{"id":9111,"date":"2019-11-29T14:37:22","date_gmt":"2019-11-29T09:07:22","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9111"},"modified":"2021-07-02T15:44:28","modified_gmt":"2021-07-02T10:14:28","slug":"1st-puc-chemistry-question-bank-chapter-7","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-question-bank-chapter-7\/","title":{"rendered":"1st PUC Chemistry Question Bank Chapter 7 Equilibrium"},"content":{"rendered":"

You can Download Chapter 7 Equilibrium Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Question Bank Chapter 7 Equilibrium<\/h2>\n

1st PUC Chemistry Equilibrium One Mark Questions and Answers<\/h3>\n

Question 1.
\nMention factors affecting rate (velocity) of reaction.
\nAnswer:<\/p>\n

    \n
  1. Nature of reactants<\/li>\n
  2. Catalyst<\/li>\n
  3. Temperature<\/li>\n
  4. Pressure<\/li>\n
  5. Concentration of reactants.<\/li>\n<\/ol>\n

    Question 2.
    \nDefine Law of mass action.
    \nAnswer:
    \n\u201cAt a given temperature the rate of chemical reaction is directly proportional to the product of active masses of the reactant.\u201d<\/p>\n

    Question 3.
    \nDefine equilibrium constant of reaction.
    \nAnswer:
    \n\u201cIs the ratio of velocity constant of forward reaction to velocity constant of backward reaction.\u201d Also defined as, \u201cIt is the ratio of product to molar concentration of product to the product of molar concentration of the reactants\u201d.<\/p>\n

    \"KSEEB<\/p>\n

    Question 4.
    \nWrite expression for Kc & KP for the reaction.
    \n\"1st
    \nAnswer:
    \n\"1st<\/p>\n

    Question 5.
    \nWrite the relationship between Kp<\/sub> & Kc<\/sub>.
    \nAnswer:
    \nKp<\/sub> = Kc <\/sub>[RT] \u0394n<\/sup>
    \nWhere \u0394n is = (number of gaseous product – number of gaseous reactant)<\/p>\n

    Question 6.
    \nDefine Le-Chatelier\u2019s principle.
    \nAnswer:
    \nDefinition : \u201cwhen a constraint applied to a system at equilibrium in a reversible reaction, the equilibrium shifts so as to nullify the constraint\u201d.
    \n[Constraint is change in temperature or pressure, addition of reactant or product]<\/p>\n

    Question 7.
    \nWhat is active mass ?
    \nAnswer:
    \nActive mass means molar concentration i.e., concentration expressed in mol\/dm3.<\/p>\n

    Question 8.
    \nWhat happens when the temperature of a reversible reaction at equilibrium is increased, if enthalpy change is positive?
    \nAnswer:
    \nEquilibrium shifts to the right.<\/p>\n

    \"KSEEB<\/p>\n

    Question 9.
    \nGive an example for a reversible reaction in which Kp<\/sub> = Kc<\/sub>RT.
    \nAnswer:
    \nPCI5(g) <\/sub>\u21cc PCI3(g)<\/sub>+CI2(g)<\/sub><\/p>\n

    Question 10.
    \nWrite an expression for Kp<\/sub> for the following reaction: A + B \u21cc C + 2D.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 11.
    \nHow are KP and Kc related ? Mention the condition under which KP<\/sub>=KC<\/sub>.
    \nAnswer:
    \nKp<\/sub> = Kc<\/sub> (RT)\u0394n<\/sup> when \u0394n = 0 i.e. no. of moles of gaseous products = no. of moles of
    \ngaseous reactants, e.g., H2(g)<\/sub> + I2(g)<\/sub> \u21cc 2HI(g)<\/sub>Kp<\/sub> = Kc<\/sub><\/p>\n

    Question 12.
    \nGive an example of a reaction where Kp<\/sub> \u2260 Kc<\/sub>.
    \nAnswer:
    \nPCI5(g)<\/sub> \u21cc PCI3(g)<\/sub> + CI2(g)<\/sub><\/p>\n

    Question 13.
    \nHow does a catalyst influence the equilibrium constant of a reversible \u2018 reaction ?
    \nAnswer:
    \nA catalyst does not influence the equilibrium constant of reversible reaction.<\/p>\n

    Question 14.
    \nWhich measurable property becomes constant in water , ~~ water vapour
    \nequilibrium at constant temperature?
    \nAnswer:
    \nVapour pressure.<\/p>\n

    Question 15.
    \nGive one example of everyday life in which there is gas y..’ >. solution
    \nequilibrium.
    \nAnswer:
    \nSoda-water bottle.<\/p>\n

    Question 16.
    \nUnder what condition, reversible process becomes irreversible?
    \nAnswer:
    \nIf one of the products (gaseous) is allowed to escape out (i.e., in open vessel).<\/p>\n

    Question 17.
    \nWhat is the effect on the value of equilibrium constant on adding catalyst?
    \nAnswer:
    \nNo effect.<\/p>\n

    \"KSEEB<\/p>\n

    Question 18.
    \nWhat is the effect of increasing pressure on the equilibrium?
    \nN2<\/sub> + 3H2<\/sub> \u21cc 2NH3<\/sub> ?
    \nAnswer:
    \nEquilibrium will shift in the forward direction forming more of ammonia.<\/p>\n

    Question 19.
    \nWhat are the conditions for getting maximum yield of NHs by Haber\u2019s process?
    \nAnswer:
    \nHigh concentrations of N2 and H2, low temperature, high pressure.<\/p>\n

