1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 12 InterviewSolution

551.	In the following questions, two statements are given; one in assertion (A) column while the other in reason (R ) column. Examine the statements carefully and mark the correct answer according to the instructions given below: If both (A) and (R) are correct and (R) is the correct explanation of (A) If both (A) and (R) are correct but (R) is not the correct explanation of (A)If (A) is correct but (R) is wrongIf (A) is wrong but (R) is correct(A) PH of water increases with increases in temperature.(R) Kw of water increase with increase with increase in temperature
Answer» If both (A) and (R) are correct but (R) is not the correct explanation of (A)

Discussion

552.	In the following questions, two statements are given; one in assertion (A) column while the other in reason (R ) column. Examine the statements carefully and mark the correct answer according to the instructions given below: If both (A) and (R) are correct and (R) is the correct explanation of (A) If both (A) and (R) are correct but (R) is not the correct explanation of (A)If (A) is correct but (R) is wrongIf (A) is wrong but (R) is correct(A) Acetic acid is a weak acid(R) It has weak conjugate base
Answer» If (A) is correct but (R) is wrong

Discussion

553.	The reaction M `2A(g) + B (g) hArr 3 C (g) + D (g)` is begun with the concentrations of A and B both at intial value of 1.00 M. Whenequilibrium is reached , the concentrations of D is measured and is reached , the concentration of D is measured and found to be `025` M . The value for the equilibrium constant for this reaction is given by the expresionA. `[(075)^(3)(025)] div [(1.00)^(2)(1.00)]`B. `[(075)^(3) (025)]div [(050)^(2) (075)]`C. `[(075)^(3) (025)]div [(050)^(2) (075)]`D. `[(075)^(3) (025)] div [(075)^(2) (0*25)]`
Answer» Correct Answer - B ` 2 A (g) + B (g) hArr 3 C (g) + D (g)` `{:(,2A(g),+,B(g),hArr,3 C(g),+,D(g)),("Intial",1.0 M,,1.0 M,,0,,0),("conc.",,,,,,,),("Eqm.",1-(2xx0.25),,1-0.25,,3xx0.25,,0.25),("conc".,0.5 M,,0.75 M,,0.75 M,,):}` ` K=([C]^(3) [D])/([A]^(2) [B])= ((0.75)^(3) (0.25))/((0.5)^(2) (0.75))`

Discussion

554.	` K_(c) "for" CO(g) + H_(2) O (g) hArr CO_(2) (g) + H_(2) (g) " at " 986 ^(@)C " is " 063 . " A mixture of 1 mole of" H_(2)O (g) and" 3 moles " CO (g) " is allowed to react to an equilibrium . The equilibrium pressure is " 20 atm .`
Answer» `{:(,CO(g),+,H_(2)O,hArr,CO_(2)(g),+,H_(2)(g)),("Intial moles",3,,1,,0,,0),(" Moles at eqm.",3-x,,1-x,,x,,x):}` Total no. of moles at equilibrium `= 3-x x+1 - x + x + x = 4 ` ` K_(c) = (x xxx)/((3-x)(1-x)) , i.e., 063 = (x^(2))/(3 + x^(2) - 4x ) ` On solving , it gives `x = 0681 " " ( x =-(bpm sqrt(b^(2) - 4ac))/(2a) )` ` :. " Moles of " H_(2) " present at eqm . " = 0* 681 "mole" ` Total pressure at eqm. = 2 atm ( Given ) Total moles at eqm . = 4 ` :. p_(CO) = ( 3- 0681 )/4 xx 2 = 116 "atm " , p_(H_(2)O) = (1-0681)/4 xx 2 = 116 " atm " , p_(CO_(2)) = p_(H_(2)) = (0* 681)/4 xx 2 = 0* 34 "atm" `

Discussion

555.	The pH of a solution obtained by mixing 50 mL of 0.2 M HCl with 50 mL of 0.20 M `CH_(3) C O OH` isA. `0.30`B. `0.70`C. `1.00`D. `2.00`
Answer» Correct Answer - C Acetic acid being a weak acid ionizes to a small extent Its ionization is further suppressed in the presence of HCl. Hence, `H^(+)` ions are obtained mainly from HCl. 50 ml of 0.2 M HCl=`50xx0.2` millimoles = 10 millimoles Volume of solution after mixing = 100 mL `:.` Molar conc.of HCl `=(10)/(100)=0.1M = 10^(-1)M` `:. pH=1`

