1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 1 InterviewSolution

1.	`2.5 ` mL of `(2)/(5)` M weak monoacidic base `(K_(b)=1xx10^(-12) "at" 25^(@)C)` is titrated with `(2)/(15)` M HCl in water at `25^(@)` C. The concentration of `H^(+)` at equivalence point is `(K_(w) = 1 xx 10^(-14) "at" 25^(@)C)`A. `3.7xx10^(-13)M`B. `3.2xx10^(-7) M`C. `3.2xx10^(-2)M`D. `2.7xx10^(-2)M`
Answer» Correct Answer - C `{:(N_(1)V_(1),=,N_(2)V_(2)),("(Base)",,"(Acid)"):}` `2.5xx(2)/(5)=(2)/(15)=(2)/(15)xxV_(2) "or" V_(2)=(15)/(2) mL = 7.5 mL ` `BOH+HClrarrBCl+H_(2)O` 2.5 mL of 2 M base contain base `=2.5xx(2)/(5) = 1` mmol `:. ` Salt BCl formed = 1 mmol Volume of solution = 2.5 mL + 7.5 mL = 10 mL `:.` Conc. of salt [BCl] in the solution `=(1)/(10)M=0.1M` For salt of weak base and strong acid `[H^(+)]=sqrt((K_(w)C)/(K_(b)))=sqrt((10^(-14)xx0.1)/(10^(-12)))=3.2xx10^(-2)M`

Discussion

2.	Which of the following equimolar solutions can act as a buffer solution?(a) NH4Cl and NH4OH(b) HCl and NaCl(c) HCOOH and HCOONa(d) HNO3 and NH4NO3
Answer» (c) HCOOH and HCOONa

Discussion

3.	The range of pH for acidic and basic buffer is where `K_(a)` and `K_(b)` are the acid base dissociation constants, respectively.A. form `pH=pK_(a)+-2 to pH=pK_(b)+-2`B. from `pH=pK_(a)+1to pH=pK_(b)+1`C. from `pH=pK_(a)+-1to pH=pK_(b)+-1`D. from `pH=pK_(a)+1to pH=pK_(b)-1`
Answer» Correct Answer - C When the pH of the acidic buffer does not differ greatly from the `pK_(a)` of the acid, the buffering action is the most efficient. This happens when the ration `[A^(-)]//[HA]` is between `0.1` and 10. Thus, `pH=pK_(a)+-1`.

Discussion

4.	Which of the following expression tepresents the Henderson equation for an acidic buffer ?A. `pH=(1)/(2)pK_(a)-(1)/(2)logC`B. `pH=pK_(a)-l"og"(["Conjugate base"])/(["Acid"])`C. `pH=pK_(a)+"log"(["Conjugate base"])/(["Acid"])`D. `pH=pK_(a)`
Answer» Correct Answer - C `underset("acid")underset("Weak")(HA)hArr underset("base")underset("Conjugate")(H^(+)+A^(-))` `K_(a)= ([H^(+)][A^(-)])/([HA])` or, rearrangning terms, `[H^(+)] = K_(a)([HA])/([A^(-)])` Taking the negative logarithm of both sides of this equation gives `- log[H^(+)]= -"log"(K_(a)([HA])/([A^(-)]))` `= -log K_(a)+"log"([A^(-)])/([HA])` or `pH= pK_(a)+"log" ([A^(-)])/([HA])`

Discussion

5.	An acidic buffer solution can be prepared by mixing equimolar amounts ofA. `B(OH)_(3)` and `Na_(2)B_(4)O_(7).10H_(2)O`B. `NH_(3)` and `NH_(4)CI`C. `HCI` and `NaCI`D. `CH_(3)COOH` and `CH_(3)COONa`
Answer» Correct Answer - D pH of the buffer solution formed by acetic acid and sodium acetate taken in equal molar concentrations will be around `4.76`.

Discussion

6.	Through a solution containing `Cu^(2+) and Ni^(2+),H_(2)S` gas is passed after adding dil HCl, which will precipitate out and why ?
Answer» `Cu^(2+)` will precipitate out because in the acidic medium, only ionic product `[Cu^(2+)][S^(2-)]` exceeds the solubility product of CuS.