    Question 20.
    \nWhat happens to the dissociation of PCI5 in a closed vessel if helium gas is introduced into it at the same temperature?
    \nAnswer:
    \nNo effect.<\/p>\n

    Question 21.
    \nWhat is the law called which gives relationship between degree of dissociation of a weak electrolyte and its concentration in the solution?
    \nAnswer:
    \nOstwald\u2019s dilution law.<\/p>\n

    Question 22.
    \nWrite the expression for comparison of relative strengths of two weak acids in terms of their ionization constants.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 23.
    \nWhat do you mean by reversible process?
    \nAnswer:
    \nReversible process is a process in which products can react to form reactants back.<\/p>\n

    Question 24.
    \nDefine equilibrium.
    \nAnswer:
    \nEquilibrium is a stage at which rate of forward reaction is equal to rate of backward reaction.<\/p>\n

    Question 25.
    \nWhat is physical equilibrium? Give one example.
    \nAnswer:
    \nPhysical equilibrium is am equilibrium between two different physical states of same
    \nsubstance, e.g., H2<\/sub>O(s) \u21cc H2<\/sub>O(I)<\/p>\n

    Question 26.
    \nGive two examples for Acedic buffer.
    \nAnswer:
    \nA solution containing a weak acid and its salt with a strong base.
    \n(i) CH3<\/sub>COOH + CH3<\/sub>COONa
    \n(ii) HCOOH + HCOOK<\/p>\n

    \"KSEEB<\/p>\n

    Question 27.
    \nWrite expression of Kc for reaction N2 <\/sub>(g) + 3H2<\/sub> (g) \u21cc 2NH3<\/sub> (g). Give units of K.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 28.
    \nName the factors that affect equilibrium position of a reversible reaction.
    \nAnswer:
    \n(i) Temperature (ii) Pressure, (iii) Concentration, (iv) Catalyst.<\/p>\n

    Question 29.
    \nAt what temperature the solid, and liquid are in equilibrium under 1 atm. pressure?
    \nAnswer:
    \nMelting point or freezing point.<\/p>\n

    Question 30.
    \nDefine solubility of substance.
    \nAnswer:
    \nSolubility of substance is maximum amount of solute that can be dissolved lOOg of solvent at a given temperature.<\/p>\n

    Question 31.
    \nState Henry\u2019s law.
    \nAnswer:
    \nHenry\u2019s law states the solubility of gas is directly proportional to the partial pressure of the gas above the solution, i.e, mass of gas dissolved is directly proportional to pressure applied on gas.
    \nm = k x p where \u2018p\u2019 is pressure applied on the gas.<\/p>\n

    Question 32.
    \nWhat is the effect of temperature on solubility of gases in liquids?
    \nAnswer:
    \nSolubility of gases in liquids decreases with increase in temperature.<\/p>\n

    Question 33.
    \nWhat is the effect of increasing pressure in the reactions?
    \n(i) PCI5<\/sub> (g) \u21cc PCI3<\/sub> (g) + CI2<\/sub> (g)
    \n(ii) N2<\/sub> (g)+ O2<\/sub>(g) \u21cc 2NO(g).
    \nAnswer:
    \n(i) The equilibrium will shift in backward reaction.
    \n(ii) No effect.<\/p>\n

    \"KSEEB<\/p>\n

    Question 34.
    \nWhat is the effect of temperature on the reactions?
    \n(i) N2<\/sub>(g) + 3H2<\/sub>(g) \u21cc 2NH3<\/sub>(g) + Heat
    \n(ii) N2<\/sub>(g) + O2<\/sub>(g) \u21cc 2NO3<\/sub>(g) – Heat
    \nAnswer:
    \n(i) The increase in temperature will favour backward reaction because the reaction is exothermic.
    \n(ii) The increase in temperature favours forward reaction because the reaction is endothermic.<\/p>\n

    Question 35.
    \nWrite expression of KP for the reaction.
    \nN2<\/sub>O(g) \u21cc 2NO2<\/sub>(g)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 36.
    \nN2 <\/sub>(g) + 3H2 <\/sub>(g) \u21cc 2NH3 <\/sub>(g),K = 0.50 at 673K.
    \nWrite the equilibrium expression and equilibrium constant for reverse reaction.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 37.
    \nWhat is difference between strong electrolyte, and weak electrolyte?
    \nAnswer:
    \nStrong electrolytes can dissociate into ions completely in aqueous solution whereas weak electrolytes do not dissociate into ions completely and there exist an equilibrium between ioins and unionized molecules, called Ionic Equilibrium.<\/p>\n

    Question 38.
    \nDefine degree of ionization or dissociation.
    \nAnswer:
    \nIt is the ratio of number of ions or molecules produced to the total number of molecules.<\/p>\n

    Question 39.
    \nState Ostwald\u2019s dilution law.
    \nAnswer:
    \nOstwald\u2019s dilution law states that the degree of dissociation of weak electrolyte is inversely proportional to square root of its concentration.
    \n\"1st
    \nWhere Ka<\/sub> is acid dissociation constant and Kb is base dissociation constant.<\/p>\n

    \"KSEEB<\/p>\n

    Question 40.
    \nDefine ionic equilibrium.
    \nAnswer:
    \nThe equilibrium between ions and unionized molecules is called ionic equilibrium.<\/p>\n

    Question 41.
    \nUnder what conditions a weak electrolyte can \u00bbhave high degree of ionization?
    \nAnswer:
    \nOn dilution, i.e., at a very low concentration, weak electrolyte has high degree of ionization.<\/p>\n

