of reactants =2 – (1+3)=-2
Hence
Kp=Kc x (RT)-2
Or Kp = Kc x 
Thus if T is such that RT > 1, then Kp < Kc
If T is such that RT< 1, then Kp > Kc
Dissociation of PCI5
PCI5(g) PCI3(g)+ CI2(g)
PCI3(g)+ CI2(g) Kc is given by according to law of Mass of Action
Kc =
Kp is given by
Kp =
For this reaction, change in number of moles is given by
n =number of mole of product-number of moles of reactants =2 – 1 =1
Kp =Kc x (RT)
Thus If T is such that RT > 1, then Kp > Kc
If T is such that RT< 1, then Kp <Kc
Q8. (a) Write down KC for the following reversible reactions. Suppose that the volume of reaction mixture in all the cases is ‘V’ dm3 at equilibrium stage.
(i) CH3 COOH+ CH3 CH2 OH CH3COOC2H5+ H2O
CH3COOC2H5+ H2O (ii) H2 +12 2HI
2HI (iii) 2HIH2 +I2
H2 +I2 (iv) PCI5 PCI3 + CI2
PCI3 + CI2 (v) N2 + 3H2 2NH3
2NH3 See section 8.1.3
(b) How do you explain that some of the reactions mentioned above are affected by change of volume at equilibrium stage?
Q9. Explain the following two applications fo equilibrium constant. Give examples.
(i) Direction of reaction
(ii) Extent of reaction
See section 8.1.5
Q10. Explain the following with reasons.
(a) The change of volume disturbs the equilibrium position for some of the gaseous phase reactions, but not the equilibrium constant.
(b) The change of temperature disturbs both the equilibrium position and the equilibrium constant of a reaction.
(c) The solubility of glucose in water is increased by increasing the temperature.
Q11. (a) What is ionic product of water? How does this value vary with the change in T? Is it true that this value is 75 times when the T of water increased form 0oC to 100oC.
Ionic Product of water is given by the equation
Kw [H+][OH-
Value of Kw increases with increase in temperature. It is because increase in temperature increase the ionization of H2O. Thus, more H+ or OH-
e.g.
At 25oC Kw =1 x 10-14
and At 100oC Kw =7.5 x 10-14
Further
At 0oC (Kw)o =0.1 x 10-14_____(1)
At 100oC (Kw)100 =7.5 x 10-14_____(2)
Divide eq (2) by eq (1)
=
=75 or (KW) 100 =75 x (Kw)0
Hence, KW at 100oC is 75 times more than at 0oC
(b) What is the Justification for the increase of ionic product with temperature?
Value of Kw increase with increase in temperature. It is because increase in temperature increase the ionization of H2O. Thus, more H+ or OH-
(c) How do you prove that at 25OC in 1 dm3 of water, there are 10-7 moles of H3O+?
At 25OC
Kw =[H3O+][OH-14 ____(1)
Since ionization of water gives equal no. of H3O+ and OH-
therefore
[H3O+]=[OH-
Hence, eq (1) can be written as
Kw =[H3O+][H3O+]=10-14
Or [H3O+]2=10-14
Taking square root on both sides
[H3O+]2=10-7 mol/dm3
Hence, at 25oC, water has 10-7 mole/dm3 of H3O+ ions.
Q12. (a) Define pH and pOH. How are they related with pKw.
(b) What happens to the acidic and basic properties of aqueous solutions when pH varies form 0 to 14.
(c) Is it true that the sum of pHa and pKb is always equal to 14 to all temperatures for any acid? If not why?
Q13. (a) What is Lowry bronsted idea of acids and bases? Explain conjugate acids and bases.
(b) Acetic acid dissolves in water and gives proton to water. But when dissolves in H2SO4, it accepts proton. Discuss the role of acetic acid in both cases.
CH3 COOH + H2O CH3 COO- + H3O+
CH3 COO- + H3O+However, H2SO4 is a stronger acid than acetic acid, therefore, H2SO4 donates proton and acts as an acid while acetic acid accepts proton and acts as a base.
H2SO4+ CH3 COOH HSO4- + CH3 COOH2+
HSO4- + CH3 COOH2+Q14. In the equilibrium PCI5(g) PCI3(g) + CI2(g) H=90 kJ/ mol
PCI3(g) + CI2(g) H=90 kJ/ mol (a) The position of equilibrium
(b) Equilibrium constant?
If (i) Temperature is increased
(ii) Volume of the container is decrease.