Discussion

556.	What is the pH of a solution obtained by mixing 10 mL of 0.1 M HCl and 40 mL 0.2 M `H_(2)SO_(4)` ?A. 0.74B. 7.4C. 4.68D. 0.468
Answer» Correct Answer - D Millimoles of `H^(+)` from `HCl=0.1xx10=1` Millimoles of `H^(+)` from `H_(2)SO_(4)=0.2xx40xx2=16` Conc. Of `H^(+)=("Millimoles")/("Volume")=(16+1)/(40+10)=(17)/(50)=0.34` `pH=-log[H^(+)]=-log(0.34)=0.468`

Discussion

557.	The pH of a solution obtained by mixing 100 mL of a solution pH=3 with 400 mL of a solution of pH=4 isA. 3- log 2.8B. 7- log 2.8C. 4- log 2.8D. 5- log 2.8
Answer» Correct Answer - C `N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))` `10^(-3)xx100+10^(-4)xx400=N_(3)(100+400)` or, `N_(3)=(0.1+0.04)/(500)=(0.14)/(500)=2.8xx10^(-4)M` `pH =-log (2.8xx10^(-4))=4-"log"2.8`

Discussion

558.	Reaction between nitrogen and oxygen takes place as follows : `2 N_(2) (g) + O_(2) (g) hArr 2 N_(2) O (g) ` If a mixture of` 0482` mol of `N_(2)` and` 0933` mol of `O_(2)` is placed in a reaction vessel of volume 10 L and allowed to form `N_(2)O ` at a temperature for which `K_(c) = 2*0 xx 10^(-37)` . Determine the composition of the equilibrium mixture.
Answer» ` {:(,2 N_(2) (g) ,+,O_(2)(g),hArr,2 N_(2)O(g)),("Intial",0482 "mol",,0933"mol",,),("At eqm.",0482 "mol",,0.933-x//2,,x),("Molar conc." ,(0482-x)/10,,(0933-x//2)/10,,x/10):}`, As K ` = 20 xx 10 ^(-37)` is very small , this means that the amount of `N_(2) and O_(2) ` reacted (x) is very very small . Hence , at equilibrium , we have `[N_(2)] = 0* 0482 "mol"L^(-1) , [O_(2)] = 0* 0933 "mol" L^(-1) , [N_(2)O] = 01 x ` ` :. K_(c) = (01 x)^(2) /((00482)^(2) (00933)) = 2* 0 xx 10^(-37) ` (Given) On solving , this gives `x = 66 xx 10^(20)` ` :. [N_(2)O ] = 01 x 6*6 xx 10^(-21) "mol"L^(-1)`

Discussion

559.	The value of `K_(w)` is `9.55xx10^(-14)` at a certain temperature. Calculate the pH of water at this temperature .
Answer» Here, we are given : `K_(w)= 9.55xx10^(-14)` Now, as for water, `[H_(3)O^(+)]=[OH^(-)] :. K_(w)=[H_(3)O^(+)][OH^(-)]=[H_(3)O^(+)][H_(3)O^(+)]=[H_(3)O^(+)]^(2)`, i.e., `[H_(3)O^(+)]^(2) = 9.55xx10^(-14) or [H_(3)O^(+)]=sqrt(9.55xx10^(-14))=3.09xx10^(-7)M` `:. pH = - log [H_(3)O^(+)]=-log (3.09xx10^(-7))=-[log 3.09+ log 10^(-7)]=- [ 0.49-7]=6.51`.