Discussion

7.	Give an example of acidic buffer?
Answer» CH₃COOh + CH₃COONa.

Discussion

8.	How does dilution with water affect the pH of a buffer solution?
Answer» Dilution with water has no effect on the pH of any buffer. This is because pH of a buffer depends on the ratio of the salt, acid or salt base and dilution does not affect this ratio.

Discussion

9.	Through a solution containing Cu2+and Ni2+,H2S gas is passed after adding dill HCl, which will precipitate out and why?
Answer» Cu²⁺ will precipitate out because in acidic medium, only ionic product [Cu²⁼][ S^2- ] exceeds the solubility product of CuS.

Discussion

10.	Write the expression for Ksp for AB3 type salt With solubility s.
Answer» Ksp:-27S⁴ ..........

Discussion

11.	Write the expression for Ksp for AB type salt With solubility s
Answer» Ksp:-S² ..............

Discussion

12.	What would be the expression for solubility product of calcium phosphate in terms of its molar solubility, S.
Answer» K_sp = (3 S)² (2 S)³ = 72 S⁵

Discussion

13.	The solubility product of AgCl is `1.8xx10^(-10)`. Precipitation of AgCl will occur only when equal volumes of solutions of :A. `10^(-8)"M "Ag^(+)and 10^(-8)"M "Cl^(-)` ionsB. `10^(-3)"M "Ag^(+)and 10^(-3)"M "Cl^(-)` ionsC. `10^(-6)"M "Ag^(+)and 10^(-6)"M "Cl^(-)` ionsD. `10^(-10)"M "Ag^(+)and 10^(-10)"M "Cl^(-)` ions
Answer» Correct Answer - B For precipitation, ionic product `gt` solubility product. In all other options, ionic product is less than solubility product.

Discussion

14.	The solubility product of `AgI` at `25^(@)C` is `1.0xx10^(-16) mol^(2) L^(-2)`. The solubility of `AgI` in `10^(-4) N` solution of `KI` at `25^(@)C` is approximately ( in `mol L^(-1)`)A. `1.0xx10^(-16)`B. `1.0xx10^(-12)`C. `1.0xx10^(-10)`D. `1.0xx10^(-8)`
Answer» Correct Answer - B Equilibrium expression is `AgI(s)hArr Ag^(+)(aq.)+I^(-)(aq.)` Let the solubility of AgI in `10^(-4)` N or `10^(-4)` M solution of KI be x mol `L^(-1)`. The total concentration of `I^(-)(aq.)`. Will be `C_(I^(-))=x+10^(-4)` Since the solubility product of AgI is very small and due to the presence of KI, equilibrium shifts backward (common ion effect), we have `C_(I^(-))~~10^(-4)mol L^(-1)` Now, `K_(sp)=C_(Ag^(+)C_(I^(-))` `1.0xx10^(-16)=(x)(10^(-4))` `x=1.0xx10^(-12)`

Discussion

15.	40 ml of 0.1 M ammonia is mixed with 20 ml of 0.1 M HCl. What is the pH of the mixture ? (` pK_(b)` of ammonia solution is 4.74)A. 4.74B. 2.26C. 9.26D. `5.00`
Answer» Correct Answer - C 40 ml of 0.1 M `NH_(3)` solution `= 40xx0.1` millimole = 4 millimoles `NH_(4)OH+HCl rarr NH_(4)Cl + H_(2)O` 2 millimole of HCl will neutralize 2 millimoles of `NH_(4)OH` to form 2 millimoles of `NH_(4)Cl`. `NH_(4)OH` left = 60 ml `:. [NH_(4)OH]=(2)/(60)M, [NH_(4)Cl]=(2)/(60)M` `pOH=pK_(b)+log .([NH_(4)Cl])/([NH_(4)OH])` `=4.74 + log .(2//60)/(2//60)=4.74` `:. pH=14-4.4.74=9.26`

Discussion

16.	Calculate the pH of `0.033` M ammonia solution if `0.033 M NH_(4)Cl` is introduced in this solution at the same temperature `(K_(b)` for `NH_(3)=1.77xx10^(-5)`)
Answer» Correct Answer - `9.25`

Discussion

17.	What is the relationship between `pK_(a) and pK_(b)` values where `K_(a) and K_(b)` represent constants of the acid and its conjugate base respectively ?
Answer» `pK_(a) + pK_(b) = pK_(w)= 14`.