(iii) Catalyst is added
(iv) CI2 is added
See section 8.2
Explain your answer.
Q15. Synthesis of NH3 by Haber’s process is an exothermic reaction.
N2 + 3H2 2NH3 H=92.46 kJ/ mol
2NH3 H=92.46 kJ/ mol(a) What should be the possible effect of change of temperature at equilibrium stage?
(b) How does the change of pressure or volume shifts the equilibrium position of this reaction?
(c) What is the role of the catalyst in this reaction?
(d) What happens to equilibrium position of this reaction if NH3 is removed form the reaction vessel form time to time.
See Section 8.2.1
Q16. Sulphuric acid is the king of chemicals. It is produced by the burning of SO2 to SO3 through an exothermic reversible process.
(a) Write the balanced reversible reaction
(b) What is the effect of pressure change on this reaction?
(c) Reaction is exothermic but still the temperature of 400-500oC is required to increase the yield of SO3. Give reasons.
See Section 8.2.2
Q17. (a) What are buffer solutions? Why do we need them in daily life?
See Section 8.2
(b) How does the mixture of sodium acetate and acetic acid give us the acidic buffer?
See Section 8.7
(c) Explain that a mixture of NH4OH and NH4CI gives us the basic buffer?
See Section 8.7
(d) How do you justify that the greater quantity of CH3COONa in acetic acid decrease the dissociating power of acetic acid and so the pH increases.
CH3COOH is a weak acid and ionizes very small, while CH3COONa is a strong electrolyte and it ionizes in water to greater extent and provides acetate ions.
CH3COOH +H2O CH3COO- + H3O+
CH3COO- + H3O+CH3COONa CH3COO- + Na+
CH3COO- + Na+ Thus CH3COONa decreases the ionization of CH3COOH due to common CH3COO- ion and pH of solution increases.
(d) Explain the term buffer capacity.
See Section 8.7.1
Q18. (a) What is the solubility product? Derive the solubility product expression for sparingly soluble compounds, AgCI, Ag2CrO4 and PbCI2.
See Section 8.8
(b) How do you determine the solubility product of a substance when its solubility is provided in grams/100 g of water?
See Section 8.8
(c) How do you calculate the solubility of a substance from the value of solubility product.
See Section 8.8
Q19. Kc value for volume for the following reaction is 0.076 at 520oC
2HI H2 + I2
H2 + I2 Equilibrium mixture constants [HI] =0.08 M, [H2] =0.01 M. To this mixture more HI is added so that its new concentration is 0.096 M . what will be the concentration of [HI], [H2] and [I2] when equilibrium is re-established.
Solution
2HI H2 + I2
H2 + I2 Equilibrium conc. 0.08 0.01 0.01
(mol/dm3)
Initial conc. 0.096 0.01 0.01
After adding more HI
(mol/dm3)
Equilibrium conc. 0.096 – 2x 0.01 0.01
When equilibrium is re-established
(mol/dm3)
According to law of mass action
Kc =
Kc = 
=0.016
=0.016 =
=0.016
=0.016 Taking square root on both sides
=
=0.126
=0.1260.01 + x=0.126 (0.096 – 2x)
0.01 + x =0.0121 – 0.252 x
x + 0.252 x=0.0121 – 0.01
1.252 x=0.0021
x ==
x=0.00168 mol/dm-3
Thus
Concentrations when equilibrium is re-established are
[H2] =0.01 + x =0.01 +0.00168 =0.01168 mol dm-3
[I2] =0.01 + x =0.01 +0.00168 =0.01168 mol dm-3
[HI] =0.096 - 2 x =0.096 -2x0.00168 =0.0926 mol dm-3
Q20. The equilibrium constant for the reaction between acetic and ethyl alcohol is 4. A mixture of 3 moles of acetic acid and one mole of C2H5OH is allowed to come to equilibrium. Calculate the amount of ethyl acetate at equilibrium stage in number of moles and grams. Also, calculate the masses of reactants left behind.