Discussion

560.	Calculate the pH of the solution obtained by mixing 100 `cm^(3)` of solution with pH = 3 with `400 cm^(3)` of solution with pH = 4 .
Answer» `100 cm^(3) ` of solution with pH = 3 contains `H^(+)=(10^(-3))/(1000)xx100=10^(-4)`mole `400cm^(3) ` of solution with pH = 4 contains `H^(+)=(10^(_4))/(1000)xx400=4xx10^(-5) ` mole Total `H^(+)=10^(-4)+4xx10^(-5)=10^(-4)(1+0.4)=1.4xx10^(4)`. Total volume = `500 cm^(3)` `:. [H^(+)]=(1.4xx10^(-4))/(500) xx1000M = 2.8xx10^(-4)M` `pH = - log (2.8xx10^(-4))=4-0.4472 ~~ 3.55`

Discussion

561.	Calculate the concentration of `H^(+)` (aq) in 0.2 M solution of HCN. Given that the dissociation constant of HCN in water is `4.9 xx 10^(-10)`.
Answer» Correct Answer - `9.9xx10^(-6)M`

Discussion

562.	If hydrogen ion concentration in a solution is `1xx10^(-5)` moles/litre, calculate the concentration of OH ion in this solution `(K_(w)=10^(-14)"moles"^(2)L^(-2)).`
Answer» `K_(w)=[H^(+)][OH^(-)]=10^(-14)` `[OH^(-)]=(10^(-14))/(10^(-5))=10^(-9)mol L^(-1)`

Discussion

563.	If hydrogen ion concentration in a solution is `1xx10^(-5)` moles/litre, calculate the concentration of `OH^(-)`ion in this solution `(K_(w)=10^(-14) "moles"^(2)//"litre"^(2))`
Answer» Correct Answer - `10^(-9)` moles/litre

Discussion

564.	Define strong and weak electrolyte.
Answer» Those electrolytes which dissociate almost completely into ions in aqueous solutions are Known as strong electrolytes while those which show poor dissociation into ions in aqueous solutions are called weak electrolytes.

Discussion

565.	What is the pH of a solution whose `H^(+)` ion concentration is `2xx10^(-5)g` ions/litre ?
Answer» Correct Answer - 4.699

Discussion

566.	If the equilibrium constant for a reaction is `4*0` , what will be the equilibrium constant for the reverse reaction.
Answer» Correct Answer - `1//4 = 0*25`

Discussion

567.	What is the value of ∆n for the given reaction.2NOCl (g) -------> 2NO (g) + Cl2 (g)
Answer» ∆ⁿ =n_p-n_r =1

Discussion

568.	On what factors does the value of the equilibrium constant of a reaction depend?
Answer» The equilibrium constant of a reaction depends upon (i) Temperature (ii) Pressure, & (iii) Stoichiometry of the reaction

568.

On what factors does the value of the equilibrium constant of a reaction depend?

Answer»

The equilibrium constant of a reaction depends upon

(i) Temperature

(ii) Pressure, &

(iii) Stoichiometry of the reaction

Discussion

569.	If the equilibrium constant for the reaction is 4.0, what will be the equilibrium constant for the reverse reaction.
Answer» Kc=1/4..........

Discussion

570.	What happens to the dissociation of PCl5 in a closed vessel if helium gas is introduced into it at the same temperature?
Answer» There is No effect.

Discussion

571.	What happens to the dissociation of PCl5 in a closed vessel if He gas is introduced into it at the same temperature?
Answer» No effect.........

Discussion

572.	Asseration : In the dissociation of `PCI_(5)` at constant pressure and temperature addition of helium at equilibrium increases the dissociation of `PCI_(5)`. Reason : Helium removes `CI_(2)` from the field of action.A. Statements -1 is true , statement -2 is also true, statement -2 is the correct explanation of statement -6B. Statement -1 is true , statement-2 is also true, statement-2 is not the correct explanation of statement-6C. Statement -1 is true, statement -2 is false.D. Statement -1 is false, satement-2 is true.
Answer» Correct Answer - C Correct satement-2 More of `PCI_(5)` dissociates since the volume of equilibrium mixture has increased.

Discussion

573.	In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.Assertion (A): In the dissociation of PCl5 at constant pressure and temperature addition of helium at equilibrium increases the dissociation of PCl5.Reason (R) : Helium removes Cl2 from the field of action.(i) Both A and R are true and R is correct explanation of A.(ii) Both A and R are true but R is not correct explanation of A.(iii) A is true but R is false.(iv) Both A and R are false.
Answer» (iv) Both A and R are false.