Discussion

18.	The `pH` of milk, black coffee, tomato juice, lemon juice and egg white are `6.8, 5.0, 4.2, 2.2` and `7.8` respectively. Calculate corresponding hydrogen ion concentration in each.
Answer» `(a) [H^(+)]` of milk `pH =6.8 " or " -log [H^(+)] =6.8 " or " log [H^(+)] =- 6.8` `[H^(+)] = "Antilog " (-6.80) = " Antilog " (bar(7).20) =1.58 xx 10^(-7) M` `(b) [H^(+)]` of block coffe `pH =5.0 " or " - log [H^(+)] =5.0 " or " log [H^(+)] =- 5.0` `[H^(+)] = "Antilog " (-5.0) = " Antilog " bar(5) = 10^(-5) M` (c) `[H^(+)]` of tomato juice `pH =4.2 " or " - log [H^(+)] =4.2 " or " log [H^(+)] =- 4.2` `[H^(+)] = "Antilog " (-4.2) = " Antilog" (bar(5).80) =6.31 xx 10^(-5)M` (d) `[H^(+)]` of lemon juice `pH =2.2 " or " - log [H^(+)] =2.2 " or " log [H^(+)] =- 2.2` `[H^(+)] = " Antilog " (-2.2) = " Antilog" (bar(3).80) =6.310 xx 10^(-3)M` `(e) [H^(+)]` of egg white `pH =7.8 " or " - log [H^(+)] =7.8 " or " log [H^(+)] =- 7.8` `[H^(+)] = " Antilog" (-7.8) = " Antilog " (bar(8).20) = 1.58 xx 10^(-8) M`

Discussion

19.	Lemon juice has a pH = 2.1 . If all the acid in lemon is citric acid `(H "Cit." hArr H^(+)+"Cit"^(-1))` and `K_(a)` for citric acid is `8.4xx10^(-4)` mole/litre, what is the concentration of citric acid in lemon juice ?
Answer» Correct Answer - `7.5xx10^(-2)M` `pH = 2.1 , i.e., - log [H^(+)]=2.1 or log [H^(+)]= - 2.1 = bar(3). 9 or [H^(+)] = 7.943xx10^(-3)` `"HCit" hArr H^(+) + "Cit"^(-1)` `K_(a) = ([H^(+)] [Cit^(-1)])/(["H Cit"])` `8.4xx10^(-4)=(7.943xx10^(-3))(7.943xx10^(-3))/(["H Cit"]) or ["H Cit" ] = 7.5 xx 10^(-2)M`

Discussion

20.	At 298 K, the pH of a lemon juice is 2.32. Calculate its `[H_(3)O^(+)]" and "[OH^(-)].`
Answer» pH of the solution `=2.32 , - log [H_(2)O^(+)] =2.32 or log [H_(3)O^(+)] =-2.32` `[H_(3)O]^(+) "Antilog" (-2.32) = "Antilog" (overset(-)(3).68) = 4.79 xx 10^(-3)` `[OH^(-)] = (K_(w))/[[H_(3)O^(+)]]=((1.0xx10^(-14)M^(-2)))/((4.79xx10^(-3)M))=2.09 xx 10^(-12)M`

Discussion

21.	`{:(,"Column I",,"Column II (pH)"),((A),"MilK",(p),2.2),((B),"Human saliva",(q),6.4),((C),"Human blood",(r),6.8),((D),"Lemon juice",(s),7.4):}`A. A-p, B-q, C-s, D-rB. A-r, B-q, C-s, D-pC. A-r, B-s, C-q, D-pD. A-q, B-r, C-s, D-p
Answer» Correct Answer - B

Discussion

22.	The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.
Answer» Given, pK_a = 4.76 pK_b = 4.75 ∵ pH = 7 + \(\frac{1}{2}\) [pK_a − pK_b] = 7 + \(\frac{1}{2}\) [4.76 - 4.75] = 7 + \(\frac{1}{2}\) [0.01] = 7 + 0.005 = 7.005

22.