    Question 42.
    \nWhat is the effect of temperature on degree of dissociation?
    \nAnswer:
    \nThe degree of Dissociation increases with increase in temperature.<\/p>\n

    Question 43.
    \nDefine pH. Give its mathematical expression.
    \nAnswer:
    \npH is defined as negative logarithm of [H2<\/sub>O+<\/sup>]concentration.<\/p>\n

    Question 44.
    \nWhat is meant by ionic product of water (Ka<\/sub>)?
    \nAnswer:
    \nIt is product of concentration of [H3<\/sub>O+<\/sup>] and [OH–<\/sup>] at a specific temperature.
    \nKw<\/sub> = [H3<\/sub>O+<\/sup> ][OH–<\/sup>] = 1.0 \u00d7 10-14<\/sup> at 298K<\/p>\n

    Question 45.
    \nIn qualitative analysis, on what basis cations are grouped?
    \nAnswer:
    \nCations are grouped on the basis of their solubility product (Ksp<\/sub>).<\/p>\n

    \"KSEEB<\/p>\n

    Question 46.
    \nWhat is the range of a pH indicator in terms of its dissociation constant (Kin)?
    \nAnswer:
    \npH = pKin<\/sub> \u00b11.<\/p>\n

    Question 47.
    \nWhat happens to the solubility of Agcl in water if NacI solution is added to it?
    \nAnswer:
    \nSolubility of AgCI decreases due to common ion effect.<\/p>\n

    Question 48.
    \nWhat is the realationship between pKa<\/sub> and pKb<\/sub> values where Ka<\/sub> and Kb<\/sub> represent ionization constants of the acid and its conjugate base respectively?
    \nAnswer:
    \npKa<\/sub> + pKb<\/sub> = pKw<\/sub> = 14.<\/p>\n

    Question 49.
    \nWhat is the relationship between pH and pOH?
    \nAnswer:
    \npH + pOH = pKw<\/sub> = 14.<\/p>\n

    Question 50.
    \nWrite the demerits of Brownsted-Lowry theory.
    \nAnswer:
    \nIt does not explained why CaO, NH3<\/sub> acts as base and BF3<\/sub> and CO2<\/sub> acts as acid.<\/p>\n

    Question 51.
    \nDefine acidolysis.
    \nAnswer:
    \nAcidolysis is the process where acidity of blood decreases from 7.3.<\/p>\n

    Question 52.
    \nDefine Alkolysis.
    \nAnswer:
    \nAlkolysis is the process where acidity of blood increases from 8.<\/p>\n

    \"KSEEB<\/p>\n

    Question 53.
    \nWhat do you mean by buffer solution?
    \nAnswer:
    \nThe solution which resist the pH or pOH when an acid or base is added to it. Buffer solution is a mixture of weak electrolyte and its salt.<\/p>\n

    Question 54.
    \nDefine buffer action.
    \nAnswer:
    \nThe ability or capacity of a buffer solution to resist the pH by addition of acid or base.<\/p>\n

    Question 55.
    \nMention the types of buffer solution.
    \nAnswer:
    \nacidic buffer
    \nbasic buffer
    \nneutral buffer.<\/p>\n

    Question 56.
    \nMention the uses of buffer solution.
    \nAnswer:
    \nIt is used in chemical industries like paper, sugar, pharmaceuticals.
    \nIt is used in all biological fluids to maintain constant pH by buffer action.<\/p>\n

    \"KSEEB<\/p>\n

    1st PUC Chemistry Equilibrium Two Marks Questions and Answers<\/h3>\n

    Question 1.
    \nExplain rate of reaction with its mathematical forms.
    \nRate (Velocity) of reaction:
    \nAnswer:
    \nIt is the ratio of change in concentration to unit interval of time.
    \nMathematically: Let us consider a general reaction, A (reactant) \u2192 B(products
    \n\"1st
    \nWhere dc = change in concentration : dt = change in time.
    \n-ve sign = forward reaction where concentration of reactant decreases.
    \n+ve sign = backward reaction where concentration of reactant increases.
    \nMeaning of [A] = – concentration of reaction A
    \n[B] = – concentration of Product B.<\/p>\n

    Question 2.
    \nDefine Rate equation with example.
    \nAnswer:
    \nAn expression which relates the rate (velocity) of the reaction with change in concentration of reactant in a given interval of time.
    \n\"1st<\/p>\n

    Question 3.
    \nDefine Irreversible reaction with example.
    \nAnswer:
    \nA Chemical reaction in which product can not able to give back the reactant are called Irreversible reaction.
    \nExample : C(s) + O2<\/sub> (g) \u2192 CO2<\/sub> (g)<\/p>\n

    Question 4.
    \nDefine Reversible reaction with example.
    \nAnswer:
    \nA Chemical reaction in which an infinitesimally small change in any one of the variable (Pressure, Temperature, Concentration of reactant or product) of the reaction is changed the direction (forward or backward) of the reaction changes.\u201d
    \nIt is repreented by half arrow mark. An Example is H2<\/sub> (g) +I2<\/sub> (g) \u21cc 2HI(g)<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nDefine Chemical Equilibrium with an example.
    \nAnswer:
    \nIs a state where the rate of forward reaction is equal to rate of backward reaction.\u201d OR is also defined as \u201c A state where concentration of reactant and product remains
    \nconstant. \u201d Example : H2<\/sub> (g) +I2<\/sub> (g) \u21cc HI(g)<\/p>\n