Discussion

574.	`{:("Column I",,"Column II"),((A) Fe(OH)_(3),,(i)K_(sp)=s^(2)),((B) Ag_(2)CrO_(4) ,, (ii)K_(sp) =27s^(4)),((C) CH_(3)COOAg,,(iii) K_(sp) =108s^(5)),((D)Ca_(3)(PO_(4))_(2),,(iv) K_(sp) =4s^(3)):}`A. `(A) to (III), (B) to (ii),(C) to (iii), (D) to (i)`B. `(A) to (ii), (B) to (iv),(C) to (i),(D) to (iii)`C. `(A) to (i), (B) to (iii), (C) to (ii) ,(D) to (iv)`D. `(A) to (iv) ,(B) to (i), (C) to (iii), (D) to (ii)`
Answer» Correct Answer - B `(A) Fe(OH)_(3) hArr underset((S))(Fe^(3+)) + underset((3S))(3OH^(-)) , K_(sp) =27s^(4)` `(B) Ag_(2)CrO_(4) hArr underset((2S))(2Ag^(+))+underset((3S))((CrO_(4))^(2-)) , K_(sp) =4s^(3)` ` (C) CH_(3)COOH hArr underset((S))(CH_(3)COO^(-)) +underset((S))(H^(+)) , K_(sp) =s^(2)` `(D) Ca_(3)(PO_(4))_(2) hArrunderset((3S))(3Ca^(2+))+ underset(2(S))(2(PO_(4))^(3-)),K_(sp) =108 s^(5)`

Discussion

575.	Asseration : In the dissociation of `PCI_(5)` at constant pressure and temperature addition of helium at equilibrium increases the dissociation of `PCI_(5)`. Reason : Helium removes `CI_(2)` from the field of action.A. Both A and R are true and R is correct explanation of A.B. Both A and R are ture but R is not correct explanation of A.C. A is true but R is false.D. Both A and are false.
Answer» Correct Answer - C Correct reason: Addition of a gas at constant pressure increases the volume. As a result, the number of moles per unit volume decreases. To increase, it more. Of `PCI_(5)` dissociates.

Discussion

576.	Asseration (A) : Increasing order of acidity of hydrogen halides is `HFltHCIltHBrltHI` Reason (R): While comparing acids formed by the elements belonging to the same group of periodic table, H-A bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond.A. Both A and B are true and R is the correct explanation of A.B. Both A and R are ture but R is not the correct explanation of A.C. A is the ture R is false.D. Both A and R are false.
Answer» Correct Answer - A Reason is the correct explanation for assertion

Discussion

577.	Assertion (A) : Increasing order of acidity of hydrogen halides is `HF lt HCl lt HBr lt HI`. Reason (R) : While comparing acids formed by the elements belonging to the same group of periodic table, H-A bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true but R is not the correct explanation of A.C. A is true but R is false.D. Both A and R are false.
Answer» Correct Answer - A

Discussion

578.	Which of the following is not an application of solubility product ?A. Predicting precipitation formationB. Predicting solubility of sparingly soluble saltC. Predicting pH of a buffer solutionD. Qualitative analysis
Answer» Correct Answer - C

Discussion

579.	Define solubility of substance.
Answer» Solubility of substance is maximum amount of solute that can be dissolved 100g of solvent at a given temperature.

Discussion

580.	What is the effect of temperature on solubility of gases in liquids?
Answer» Solubility of gases in liquids decreases with increase in temperature.

Discussion

581.	In a reversible reaction, the two substances are in equilibrium. If the concentration of each one is reduced to half, then what is the effect on the equilibrium constant?
Answer» The equilibrium constant remains the same because its value does not depend on initial concentration of the reactants.

Discussion

582.	State Henry’s law.
Answer» Henry’s 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. m = k x p where ‘p’ is pressure applied on the gas.

582.

State Henry’s law.

Answer»

Henry’s 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.

m = k x p where ‘p’ is pressure applied on the gas.

Discussion

583.	What happens to the boiling point of water at high altitude?
Answer» Boiling point of water depends on the altitude of the place. At high altitude the boiling point decreases.

Discussion

584.	State Henry's Law.
Answer» According to this law, "the mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent.’’

Discussion

585.	State Henry’s law.
Answer» The mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the gas above the solvent.

Discussion

586.	On what factor does the boiling point of the liquid depends?
Answer» Boiling point depends on the atmospheric pressure.