The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.

Answer»

Given,

pK_a = 4.76

pK_b = 4.75

∵ pH = 7 + \(\frac{1}{2}\) [pK_a − pK_b]

= 7 + \(\frac{1}{2}\) [4.76 - 4.75]

= 7 + \(\frac{1}{2}\) [0.01]

= 7 + 0.005

= 7.005

Discussion

23.	The ionization constant of phenol is `1.0xx10^(-10)`. What is the concentration of phenate ion in 0.05 M solution of phenol ? What will be its degree of ionization if the solution is also 0.01 M in sodium phenate ?
Answer» `{:(,C_(6)H_(5)OH,hArr,C_(6)H_(5)O^(-),+,H^(+),,),("Initial",0.05M,,,,,,),("After disso.",0.05-x,,x,,x,,):}` `:. K_(a) = (x xx x)/(0.05-x) = 1.0xx10^(-10)` (Given) or `(x^(2))/(0.05) = 1.0xx10^(-10)` or `x^(2)=5xx10^(-12) or x = 2.2 xx 10^(-6) M` In presence of 0.01 `C_(6)H_(5)ON a`, suppose y is the amount of phenol dissociated, then at equilibrium `[C_(6)H_(5)OH]=0.05-y~=0.05, [C_(6)H_(5)O^(-)]=0.01+y~=0.01M, [H^(+)]=y M` `:. K_(a)=((0.01)(y))/(0.05)=1.0xx10^(-10) ` (Given) or `y=5xx10^(-10)` `:. alpha = (y)/(c) = (5xx10^(-10))/(5xx10^(-2))=10^(-8)`.

Discussion

24.	An aqueous solution contains 0.10 M `H_(2)S " is "1.0 xx 10^(-7)" and that of " S^(2-) "ion from " HS^(-) "ions is " 1.2 xx 10^(-3) ," then the concentration of " S^(2-) " ions in aqueous solution is "`A. ` 5 xx 10^(-8)`B. `3 xx 10^(-20)`C. `6xx 10^(-21)`D. `5xx 10^(-19)`
Answer» Correct Answer - B Given : (i) ` H_(2) S hArr HS^(-) + H^(+),` `K_(a_(1)) = 1.0 xx 10^(-7)` (ii) `HS^(-) hArr S^(2-) + H^(+), K_(a_(2))= 1.2 xx 10^(-13)` For the reaction , ` H_(2)S hArr 2 H^(+) + S^(2-)` (obtained by adding (i) and (ii) ) `K_(a) = K_(a_(1)) xx K_(a_(2))` ` :. (1.0 xx 10^(-7)) (1.2 xx 10^(-13))= ([H^(+)]^(2) [S^(2-)])/([H_(2)S]) ` In presence of HCl, `[H^(+)]` are obtained mainly from HCl. Hence , ` [H^(+)] = 0.20 M, [H_(2) S]= 0.1 M ` `:. 1.2 xx 10^(-20) = ((0.2 )^(2)[S^(2-)])/0.1` or `[S^(2-)] = 3 xx 10^(20)`

Discussion

25.	For the following three reaction 1, 2 and 3, equilibrium constants are given: (1) `CO_((g)) + H_(2)O_((g))hArrCO_(2(g)) + H_(2(g)), K_(1)` (2) `CH_(4(g)) + H_(2)O_((g))hArrCO_((g)) + 3H_(2(g)), K_(2)` (3) `CH_(4(g)) + 2H_(2)O_((g))hArrCO_(2(g)) + 4H_(2(g)), K_(3)` Which of the following relations is correct ?A. `K_(2)K_(3)=K_(1)`B. `K_(1)sqrt(K_(2))=K_(1)`C. `K_(3)=K_(1)K_(2)`D. `K_(2)K_(3)=K_(1)`
Answer» Correct Answer - C We can notice that Eq. (C ) can be obtained by adding Eqs. (A) and (B): `CO(g)+H_(2)(g)hArrCO_(2)(g)+H_(2)(g)` `(CH_(4)(g)+H_(2)O(g)hArr CO(g)+3H_(2)(g))/(CH_(4)(g)+2H_(2)O(g)hArrCO_(2)+4H_(2)(g))` Whenever we add two or more equations to create another equation, the equilibrium constants are multiplied: `K_(3)=K_(1)K_(2)`