    Question 6.
    \nChemical equilibrium is dyanamic. Justify.
    \nAnswer:
    \nChemical equilibrium is dynamic : Because the velocities of forward and backward reactions are equal also the concentration of reactant and product unchanged at this equilibrium state, (but may not be equal). The equilibrium can be made to shift in either direction by changing the concentration. Pressure, Volume etc., Hence is dynamic in nature.<\/p>\n

    Question 7.
    \nWrite the characteristics of chemical equilibrium.
    \nAnswer:<\/p>\n

      \n
    • Chemical equilibrium is dynamic.<\/li>\n
    • Equilibrium reaction will not come to an end.<\/li>\n
    • Equilibrium is attained only in a closed system.<\/li>\n
    • Catalyst does not alter equilibrium state.<\/li>\n<\/ul>\n

      Question 8.
      \nThe equilibrium constant for a reaction is 26.66. If the rate constant of the forward reaction is 3.2 \u00d7 10-4<\/sup> s-1<\/sup>, what is the rate constant of the backward reaction?
      \nAnswer:
      \nData given,
      \nEquilibrium constant for a reaction Kc – 26.66
      \nRate constantof the forward reaction = k1<\/sub>= 3.2 \u00d7 10-4<\/sup> \/sec.
      \n\u2234 Rate constant of the backward reaction = k2<\/sub> = ?
      \nEquilibrium constant of a reaction \u2018Kc<\/sub>\u2019 is given by the formula.
      \n\"1st<\/p>\n

      Question 9.
      \nDefine Arrhenius acid – base theory with one example.
      \nAnswer:
      \nAn electrolyte which when dissolved in water, produces H+<\/sup> ion is called acid.
      \nAn electrolyte which when dissolved in water, produces OH– <\/sup> ion is called base.
      \n\"1st<\/p>\n

      Question 10.
      \nDefine Bronsted Lowry theory OR Protonic theory with one example.
      \nAnswer:
      \nA substance or molecule which donate proton is called acid [Protogenic]
      \nA substance or molecule which accepts proton is called base [Protophillic]
      \nExample:
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 11.
      \nWhat are amphoteric substance? Give one example.
      \nAnswer:
      \nA substance or molecule which acts both as acid as well as base are called amphoteric.
      \nExample:
      \n\"1st<\/p>\n

      Question 12.
      \nWhat are conjugate acid-base pair? Give one example.
      \nAnswer:
      \nAcid – base pairs which differed by one proton is called conjugate acid – base pair.
      \nNH3<\/sub>+H2<\/sub>O \u2192 NH4<\/sub>+<\/sup> + OH–<\/sup>
      \nConjugate base of acid H2<\/sub>O is OH–<\/sup> ; H2<\/sub>O + H2<\/sub>O \u2192 H3<\/sub>O+<\/sup> + OH–<\/sup>
      \nConjugate acid of base NH3<\/sub> is NH3<\/sub>+<\/sup> ; NH3<\/sub> +NH3<\/sub> \u2192 NH4<\/sub>+<\/sup> +NH2<\/sub>–<\/sup>.<\/p>\n

      Question 13.
      \nExplain Lewi’s electron acid – base concept with an example.
      \nAnswer:
      \nAccording to Lewi’s theory, acid is one which accepts pair of electron and base is a – substance which donates pair of electron.
      \n\"1st<\/p>\n

      Question 14.
      \nCalculate pH of 0.01 M HC1.
      \nAnswer:
      \npH = -log10<\/sub>[H+<\/sup>] = -log10<\/sub>[0.01] = -log10<\/sub>[10-2<\/sup>] = 2<\/p>\n

      Question 15.
      \nCalculate pH of 0.0001 M of HNO3.
      \nAnswer:
      \npH = -log10<\/sub>[H+<\/sup> ]-log10<\/sub>[0.0001] = -log10<\/sub>[10-4<\/sup>] = 4<\/p>\n

      Question 16.
      \nCalculate pH of 0.00025 M HNO3<\/sub>.
      \nAnswer:
      \npH = -log10<\/sub>[H+<\/sup>] = -Iog10<\/sub>[0.00025] – log10<\/sub> [2.5 \u00d7 10-4<\/sup>] = 4 – log2.5
      \npH =4 – 0.3979 = 3.6021<\/p>\n

      \"KSEEB<\/p>\n

      Question 17.
      \nCalculate pH of 0.1 M of H2SO4 (concentration 0.1 \u00d7 2 hydrogen = 0.2).
      \nAnswer:
      \npH = -log10<\/sub>[H+<\/sup>] = -log10<\/sub>[0.1 \u00d7 2] = -log10<\/sub>[0.2] = -log10<\/sub>[2 \u00d7 10-1<\/sup>] = 1 – log2
      \npH = 1 – 0.3010 = 0.699<\/p>\n

      Question 18.
      \nCalculate pH of 0.05 M HC1.
      \nAnswer:
      \npH = -log10<\/sub>[H+<\/sup>] = -log10<\/sub>[0.05] – log10<\/sub> [5 \u00d7 10-2<\/sup>] = 2 – log5 = 2 – 0.6990 =1.301<\/p>\n

      Question 19.
      \nCalculate pH of 0.005 M H2SO4<\/sub>.
      \nAnswer:
      \npH = -log10<\/sub>[H+<\/sup>] = -log10<\/sub>[0.005 \u00d7 2] = -log10<\/sub>[0.01] = -log10<\/sub>[ 1\u00a0\u00d7 10-2 <\/sup>]
      \npH = 2 – log 1 = 2 – 0 = 2<\/p>\n