Discussion

587.	What is meant by the statement ‘Equilibrium is dynamic in nature’?
Answer» At equilibrium, reaction does not stop rather it still continues, the equilibrium is dynamic in nature. It appears to stop because rate of forward reaction is equal to the rate of backward reaction.

Discussion

588.	What is physical equilibrium? Give an example.
Answer» Physical equilibrium is an equilibrium between two different physical states of same substance e.g. H₂O (S) ⇌ H₂O(l)

Discussion

589.	Name the three group into which chemical equilibrium can be classified.
Answer» Chemical equilibrium can be classified into three groups – (i) The reaction that proceeds nearly to completion and only negligible concentrations of the reactants are left. (ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage. (iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium.

589.

Name the three group into which chemical equilibrium can be classified.

Answer»

Chemical equilibrium can be classified into three groups –

(i) The reaction that proceeds nearly to completion and only negligible concentrations of the reactants are left.

(ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage.

(iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium.

Discussion

590.	For the reaction `Ag(CN)bar(2) hArr Ag^(+) 2 CN^(-), K_(c)" at "25^(@)C" is "4 xx 10^(-19)` . Calculate `[Ag^(+)]` in solution which was originally `01` M in KCN and `003` M in `AgNO_(3)`.
Answer» Orignally on mixing KCN and `AgNO_(3)` , the reaction is `{:(,2 KCN,+,AgNO_(3),to,"Ag"(CN)_(2)^(-),+,KNO_(3),+,K^(+)),("Intial amounts",01 M,,003 M,,0,,0,,0),("Amounts after reaction ",(01-006),,0,,003 M,,003M,,003),(,=004 M,,,,,,,,):}` Thus , in the solution , now we have `[Ag (CN)_(2^(-)) ] = 03 M , [CN^(-)] = 0 04 M .` Suppose x is the amount of `Ag(CN)_(2)^(-)` dissociated at equilibrium . Then ` {:(,AG(CN)_(2)^(-),hArr,AG^(+),+,2CN^(-)),("Intial amounts",003,,0,,004),("Amounts at eqm.",(003 -x),,x,+,004 + 2x):}` ` K_(c) = [ AG^(+) [CN^(-)]^(2))/([Ag(CN)_(2)^(-)])= (x (004 + 2x)^(2))/(003 - x) = 4 xx 10^(-19) ("Given")` As `K_(c)` is very small , dissociation of `Ag(CN)_(2)^(-)` is very small i.e., x is very small . Hence , `004 + 2x approx and 03 - x approx 003 .` ` :. (x( 004)^(2))/(003 )= 4 xx 10^(-19) or x = 75 xx 10^(-18)` Thus at equilibrium , `[Ag^(+)] = 7* 5 xx 10^(-18)M`

Discussion

591.	A sample of air consisting of `N_(2) and O_(2)` was heated to 2500 K until the equilibrium `N_(2) (g) + O_(2) (g) hArr 2 NO (g)` was established the intial composition of air in mole fraction of `N_(2) and O_(2).`
Answer» `{:(,N_(2)(g),+,O_(2),hArr,2NO,),("Intial ",a,,b,,0,),(,,,,,,a+b = 100" " ...(ii)),("At eqm",a-x,,b-x,,2x,):}` ` K_(c) = (2x)^(2) /((a-x)(b-x)) = (4x^(2))/((a-x)(b-x))` In the question , we are given `2x = 18 ("because total moles " =100)` or ` x= 09 and K_(c) = 21 xx 10^(-3) ` ` :. 21 xx 10^(-3) = (18)^(2)/((a-09 )(b-09)) ` ` ab - 09 a- 09 b + 081 = 1620` ` ab - 09 (a+b) + 081 = 1620 ` ` ab- 09 xx 100 + 081 = 1620 ` ` or ab = 1909 * 19 = 1709` ` Now (a-b)^(2) = (a+b)^(2) - 4 ab = (100)^(2) - 4 xx 709 = 3164 ` or ` a-b = sqrt (3164)= 562 " " `...(ii) Solving (i) and (ii), `a= 781 "moles"` Moles fraction of ` N_(2) = (781 )/100 = 0 781 ` Mole fraction of ` O_(2) = 1 - 0* 781 = 0* 219 `