Solution
CH3COOH + C2H5OH CH3 COOC2H5 +H2O
CH3 COOC2H5 +H2O Initial conc. 3 1 0 0
(mol/dm3)
Equilibrium conc. 3- x 1 – x x x
(mol/dm3)
According to law of mass action
Kc=
Kc= 
X2 =4(3 – x) (1 – x)
X2 =4(3 – 3x) (1 – x)
X2 =4(3 –4x + – x2)
X2 =12 –16x + 4 x2)
Or 12 – 16x + 4x2 – x2 =0
3x2 – 16x + 12=0
It is quadratic equation and can be solved by using quadric formula
Here a = 3 , b = 12 , c = 12
Thus
x=
x=
x=
x=
x=
Either x= or x=
or x= x= 4.43 mol dm-3 or x= 0.9 mol dm-3
x= 4.43 is not possible as it is greater than the initial concentrations of reactants, therefore , x= 0.9 mol dm-3
Therefore
Moles of ethyl acetate =x=0.9 moles
Mass of ethyl acetate = 0.9x88=79.46g.
Moles of water =x=0.9 moles
Mass of water =0.9 x 18=16.2 g.
Moles of acetic acid = 3 – x =3 – 0.9 =2.1 moles
(left behind)
Moles of ethyl alcohol = 1 – x =1 – 0.9= 0.1 moles
(Left behind)
Mass of ethyl alcohol =0.1 x 46=4.6g.
(left behind)
Q21. study the equilibrium
H2O + CO H2 + CO2
H2 + CO2 (a) Write an expression of Kp
Kp= 
(b) When 1 mole of steam and 1 mole of CO are allowed to reach equilibrium, 33.3% of the equilibrium mixture is hydrogen. Calculate the value of Kp. state the units of Kp.
Solution
H2O + CO H2 + CO2
H2 + CO2 Initial conc. 1 1 0 0
(mol/dm3)
Equilibrium conc. 1 – x 1 – x x x
(mol/dm3)
Total no. of moles at equilibrium =1 – x + 1 – x + x + x=2
Hence
% of H2 =
33.3=
or no of moles of H2 
=0.67 moles
=0.67 molesHence
At equilibrium
Moles of H2 =x=0.67 moles
Moles of CO2 =x=0.67 moles
Moles of H2O =1 – x = 1 - 0.67= 0.33 moles
Moles of CO =1 – x = 1 - 0.67= 0.33 moles
Hence, Kc is given as
Kc = 
Kc = 
Since Kp =Kc (RT) andn =n products -n rectants =0, therefore
n =n products -n rectants =0, therefore Kp=Kc=4
Q22. Calculate the pH of
(a) 10-4 moles/ dm3 of HCI
HCI ionizes as
HCI H+ + CI-
H+ + CI- Since HCI is a strong acid, and it is 100% dissociated. Hence 10-4 mol/dm3 of HCI produces 10-4 mol/dm3 of H+ ions.
Thus
[H+] = 10-4 mol/dm3
So
pH = - log [H+]
pH = - log [10-4]
pH = 4
(b) 10-4 moles/dm3 of Ba(OH)2
Ba(OH)2 ionizes as
Ba(OH)2 Ba+2 + 2 OH-
Ba+2 + 2 Since Ba(OH)2 is a strong base it is 100% dissociated.
Hence 10-4 mol/dm3 of Ba(OH)2 produces 2 x 10-4 mol/dm3 of OH-
Thus
[OH-
So
pOH= - log[OH-
pOH= - log[2 x 10-4]
pOH= 3.699
Since
pH + pOH=14
Therefore
pH= 14 – pOH
=14 – 3.699 =10.301
(c) 1 mol/dm3 of H2X, which is 50% dissociated
H2X ionizes as
H2X 2H+ + X-
2H+ + X- 1mole of H2X produces 2 moles of H+ ions if 100% dissociated
However, since H2X is 50% dissociated therefore 1 mole of H2X produces 1 mole of H+ ion
Thus
[H+]=1 mol/dm3
So
pH= -log [H+]
pH= -log [1]
pH= 0
(d) 1 mol/dm3 of NH4 OH which is 1% dissociated
NH4 OH ionizes as
NH4OH NH4+ + OH-
NH4+ + It shows that 1 mole of NH4OH produces 1 mole of OH-
NH4OH is only 1% dissociated
Hence
% dissociation =
x 100
x 100 1 =
x 100
x 100 mol of OH- 
x1=0.01 mol/dm3
x1=0.01 mol/dm3 Thus
[OH-
So
pOH = - log [OH-
pOH = - log [0.01]
pOH = 2
Since
pH + pOH =14
Therefore
pH=14 – pOH
=14 – 2=12
Q23. (a) Benzoic aicd C6H5COOH is weak mono-basic acid (Ka=6.4 x10-5mol/dm3). What is the pH of a solution containing 7.2 g of sodium benzoate in one dm3 of o.02 mol/dm3 benzoic acid.