Discussion

26.	Calculate the pH at which `Mg(OH)_(2)` begins to precipitate from a solution containing 0.10 M `Mg^(2+)` ions. `K_(sp)` of Mg`(OH)_(2)=1xx10^(-11)`.
Answer» Minimum `[OH^(-)]` after which `Mg (OH)_(2)` begins to precipitate can be calculated from `[Mg^(2+)][OH^(-)]^(2)=K_(sp) ` of `Mg(OH)_(2)` `(0.10)[OH^(-)]^(2) = 10^(-11) or [OH^(-)]^(2) = 10^(-10) or [OH^(-)]=10^(-5) or [H^(+)]=10^(-9) or pH = 9`

Discussion

27.	Calculate pH at which `Mg(OH)_(2)` begins to precipitate from a solution containing `0.10M Mg^(2+)` ions. `(K_(SP)of Mg(OH)_(2)=1xx10^(-11))`
Answer» `[Mg^(2+)] =0.1 M " and " K_(sp) " for " Mg(OH)_(2) =1.0 xx 10^(-11)` the equilibrium in the saturated solution `Mg(OH)_(2) hArr Mg^(2+) + 2OH^(-)` `[OH^(-)]^(2) =(K_(sp))/([Mg^(2+)]] =(1.0 xx 10^(-11))/(0.1) =1.0 xx 10^(-10)` `:. [OH]^(-) =(1.0 xx 10^(-10))^(1//2) =10^(-5) M` We know that `[H_(3)O^(+)][OH^(-)] =K_(w) = 10^(-14) :. [H_(3)O^(+)] = (10^(-14))/([OH^(-)]] =(10^(-14))/(10^(-5)) =10^(-9) M` `pH =- log [H_(3)O^(+)] =- log [10^(-9)] =(-) (-9) log 10 =9`

Discussion

28.	Calculate pH at which `Mg(OH)_(2)` begins to precipitate from a solution containing `0.10M Mg^(2+)` ions. `(K_(SP)of Mg(OH)_(2)=1xx10^(-11))`A. 4B. 6C. 9D. 7
Answer» Correct Answer - C `K_(sp)" for "Mg(OH)_(2)=1.0xx10^(-11)` `[Mg^(2+)][OH^(-)]^(2)=(0.1)xx10[OH^(-)]^(2)=1.0xx10^(-11)` `[OH^(-)]^(2)=(1xx10^(-11))/(0.1)=10^(-10)` `[OH^(-)]=10^(-5),[H^(+)]=(10^(-14))/(10^(-5))=10^(-9)` `pH=9`

Discussion

29.	The `K_(sp)` of `Mg(OH)_(2)` is `1xx10^(-12). 0.01M Mg^(2+)` will precipitate tate at the limiting pH ofA. `1.3`B. `9`C. `3.5`D. `8`
Answer» Correct Answer - B `Mg(OH)_(2)(s)hArr Mg^(2+)(aq.)+2OH^(-)(aq.)` `:. K_(sp)= C_(Mg^(2+))C_(OH^(-))^(2)` `10^(-12)=(10^(-2))C_(OH^(-))^(2)` or `C_(OH^(-))^(2)=10^(-10)` `C_(OH^(-))=10^(-5)M` For any aqueous solution at `25^(@)C`, `C_(H^(+))C_(OH^(-))=10^(-14)` Hence, `C_(H^(+))=(10^(-14))/(C_(OH^(-)))=(10^(-14))/(10^(-5))=10^(-9)` Since precipitation occurs when ionic product just exceeds `K_(sp)`, we have the limiting pH of `9(C_(H^(+))=10^(-pH)mol L^(-))`.

Discussion

30.	In the reaction `A+B hArrC+D` what will happen to the equlibrium if concentration of A is increased?
Answer» It will shift to the right.