Discussion

592.	The answer to each of the folowing questions is a single digit integar, ranging from 0 to 9. If the correct answers to the question numbers A, B, C and D (say) are 4,0,9 and 2 respectively , then the correct darkening of bubbles should be as shown on the side : For the reaction involving oxidation of ammonia by oxygen to form nitric oxide and water vapour , the equilibrium constant has the units `("bar")^(n)` . Then n is
Answer» Correct Answer - 1 ` 4 NH_(3) (g) + 5 O_(2) (g) hArr 4 NO(g) + 6H_(2)O (g) ` ` K = (p_(NO)^(4)xxp_(H_(2)O)^(6))/(p_(NH_(3))^(4)xxp_(O_(2)^(5)))=("bar")^(10)/("bar")^(10) = ("bar")^(1) :. n = 1 .`

Discussion

593.	Assertion (A): `pH` of `HCI` solution is less than that of acetic acid of the some concentartion. Reason (R) : In equimolar solution, the number of titrable protons present in `HCI` is less than that present in acetic acid.A. if both assertion and reason are correct and reason is correct explanation for assertion.B. if both assertion and reason are correct but reason is not correct explanation for assertion.C. if assertion is correct but reason is incorrect.D. if assertion and reason boht are incorrect.
Answer» Correct Answer - C Correct reason. In equimolar solutions the number of ionisable `H^(+)` ions in HCI is more than in acetic acid `(CH_(3)COOH)`.

Discussion

594.	Calculate the pH of a solution obtained by mixing 100 ml of 0.10 M CH3COOH and 100 ml of 0.05 M NaOH solution. Ka for CH3COOH is 1∙8 x 10-5 ?
Answer» NaOH + CH₃COOH → CH₃COONa + H₂O 100 ml of 0 ∙ 10 M CH₃COOH contains = 100 × 0∙1 = 10 m mol CH₃COOH 100 ml of 0∙5 M NaOH contains = 100 × 0∙05 = 5 m mol of NaOH 5 m mol NaOH react with 5 m mol CH₃COOH to form 5 mol of CH₃COONa. 5 m mol of CH₃COOH remain unreacted. After mixing, [CH₃COOH] = \(\frac{5}{200}\) M = 2.5 x 10^-2 M [CH₃COOH] = \(\frac{5}{200}\) = 2.5 x 10^-2 M pH = pK_a + log\(\frac{[CH_3COO^-]}{[CH_3COOH]}\) pH = pK_a+ log\(\frac{[2.5\times10^{-2}]}{[2.5\times10^{-2}]}\) K_a= 1∙8 × 10⁻⁵ pH = pK_a = − log (1 ∙ 8 × 10⁻⁵) = 4∙74.

594.

Calculate the pH of a solution obtained by mixing 100 ml of 0.10 M CH3COOH and 100 ml of 0.05 M NaOH solution. Ka for CH3COOH is 1∙8 x 10-5 ?

Answer»

NaOH + CH₃COOH → CH₃COONa + H₂O

100 ml of 0 ∙ 10 M CH₃COOH contains

= 100 × 0∙1

= 10 m mol CH₃COOH

100 ml of 0∙5 M NaOH contains

= 100 × 0∙05

= 5 m mol of NaOH

5 m mol NaOH react with 5 m mol CH₃COOH to form 5 mol of CH₃COONa.

5 m mol of CH₃COOH remain unreacted. After mixing,

[CH₃COOH] = \(\frac{5}{200}\) M = 2.5 x 10^-2 M

[CH₃COOH] = \(\frac{5}{200}\) = 2.5 x 10^-2 M

pH = pK_a + log\(\frac{[CH_3COO^-]}{[CH_3COOH]}\)

pH = pK_a+ log\(\frac{[2.5\times10^{-2}]}{[2.5\times10^{-2}]}\)

K_a= 1∙8 × 10⁻⁵

pH = pK_a

= − log (1 ∙ 8 × 10⁻⁵)

= 4∙74.

Discussion

595.	For the following reaction : `NO_((g))+O_(3(g))hArrNO_(2(g))+O_(2(g))` The value of `K_(c)` is `8.2xx10^(4)`. What will be the value of `K_(c)` for the reverse reaction ?A. `8.2xx10^(4)`B. `(1)/(8.2xx10^(4))`C. `(8.2xx10^(4))`D. `sqrt(8.2xx10^(4))`
Answer» Correct Answer - B The value of `K_(c)` for the reverse reaction will be `(1)/(8.2xx10^(-4))`.