Mass of sodium benzoate =7.2 g /dm3
| Formula of sodium benzoate is C6H5COONa |
Mol. Mass of sodium benzoate =144 g/mol
Moles of sodium benzoate =
=0.05 mol/dm3
=0.05 mol/dm3 Moles of benzoic acid =0.02 mol/dm3
Ka of benzoic acid 6.4 x 10-5 mol/dm3
Thus pKa = - log Ka= - log (6.4 x10-5 ) =4.2
Hence, according to Henderson
pH =pKa + log
pH =4.2 + log 
pH = 4.2 + 0.39
pH =4.59
(b) A buffer solution has been prepared by mixing 0.2 M CH3COONa and 0.5 M CH3 COOH in 1 dm3 of solution. calculate the pH of solution. pKa of acid is 4.74 at 25oC. How the value of pH will change by adding 0.1 mole 0.1 mole of NaOH and 0.1 mole mol HCI respectively.
Solution
[CH3 COOH] =0.5 M
[CH3 COONa] =0.2 M
pKa of CH3 COOH =4.74
pH =?
Since pH =pKa + log 
or pH =pKa + log 
pH =4.74 + log 
pH =4.74 - 0.4
pH =4.34
When 0.1 mole of NaOH is added
NaOH is strong base. It dissociates completely. Therefore, it produces 0.1 moles of OH- OH-
On the other hand, due to slat formed by the neutralization reaction, conc. of salt (CH3 COONa) is increased from 0.2 moles to 0.3 moles.
Hence, new conc. will be
[CH3 COOH] =0.4 M [CH3 COONa] =0.3M
Thus pH =pKa + log
pH =4.74 + log 
pH =4.74 - 0.12
pH =4.62
Addition of 0.1 mole of HCI
HCI is a strong acid. It dissociates completely. Therefore, it produces 0.1 moles of H+ ions. Thus, 0.1 moles of H+ ions react with 0.1 moles of CH3COO- ions. Hence, out of 0.2 moles of salt (CH3COO-Na+), 0.1 moles of salt are left behind.
On the other hand, conc. of acid (CH3COOH) is increased form 0.5 moles to 0.6 moles.
Hence, new conc. will be
[CH3 COOH] =0.6 M [CH3 COONa] =0.1M
Thus pH =pKa + log
pH =4.74 + log 
pH =4.74 - 0.78
pH =3.96
(See Section 8.7.1 for complete understanding of this numerical)
Q24. Solubility of CaF2 in water at 25OC is found to be 2.05 x 10-4 mol dm-3. What is the value of Ksp at this T.
Solubility of CaF2 = 2.05 x 10-4 mol dm-3
According to balanced chemical eq.
CaF2(aq) Ca2+(aq) + 2F-(aq)
Ca2+(aq) + 2F-(aq) At initial stage 2.05 x 10-4 0 0
(mol/dm3)
After solubility 0 2.05 x 10-4 2x2.05 x 10-4
Hence Ksp=[Ca+2][F-]
Ksp=[0.05 x 10-4][2 x 2.05 x 10-4]2
Ksp=3.446 x 10-11 mol3 dm-9
Q25. The solubility product of Ag2CrO4 is 2.6 x 120-2 at 25oC. Calculate the solubility of the compound.
Ksp of Ag2 CrO4 =2.6 x 10-2
We know
Ag2CrO4(aq) 2Ag+ CrO42-(aq)
2Ag+ CrO42-(aq)Initial stage (mol/dm3) Ag2CrO4 0 0
At equilibrium (mol/dm3) Ag2CrO4 2S S
Hence
Ksp =[Ag+]2[CrO42-]
Ksp=[2S]2[S] =2.6 x 10-2
4[S]3 =2.6 x 10-2
[S]= 
or [S]=0.1866 mol/dm3
Hence at equilibrium
[Ag+] =2 x 0.1866 mol/dm3 =0.3732 mol/dm3
and [CrO42- ] =0.1866 mol/dm3
Since
1 mole of Ag2CrO4 gives 1 mole of CrO42-ions, hence
Solubility of Ag2CrO4=[CrO42-]=0.1866 mol/dm3
7 comments:
what an effort>>>>>>GOOOD YARR.....excellent
very goooood
plz review ur short question no.5
thank u ........this is very helpful
Sir plz provide us with short question of each chapter
thnak u very very much sir ji ap ny hamary liye itna sab kuch kia thank you
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