Discussion

31.	An aqueous solution of `CuSO_(4)` is acidic while that of `Na_(2)SO_(4)` is neutral. Explain.
Answer» `CuSO_(4)` is a salt of weak base and strong acid. Upon hydrolysis it will form a weak base and strong acid. This means that the solution contains free`H^(+)` ions and will be therefore acidic in nature. But `Na_(2)SO_(4)` is a salt of strong base and strong acid. Although it dissolves in water but will not undergo any hydrolysis . The solution will be therefore neutral. `CuSO_(4) +2H_(2)O to Cu(OH)_(2) + H_(2) SO_(4)` (weak ) (Strong)

Discussion

32.	The aqeous solutions of `HCOONa, C_(6)H_(5)NH_(3)CI`, and KCN are, respectively,A. acidic, acidic, basicB. acidic, basic, neutralC. basic, neutral, neutralD. basic, acidic, basic
Answer» Correct Answer - D HCOONa is the salt of weak acid (HCOOH) and strong base (NaOH). Due to hydrolysis, its aqueous solution is basic. `C_(6)H_(5)NH_(3)Cl` is the salt of weak base `(C_(6)H_(5)NH_(2))`, aniline) and strong acid (HCI). Due to hydrolysis, its aqueous solution is acidic. KCN is the salt of strong base (KOH) and weak acid (HCN). Due to hydrolysis, its aqueous solution is basic.

Discussion

33.	One "mole" of `N_(2)O_(4)(g)` at `300 K` is kept in a closed container under `1` atm. It is heated to `600 K`, when `20%` by mass of `N_(2)O_(4)(g)` decomposes to `NO_(2)(g)`. The resultant pressure is
Answer» Correct Answer - ` K _p = 0.3136 at m ^(2) `

Discussion

34.	One "mole" of `N_(2)O_(4)(g)` at `300 K` is kept in a closed container under `1` atm. It is heated to `600 K`, when `20%` by mass of `N_(2)O_(4)(g)` decomposes to `NO_(2)(g)`. The resultant pressure isA. `1.2` atmB. `2.4` atmC. `2.0` atmD. `1.0` atm
Answer» Correct Answer - B

Discussion

35.	When `alpha-D` glucose is dissolved in water , it undergoes mutarotation to form an equilibrium mixture of `alpha - D` glucose and `beta - D` glucose containing `63*6` % of the latter. Calculate `K_(c)` for the mutarotation.
Answer» `{:(,alpha -" D glucose ",hArr,beta - " D glucose "),(" At equilibrium " ,364%,,636 %):}` ` K_(c) (636)/(364) = 1* 747`

Discussion

36.	The equilibrium constant for the reaction `H_(2) (g) + Br_(2) (g) hArr 2 HBr (g) " at " 1024 K " is " 16 xx 10^(5)` Find the equilibrium pressure of all gases if `100` bar of HBr is introduced into a sealed container at 1024 K.
Answer» ` {:(,2HBr (g),hArr,H_(2)(g),+,Br_(2)(g),K=1/(16 xx10^(5))),("Intial",10"bar",,,,,),("At eqm.",10-p,,p//2,,p//2,):}` ` K_(p) = ((p//2)( p//2))/(10-p)^(2)= 1/(16 xx10^(5))(p^(2))/(4(10-p)^(2) )= 1/(16 xx 10^(5) )` Taking square root of both sides, we get ` p/(2(10-p) )= 1/(4 xx10^(2)) or 4 xx 10^(2) p= 2 (10-p) or 402 p = 20 or p= 20/402 = 4 98 xx 10^(-2) "bar "` Hence , at equilibrium `p_(H_(2)) =pBr_(2) = p//2 = 2*5 xx 10^(-2) "bar" , p_(HBr) = 10-p ~~ 10 " bar" `

Discussion

37.	For the following, Kc = 6.3 x 1014 at 100 KNO(g) + O3(g) ⇌ NO2(g) + O2(g)Both the forward and reverse reactions in the equilibrium are elementary bi-molecular reactions. What is Kc, for the reverse reaction?
Answer» The reaction, NO₂(g) + O₂(g) ⇌ NO(g) + O₃(g) is reverse of the reaction, NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g). Therefore, the equilibrium constant of the former is reciprocal of the latter. Therefore, K (NO₂(g) + O₂ ⇌ 2NO(g) + O₃(g)) = \(\frac{1}{K(NO(g)+O_3(g)\, \rightleftharpoons\, NO_2(g)+O_2(g))}\) = \(\frac{1}{6.3\times 10^{14}}\) = 0.159 x 10^-14 = 1.59 x 10^-15

37.