Discussion

596.	For the reaction `CO(g) + Cl_(2) (g) hArr COCl_(2) (g) , K_(p)//K_(c)` is equal toA. `sqrt(RT)`B. RTC. `1/(RT)`D. `1*0`
Answer» Correct Answer - C ` K_(p) = K_(c) (RT)^(Delta n )` For the given reactions, ` Delta n = 1 - 2 = - 1` ` :. (K_(p))/(K_(c)) = (RT)^(-1) = 1/(RT)`

Discussion

597.	The solubility product of `Hg_(2)I_(2)` is equal toA. `[Hg_(2)^(++)][I^(-)]`B. `[Hg^(++)][I^(-)]`C. `[Hg_(2)^(+)][I^(-)]^(2)`D. `[Hg^(++)][I^(-)]^(2)`
Answer» Correct Answer - C is the correct answer.

Discussion

598.	Match the column-I with column-II `{:("Column_I","Column-II"),((i) "Equilibrium",(a) DeltaG=+ve),((ii) "Spontaneous reaction",(b) DeltaG=0),((iii) "Non-spontaneous reaction",(c) DeltaG=-ve):}`
Answer» `("I") -(b) , ("ii") - (C) ,("iii") -(a).`

Discussion

599.	The units of solubility product of `Ag_(2)SO_(4)` will be (connentration being expressed in mol `L^(-1)`):A. mol `L^(-1)`B. `mol^(2) L^(-2)`C. `mol^(3) L^(-3)`D. none of these
Answer» Correct Answer - C Is the correct answer.

Discussion

600.	1M NaCI and 1M HCI are present in an aqueous solution. The solution isA. not a buffer with `pH lt 7`B. not a buffer with `pH lt 7`C. a buffer with `pH lt 7`D. a buffer with `pH lt 7.`
Answer» Correct Answer - A The solution is not a buffer. If is acidic with pH less than 7.

Discussion

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` K_(c) "for" CO(g) + H_(2) O (g) hArr CO_(2) (g) + H_(2) (g) " at " 986 ^(@)C " is " 0*63 . " A mixture of 1 mole of" H_(2)O (g) and" 3 moles " CO (g) " is allowed to react to an equilibrium . The equilibrium pressure is " 2*0 atm .`

The pH of a solution obtained by mixing 50 mL of 0.2 M HCl with 50 mL of 0.20 M `CH_(3) C O OH` isA. `0.30`B. `0.70`C. `1.00`D. `2.00`

What is the pH of a solution obtained by mixing 10 mL of 0.1 M HCl and 40 mL 0.2 M `H_(2)SO_(4)` ?A. 0.74B. 7.4C. 4.68D. 0.468

The pH of a solution obtained by mixing 100 mL of a solution pH=3 with 400 mL of a solution of pH=4 isA. 3- log 2.8B. 7- log 2.8C. 4- log 2.8D. 5- log 2.8

The value of `K_(w)` is `9.55xx10^(-14)` at a certain temperature. Calculate the pH of water at this temperature .

Calculate the pH of the solution obtained by mixing 100 `cm^(3)` of solution with pH = 3 with `400 cm^(3)` of solution with pH = 4 .

Calculate the concentration of `H^(+)` (aq) in 0.2 M solution of HCN. Given that the dissociation constant of HCN in water is `4.9 xx 10^(-10)`.

If hydrogen ion concentration in a solution is `1xx10^(-5)` moles/litre, calculate the concentration of OH ion in this solution `(K_(w)=10^(-14)"moles"^(2)L^(-2)).`

If hydrogen ion concentration in a solution is `1xx10^(-5)` moles/litre, calculate the concentration of `OH^(-)`ion in this solution `(K_(w)=10^(-14) "moles"^(2)//"litre"^(2))`

Define strong and weak electrolyte.

What is the pH of a solution whose `H^(+)` ion concentration is `2xx10^(-5)g` ions/litre ?

If the equilibrium constant for a reaction is `4*0` , what will be the equilibrium constant for the reverse reaction.

What is the value of ∆n for the given reaction.2NOCl (g) -------&gt; 2NO (g) + Cl2 (g)

On what factors does the value of the equilibrium constant of a reaction depend?