      Question 20.
      \nCalculate pH of 3 \u00d7 10-9<\/sup> M NaOH.
      \nAnswer: :
      \npOH = -log10<\/sub>[OH–<\/sup>] = -log10<\/sub>[3 \u00d7 10-9<\/sup>] = 9 – log3 = 9 – 0.4771 = 8.5229<\/p>\n

      Question 21.
      \nCalculate [OH–<\/sup>] if pOH = 8.3.
      \nAnswer:
      \npOH =-log10<\/sub>[OH–<\/sup>]; 8.3 =-log10<\/sub>[OH–<\/sup>]
      \nTaking antilog on both the sides
      \n[OH–<\/sup>] = antilog (-8.3) = [OH–<\/sup>] = antilog (-9 – 8.3 + 9) = antilog (0.7) \u00d7 10-9<\/sup>]
      \n= 5.012 \u00d7 10-9<\/sup>\u00a0mol\/dm3<\/sup><\/p>\n

      Question 22.
      \nCalculate [H+<\/sup>] if pOH = 9.23.
      \nAnswer:
      \npH + pOH = 14 pH = 14 – pOH = 14 – 9.23 = 4.77
      \npH = -log10<\/sub>[H+<\/sup>] = -4.77 = log[H+<\/sup>]
      \nTaking antilog on both the sides
      \n[H+<\/sup>] = antilog (-4.77) = antilog [+5 -4.77 – 5] = antilog [0.23] \u00d7 10-5<\/sup>
      \n= 1.698 \u00d7 10-5<\/sup>] mol\/dm3<\/sup>]<\/p>\n

      Question 23.
      \nCalculate [OH–<\/sup>] if pH = 5.284
      \nAnswer:
      \npH = -log10<\/sub>[H+<\/sup>]; [H+<\/sup>] = antilog [-5.284] = antilog [6 – 5.284 – 6]
      \n= antilog (0.716 – 6) = 5.2 \u00d7 10-6<\/sup> mol\/dm3<\/sup>
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 24.
      \nWhat is the pH of a 0.05 M solution of formic acid ? (Ka<\/sub> = 1.8 \u00d7 10-4<\/sup>)
      \nAnswer:
      \n\"1st<\/p>\n

      Question 25.
      \nCalculate the [OH ] of a solution whose pH is 9.62.
      \nAnswer:
      \npH + pOH = 14 pOH = 14 – pH = 14 – 9.62 = 4.38
      \npOH = 4.38 = -log10<\/sub>[OH–<\/sup>]; 4.38 = -log10<\/sub>[OH”]
      \nlog10<\/sub>[OH–<\/sup>] = -4.38; [OH-] = antilog (-4.38) = antilog (5 – 4.38 – 5)
      \n= antilog (0.62 – 5) = antilog (0.62) \u00d7 10-5<\/sup> = 4.169 \u00d7 10-5<\/sup> mol dm-3<\/sup>.<\/p>\n

      Question 26.
      \nThe pOH of a solution is 5.725. Calculate the [H+<\/sup>].
      \nAnswer:
      \npH + pOH = 14 pH = 14 – pOH = 14 – 5.725 = 8.275
      \npH = -log10<\/sub>[H+<\/sup>] \u21d2 8.275 = log10<\/sub>[H+<\/sup>] \u21d2
      \n[H+<\/sup>] = antilog (-8.275) = antilog(9 – 8.275 – 9).
      \n[H+<\/sup>] = antilog (0.275 – 9) = 5.309 \u00d7 10-9<\/sup> mol dm-3<\/sup>.<\/p>\n

      Question 27.
      \nCalculate the pH of 0.125M H2<\/sub>SO4<\/sub>.
      \nAnswer:
      \nH2<\/sub>SO4 <\/sub>\u2192 2H+<\/sup> +SO4<\/sub>2-<\/sup>
      \n[H+<\/sup>] = 2[H2<\/sub>SO4<\/sub>] = 2 \u00d7 0.125 = 0.25
      \npH = -log10<\/sub>[H+<\/sup>] = -log10<\/sub>[0.25] = -log10 <\/sub>[2.5 \u00d7 10-1<\/sup>]
      \npH =-log2.5-log10-1<\/sup> =-0.3979+1 = 0.6021.<\/p>\n

      Question 28.
      \nSolubility product of barium sulphide (BaS04) is 2.4 \u00d7 10-9<\/sup>. Calculate it\u2019s solubility.
      \nAnswer:
      \nBaSO4 <\/sub>is AB type of salt
      \nKsp = 2.4 \u00d7 10-9<\/sup>
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 29.
      \nIf the solubility product of silver chloride is 1.8 \u00d7 10-10<\/sup>. What is the solubility of siliver ion if concentration of Cl–<\/sup> is 0.01 molar.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 30.
      \nGive any two differences between strong and weak electrolyte.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 31.
      \nWhat is common ion effect ? Give example.
      \nAnswer:
      \nSupression in degree of dissociation of a weak electrolyte by addition of a common ion is called common ion effect.
      \nExample : CH3<\/sub>COOH and CH3<\/sub>COONa<\/p>\n