For the following, Kc = 6.3 x 1014 at 100 KNO(g) + O3(g) ⇌ NO2(g) + O2(g)Both the forward and reverse reactions in the equilibrium are elementary bi-molecular reactions. What is Kc, for the reverse reaction?

Answer»

The reaction, NO₂(g) + O₂(g) ⇌ NO(g) + O₃(g) is reverse of the reaction, NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g). Therefore, the equilibrium constant of the former is reciprocal of the latter.

Therefore, K (NO₂(g) + O₂ ⇌ 2NO(g) + O₃(g))

= \(\frac{1}{K(NO(g)+O_3(g)\, \rightleftharpoons\, NO_2(g)+O_2(g))}\)

= \(\frac{1}{6.3\times 10^{14}}\) = 0.159 x 10^-14

= 1.59 x 10^-15

Discussion

38.	`1*5` moles of `PCl_(5)` are heated at constant temperature in a closed vessel of 4 litre capacity. At the equilibrium point, `PCl_(5)" is "35 % " dissociated into " PCl_(3) and Cl_(2).` Calculate the equilibrium constant.
Answer» Correct Answer - `0071` `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Intial moles",15,,,,),("At. eqm.",(15 xx 35/100 xx 15),,,,),(,=15 - 0525=0975,,0525,,0525),(" Molar cons.",0975//4,,0525//4,,0525 //4):}` `K_(c) = (( 0* 525 // 4) (0525//4))/((0975 //4))=0*071`

Discussion

39.	Calculate the pH at the equivalence point during the titration of `0.1M, 25 mL CH_(3)COOH` with `0.05M NaOH` solution. `[K_(a)(CH_(3)COOH) = 1.8 xx 10^(-5)]`
Answer» Correct Answer - pH = 8.63

Discussion

40.	15 g sample of `BaO_(2)` is heated to `794^(@)C` in a closed evacuated vessel of 5 litre capacity. How many g of peroxide are converted to `BoO(s)`? `2BaO_(2)(s)hArr2BaO(s)+O_(2)(g), K_(rho)=0.5` atm
Answer» Correct Answer - ` 9.633 g`

Discussion

41.	Formation of `ClF_(3)` from `Cl_(2)and F_(2)` is an exothermic process . The equilibrium system can be represented as `Cl_(2(g))+3F_(2(g))rArr2ClF_(3(g)),DeltaH=-329kJ` Which of the following will increase quantity of `ClF_(3)` in the equilibrium mixture ?A. Increase in temperature, decrease in pressure addition of `Cl_(2)`B. Decrease in temperature and pressure, addition of `ClF_(3)`C. Increase in temperature and pressure, removal of `Cl_(2)`D. Decrease in temperature, increase in pressure, addition of `F_(2)`
Answer» Correct Answer - D Exothermic reaction, decrease in number of moles, increase in concentration of reactants.

Discussion

42.	The exothermic formation of `ClF_(3)` is represented by thr equation: `Cl_(2)(g)+3F_(2)(g) hArr 2ClF_(3)(g), DeltaH=-329 kJ` Which of the following will increase the quantity of `ClF_(3)` in an equilibrium mixture of `Cl_(2), F_(2)`, and `ClF_(3)`?A. Increassing the temperatureB. Removing `Cl_(2)`C. Increasing the volume of the containerD. Adding `F_(2)`
Answer» Correct Answer - D

Discussion

43.	The equilibrium constant for the reaction `CaSO_(4).H_(2)O(s)hArr CaSO_(4).3H_(2)O(s)+2H_(2)O(g)` is equal toA. `([CuSO_(4).5H_(2)O][H_(2)O]^(2))/([CuSO_(4).5H_(2)O])`B. `([CuSO_(4).3H_(2)O])/([CuSO_(4).5H_(2)O])`C. `[H_(2)O]^(2)`D. `[H_(2)O]`
Answer» Correct Answer - C