If the equilibrium constant for the reaction is 4.0, what will be the equilibrium constant for the reverse reaction.

What happens to the dissociation of PCl5 in a closed vessel if helium gas is introduced into it at the same temperature?

What happens to the dissociation of PCl5 in a closed vessel if He gas is introduced into it at the same temperature?

Which of the following is not an application of solubility product ?A. Predicting precipitation formationB. Predicting solubility of sparingly soluble saltC. Predicting pH of a buffer solutionD. Qualitative analysis

Define solubility of substance.

What is the effect of temperature on solubility of gases in liquids?

In a reversible reaction, the two substances are in equilibrium. If the concentration of each one is reduced to half, then what is the effect on the equilibrium constant?

State Henry’s law.

What happens to the boiling point of water at high altitude?

State Henry's Law.

State Henry’s law.

On what factor does the boiling point of the liquid depends?

What is meant by the statement ‘Equilibrium is dynamic in nature’?

What is physical equilibrium? Give an example.

Name the three group into which chemical equilibrium can be classified.

For the reaction `Ag(CN)bar(2) hArr Ag^(+) 2 CN^(-), K_(c)" at "25^(@)C" is "4 xx 10^(-19)` . Calculate `[Ag^(+)]` in solution which was originally `0*1` M in KCN and `0*03` M in `AgNO_(3)`.

A sample of air consisting of `N_(2) and O_(2)` was heated to 2500 K until the equilibrium `N_(2) (g) + O_(2) (g) hArr 2 NO (g)` was established the intial composition of air in mole fraction of `N_(2) and O_(2).`

Calculate the pH of a solution obtained by mixing 100 ml of 0.10 M CH3COOH and 100 ml of 0.05 M NaOH solution. Ka for CH3COOH is 1∙8 x 10-5 ?

For the following reaction : `NO_((g))+O_(3(g))hArrNO_(2(g))+O_(2(g))` The value of `K_(c)` is `8.2xx10^(4)`. What will be the value of `K_(c)` for the reverse reaction ?A. `8.2xx10^(4)`B. `(1)/(8.2xx10^(4))`C. `(8.2xx10^(4))`D. `sqrt(8.2xx10^(4))`

For the reaction `CO(g) + Cl_(2) (g) hArr COCl_(2) (g) , K_(p)//K_(c)` is equal toA. `sqrt(RT)`B. RTC. `1/(RT)`D. `1*0`

The solubility product of `Hg_(2)I_(2)` is equal toA. `[Hg_(2)^(++)][I^(-)]`B. `[Hg^(++)][I^(-)]`C. `[Hg_(2)^(+)][I^(-)]^(2)`D. `[Hg^(++)][I^(-)]^(2)`

Match the column-I with column-II `{:("Column_I","Column-II"),((i) "Equilibrium",(a) DeltaG=+ve),((ii) "Spontaneous reaction",(b) DeltaG=0),((iii) "Non-spontaneous reaction",(c) DeltaG=-ve):}`

The units of solubility product of `Ag_(2)SO_(4)` will be (connentration being expressed in mol `L^(-1)`):A. mol `L^(-1)`B. `mol^(2) L^(-2)`C. `mol^(3) L^(-3)`D. none of these

1M NaCI and 1M HCI are present in an aqueous solution. The solution isA. not a buffer with `pH lt 7`B. not a buffer with `pH lt 7`C. a buffer with `pH lt 7`D. a buffer with `pH lt 7.`

` K_(c) "for" CO(g) + H_(2) O (g) hArr CO_(2) (g) + H_(2) (g) " at " 986 ^(@)C " is " 063 . " A mixture of 1 mole of" H_(2)O (g) and" 3 moles " CO (g) " is allowed to react to an equilibrium . The equilibrium pressure is " 20 atm .`

What is the value of ∆n for the given reaction.2NOCl (g) -------> 2NO (g) + Cl2 (g)

For the reaction `Ag(CN)bar(2) hArr Ag^(+) 2 CN^(-), K_(c)" at "25^(@)C" is "4 xx 10^(-19)` . Calculate `[Ag^(+)]` in solution which was originally `01` M in KCN and `003` M in `AgNO_(3)`.