      \"KSEEB<\/p>\n

      1st PUC Chemistry Equilibrium Three Marks Questions and Answers<\/h3>\n

      Question 1.
      \nSome processes are given below. What happens to the process if it subjected
      \nto a change given in the brackets?
      \n(i)
      \n\"1st
      \n(ii) Dissolution of NaOH in water (Temperature is increased)
      \n(iii) N2<\/sub> (g) + O2<\/sub> (g) \u21cc 2NO(g) -180.7kJ (Pressure is increased and temperature is decreased)
      \nAnswer:
      \n(i) Equilibrium will shift in the forward direction, i.e., more of ice will melt.
      \n(ii) Solubility will decrease because it is an exothermic process.
      \n(iii) Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction.<\/p>\n

      \"KSEEB<\/p>\n

      Question 2.
      \nWhat qualitative information can you obtain from the magnitude of equilibrium constant?
      \nAnswer:
      \n(i) Large value of equilibrium constant (>103<\/sup>) shows that forward reaction is favoured, i. e., concentration of products is much larger than that of the reactants at equilibrium.
      \n(ii) Intermediate value of K(10-3<\/sup> to 103<\/sup>) shows that the concentration of the reactants and products are comparable,
      \n(iii) Low value of K(<10-3<\/sup>)shows that backward reaction- is favoured, i.e, concentration of reactants is much larger than that of the products.<\/p>\n

      Question 3.
      \nHow do you explain detection of I group basic radicals based on application and solubility product.
      \nAnswer:
      \nGroup reagent: Dil. HCl \u2192 H+<\/sup> + Cl–<\/sup>
      \nRedicals present: Pb2+<\/sup>, Hg2+<\/sup>, Cd2+<\/sup>
      \nThe group reagent HCl is a strong electrolyte under goes complete dissociation. The metal radicals forms it’s respective chlorides [PbCl2<\/sub>, HgCl2<\/sub>, CdCl2<\/sub> ] Whose solubility product is less than ionic product. Hence, forms pricipitate.<\/p>\n

      Question 4.
      \nHow do you explain detection of H group basic radicals based on application of solubility product.
      \nAnswer:
      \nGroup reagent: HCl \u2192 H+<\/sup> + Cl–<\/sup>; H2<\/sub>S \u21cc 2H+<\/sup> + S2-<\/sup>
      \nRadicals: Cu2+<\/sup>, Bi2+<\/sup>, Sn2+<\/sup><\/p>\n

      Here HCl is a strong electrolyte under goes complete dissociation where as H2S is a weak electrolyte under goes partial dissociation and attains equilibrium between it’s ions and molecules.<\/p>\n

      In this solution, because of common ion (H+<\/sup>) effect supression of degree of dissociation of H2S takes place. As a result [S2-<\/sup>] decreases. Hence the metal radical forms it’s respective sulphides [CuS, SnS, BiS] whose solubility product is less than ionic product. Hence forms pricipitate.<\/p>\n

      \"KSEEB<\/p>\n

      Question 5.
      \nHow do you explain the detection of HI group basic radicals based on application of solubility product ?
      \nAnswer:
      \nGroup reagent: NH4<\/sub>OH \u21cc NH4<\/sub>+<\/sup> + OH–<\/sup> ; NH4<\/sub>Cl \u21cc NH4<\/sub>+<\/sup> + Cl–<\/sup>
      \nRadicals : Al3+<\/sup>, Cr3+<\/sup>, Fe3+<\/sup>
      \nHere NH4<\/sub>Cl is a strong electrolyte under goes complete dissociation where as NH4<\/sub>OH is a weak electrolyte under goes partial dissociation and attains equillibrium between it’s ions and molecules.<\/p>\n

      In this solution because of common ion (NH4<\/sub>+<\/sup>) effect supression of degree of dissociation of NH4<\/sub>OH takes place. As a result [OH–<\/sup>] decreases. Hence the metal radical forms it’s respective hydroxides [Al(OH)3<\/sub>, Cr(OH)3<\/sub>, Fe(OH)3<\/sub>] Whose solubility product is less than ionic product. Hence forms precipitate.<\/p>\n

      Question 6.
      \nExplain detection of IV group basic radicals using application of solubility product.
      \nAnswer:
      \nGroup reagent: NH4<\/sub>Cl \u2192 NH4<\/sub>+<\/sup> + Cl–<\/sup> ; NH4<\/sub>OH \u21cc NH4<\/sub>+<\/sup> + OH–<\/sup>
      \nH2<\/sub>S \u21cc 2H+<\/sup> + S2-<\/sup> ; H+<\/sup> + OH–<\/sup> \u2192 H2<\/sub>O
      \nRadicals: Zn2+<\/sup>, Mn2+<\/sup><\/p>\n

      Here, NH4<\/sub>Cl is a strong electrolyte undergoes complete dissociation, where as NH4<\/sub>OH – and H2<\/sub>S are weak electrolytes undergoes partial dissociation and attains equillibrium between their ions and molecules.<\/p>\n

      Along with common ion (NH4<\/sub>+<\/sup>) effect, the H+<\/sup> and OH–<\/sup> ion combines to form water molecules.
      \nAt the same time the metals forms their respective sulphides [ZnS, MnS] Whose solubility product is less than ionic product. Hence, forms precipitate.<\/p>\n

      Question 7.
      \nDefine Arrhenius acid – base theory with one example.
      \nAnswer:
      \nAn electrolyte which when dissolved in water, produces H+ ion is called acid.
      \nAn electrolyte which when dissolved in water, produces OH- ion is called base.
      \n\"1st<\/p>\n

      Question 8.
      \nDeduce equation for Ostwalds dilution law with respect to the weak acid.
      \nAnswer:
      \nConsider a weak acid HA with concentration C mol\/dm3<\/sup>. After a time interval it reaches equillibrium with dissociation x mol\/dm3 as follows.
      \n\"1st
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 9.
      \nDeduce equation for Ostwalds dilutions law with respect to weak base.
      \nAnswer:
      \nConsider a weak base BOH with concentration C mol\/dm3<\/sup>. After a time internal it reaches equilibrium with dissociation x mol\/dm3<\/sup> as follows.
      \n\"1st
      \n\"1st<\/p>\n

      (ii) Where, Kb is dissociation constant of weak base.<\/p>\n

      Question 10.
      \nDerive ionic product of water. Also find the value at 25 \u00b0C
      \nAnswer:
      \n\"1st
      \nWhere, Kw<\/sub> is ionic product of water.
      \nValue of Kw<\/sub> at 25\u00b0C :
      \nIt is found that at 25\u00b0C [H+<\/sup>] = [OH–<\/sup>] = 10-7<\/sup> mol\/dm3<\/sup>
      \n\u2234Kw<\/sub> = [H+<\/sup>][OH–<\/sup>] = 10-7<\/sup> \u00d7 10-7<\/sup> = 10-14<\/sup> (mol\/dm3<\/sup>)2<\/sup>
      \n(iii) At 25\u00b0C the value of Kw is 10-14<\/sup>( tmol\/dm3<\/sup>)2\u00a0<\/sup><\/sup><\/p>\n

      \"KSEEB<\/p>\n

      1st PUC Chemistry Equilibrium Five Marks Questions and Answers<\/h3>\n

      Question 1.
      \nHow do you apply law of mass action to a gaseous reversible reaction.
      \nAnswer:
      \nIf the reactants and products are in gaseous state, then for the reaction,
      \n\"1st
      \nThe total pressure P = PA<\/sub> + PB<\/sub> + PC<\/sub> + PD<\/sub>
      \nIf PA<\/sub> PB<\/sub> PC<\/sub> & PD<\/sub> are represents the partial pressure exerted by A, B, C and D respectively, then according to law of mass action,
      \nVelocity of forward reaction Vf<\/sub> \u221d proportional PA<\/sub> PB<\/sub> Or
      \nVf<\/sub>=Kf <\/sub>PA<\/sub>PB<\/sub>………….(3)
      \nWhere Kf velocity (proportionality) constant of forward reaction.
      \nVelocity of backward reaction Vb<\/sub> \u221d PC<\/sub> PD<\/sub> Or
      \nVb<\/sub> = Kb<\/sub> PC<\/sub>PD<\/sub> ……….(4)
      \nWhere Kb<\/sub> velocity (proportionality) constant of backward reaction.
      \nAt equilibrium, velocity of forward reaction is equal to velocity of backward reaction, from (3) & (4).
      \nVf<\/sub> = Vb<\/sub> Then
      \n\"1st
      \nWhere KP<\/sub> is equilibrium constant with respect topartial pressure of the reactant.<\/p>\n

      Question 2.
      \nWrite the characteristics of Equilibrium Constant (KP<\/sub> & KC<\/sub>).
      \nAnswer:<\/p>\n

        \n
      • It is independent of initial concentrations of the reactants, the direction from (reactant or product) which equilibrium attained.<\/li>\n
      • The value of equilibrium constant independent of Volume, Pressure, and presence of catalyst or any inert material.<\/li>\n
      • Equilibrium constant gives relative idea about the extent to whick the reaction takes place.<\/li>\n
      • At a given temperature, the equilibrium constant of a particular reaction has definite value.<\/li>\n
      • If its (Kc or Kp) value is large, then the forward reaction takes place nearly to completion whereas backward reaction to a small extent Vice versa.<\/li>\n
      • If the forward reaction is exothermic, then the value of Kc or KP decreases. Vice versa.<\/li>\n
      • In a reversible reaction, the value of equilibrium constant of forward reaction is reciprocal of backward reaction, ieKf=1\/Kb.<\/li>\n
      • The value of equilibrium constant depends on the stoichiometry of reactants, and products at equilibrium.
        \n\"1st<\/li>\n<\/ul>\n

        Question 3.
        \nWrite the conjugate acid for the following (a) Cl–<\/sup>, (b) NO3<\/sub>–<\/sup>, (c) HSO4<\/sub>–<\/sup>, (d) HCO3<\/sub>–<\/sup>, (e) SO4<\/sub>2-<\/sup>, (f) CO3<\/sub>2-<\/sup>, (g) NH2-<\/sup>, (h) N3-<\/sup>, (i) O2-<\/sup>, (j) H2<\/sub>O, (k) NH3<\/sub>.
        \nAnswer:
        \n\"1st<\/p>\n

        Question 4.
        \nDeduce Handersen equation for basic buffer.
        \nAnswer:
        \nConsider the basic buffer, a mixture of ammonium hydroxide and ammonium Chloride.
        \n\"1st<\/p>\n

        \"KSEEB<\/p>\n

        Question 5.
        \nDeduce Hendersons equation for acidic buffer.
        \nAnswer:
        \nConsider the acidic buffer, a mixture of acetic acid and sodium acetate as.
        \nCH3<\/sub>COOH \u21cc CH3<\/sub>COO–<\/sup> + H+<\/sup> \u2192 (1)
        \nCH3<\/sub>COONa \u2192 CH3<\/sub>COO–<\/sup> + Na+<\/sup> \u2192(2)
        \n\"1st
        \nWhere, Ka<\/sub> is dissociation constant of weak acid.
        \nIn this solution, [ CH3<\/sub>COO–<\/sup> ] \u2245 [Salt] \u2192 (4)
        \n[CH3<\/sub>COOH–<\/sup>] \u2245 [Acid] \u2192 (5)
        \n\"1st<\/p>\n

        Question 6.
        \nExplain Mechanism or working of acid buffer:
        \nAnswer:
        \nConsider acidic buffer mixture,
        \nCH3<\/sub>COOH \u21cc CH3<\/sub>COO–<\/sup> + H+<\/sup> ; CH3<\/sub>COONa \u2192 CH3<\/sub>COO–<\/sup> + Na+
        \n<\/sup>Here, acetic acid is a weak electrolyte, in it’s solution there exists equillibrium between it’s ions and molecules. Where as sodium acetate completely dissociates into it’s ions.<\/p>\n

        Therefore, the buffer mixture contains large number of CH3<\/sub>COO–<\/sup> ions followed by Na+<\/sup>, H+<\/sup>, CH3<\/sub>COOH.<\/p>\n

        Case i) When an acid added to this solution :
        \nH+<\/sup> ion of the acid combines with CH3<\/sub>COO–<\/sup> ion in buffer solution and makes equillibrium with acetic acid.
        \nCH3<\/sub>COO–<\/sup> + H+<\/sup> \u21cc CH3COOH
        \nAs a result pH remains constant.<\/p>\n

        Case ii) When a base is added to this solution :
        \nOH–<\/sup> ion of the base combines with H+<\/sup> ion in buffer solution forms water molecule.
        \nH+<\/sup>+ OH–<\/sup> \u2192 H2<\/sub>O.
        \nAs a result pH remains constant.<\/p>\n

        Question 7.
        \nExplain the Mechanism or working of basic buffer.
        \nAnswer:
        \nConsider basic buffer solution,
        \nNH4<\/sub>OH \u21cc NH4<\/sub>+<\/sup> + OH–<\/sup> ; NH4<\/sub>Cl \u2192 NH4<\/sub> + Cl–<\/sup>
        \nHere, ammonium hydroxide is a weak electrolyte, in it’s solution there exist equillibrium between it’s ions and molecules.
        \nWhere ammonium chloride completely dissociates into it’s ions.
        \nTherefore, buffer solution contains large number of NH4<\/sub>+<\/sup>ions followed by Cl–<\/sup>, OH–<\/sup> and NH4<\/sub>OH.<\/p>\n

        Case (i) When a base is added to this solution:
        \nOH- ion of the base combines with NH4<\/sub>+<\/sup> ion of the buffer solution and makes equillibrium with ammonium hydroxide.
        \nNH4<\/sub>+<\/sup> + OH– <\/sup>\u21cc NH4<\/sub>OH; As a result pOH remains constant.<\/p>\n

        \"KSEEB<\/p>\n

        Case (ii) When an acid is added to this solution :
        \nH+<\/sup> ion of acid combines with OH–<\/sup> ion of buffer solution and forms water molecule. H+<\/sup> + OH–<\/sup> \u2192 H2<\/sub>O
        \nAs a result pOH remains constant.<\/p>\n

        Question 8.
        \nThe pKa<\/sub> value of acetic acid is 4.7447 at 25 \u00b0C. How would you obtain a buffer of acetic acid and sodium acetate with pH = 4?
        \nAnswer:
        \n\"1st
        \nSo by mixing sodium acetate and acetic acid in the ratio 0.18 : 1 mole per dm3, the buffer of pH = 4 is obtained.<\/p>\n

        Question 9.
        \nA buffer solution of pH = 4.7 is prepared from CH3<\/sub>COONa and CH3<\/sub>COOH. Dissociation constant of acetic acid is 1.75 \u00d7 10-5<\/sup>. Calculate the mole proportion of sodium acetate and acetic acid.
        \nAnswer:
        \n\"1st
        \n\"1st<\/p>\n

        Question 10.
        \nA buffer solution of pH 8.3 is prepared from nmnionium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8 \u00d7 10-5<\/sup>. What Is the mole proportion of ammonium chloride and ammonium hydroxide?
        \nAnswer:
        \n\"1st<\/p>\n

        Question 11.
        \n3.0 g of pure acetic acid and 4.1 g of anhydrous sodium acetate are dissolved together in water and the solution is made up to 500 ml. Calculate the pH of the solution. Given Ka<\/sub> of acetic acid is 1.75 \u00d7 10-5<\/sup>.
        \nAnswer:
        \nCalculation of concentration of acetic acid: mass\/dm3<\/sup> = N x g eq. mass
        \n\"1st
        \nMolecular wt. of CH3<\/sup>COOH = 12 \u00d7 2 + 1 \u00d7 4 + 16 x 2 = 24 + 04 + 32 = 60.
        \n[CH3<\/sup>COOH] \\(\\frac{3.0 \\times 2}{60}=\\frac{6.0}{60}\\) = 0.1N
        \nCalculation of concentration of sodium acetate
        \n\"1st<\/p>\n

        \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

        You can Download Chapter 7 Equilibrium Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Chemistry Question Bank Chapter 7 Equilibrium 1st PUC Chemistry Equilibrium One Mark Questions and Answers Question 1. Mention …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9111"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=9111"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9111\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9111"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9111"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9111"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}