Discussion

44.	Identify the correct order of solubility of `Na_(2)S,CuS` and ZnS in aqueous solutionA. `CuS gt ZnS gt Na_(2)S`B. `ZnS gt Na_(2)S gt CuS`C. `Na_(2)S gt CuS gt ZnS`D. `Na_(2)S gt ZnS gt CuS`
Answer» Correct Answer - D `Na_(2)S` is strongly ionised and is therefore highly soluble in water . Out of CuS and ZnS the `K_(sp)` of CuS is lower and is precitated first. Therefore it is less soluble than ZnS. The correct order of solubility is `Na_(2)S gt ZnS gt CuS.`

Discussion

45.	If pH of a saturated solution of `Ba(OH)_(2)` is 12, the value of its `K_((sp))` isA. `5.00 xx 10^(-7) M^(3)`B. `4.00 xx 10^(-6) M^(3)`C. `4.00 xx 10^(-7) M^(3)`D. `5.00 xx 10^(-6) M^(3)`
Answer» Correct Answer - A pH = 12 means `[H^(+)]=10^(-12)M`. Hence, `[OH^(-)]=10^(-2)M` `Ba(OH)_(2) hArr Ba^(2+)+2OH^(-)` `[OH^(-)]=10^(-2)M` `:. Ba^(2+)=(10^(-2))/(2) M=5xx10^(-3)M` `K_(sp)=[Ba^(2+)][OH^(-)]^(2)` `=(5xx10^(-3))(10^(-2))^(2)=5xx10^(-7) M^(3)`

Discussion

46.	`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` is
Answer» pH of solution `=12 , [H^(+)] = 10^(-12) M` `[OH^(-)] = (K_(w))/[[H^(+)]]= (10^(-14) M^(2))/(10^(-12)M)= 10^(-2)M` `Ba(OH)_(2) hArr underset([s]](Ba^(2+)) + underset[[2s]](2OH^(-))` `2S = 10^(-2) M, S= 0.5 xx 10^(-2) M` `K_(sp) =[S] [2S]^(2) =4S^(3) =4 xx (0.5 xx 10^(-2))^(3)` `=5.0 xx 10^(-7) M^(3) (Mol L^(-1))^(3)`

Discussion

47.	The pH an aqueous solution of `Ba(OH)_(2)` is `10.0`. If the `K_(sp)` of `Ba(OH)_(2)` is `1.0xx10^(-9)`, the concetration of `Ba^(2+)` ions in the solution isA. `1.0xx10^(-5)M`B. `1.0xx10^(-1)M`C. `1.0xx10^(-4)M`D. `1.0xx10^(-2)M`
Answer» Correct Answer - B Unless otherwise stated, the solvent is water and the temperature is `25^(@)C`. `Ba(OH)_(2)(s)hArrBa^(2+)(aq.)+2OH^(-)(aq.)` `pH=10.0` Since `pH+pOH=14.0(at 298 K)` `pOH= 14.0-10.0` `= 4.0` `:. C_(OH^(-))= 10^(-4)mol L^(-1)` According to solubility equilibrium, `K_(sp)=C_(Mg^(2+))C_(OH^(-))^(2)` `1.0xx10^(-9)=C_(Mg^(2+))(10^(-4))^(2)` `C_(Mg^(2+))=(1.0xx10^(-9))/(10^(-8))` `= 1.0xx10^(-1)M`

Discussion

48.	What is meant by the conjugate acid-base pair ? Find the conjugate acid/base for the secies : `HNO_(2), CN^(-), HClO_(4),F^(-),OH^(-),CO_(3)^(2-) and S^(2-)`
Answer» An acid-base pair which differ by a proton is called conjugate acid base pair . `NO_(2)^(-), HCN, ClO_(4)^(-) , HF, H_(2)O ` (acid) or `O^(2-) ` (base), `HCO_(3)^(-) and HS^(-)`.

Discussion

49.	If `CO_(2)` is allowed to escape from the following reaction at equilibrium `CO_(2)+H_(2)O+H_(2)CO_(3)hArr 2H^(+) + 2HCO_(3)^(-)`A. pH will decreaseB. pH will remain constantC. pH will increaseD. forward reaction will be favoured
Answer» Correct Answer - C Equilibrium will sift in the backward direction. Hence, `H^(+)` ion concentration will decrease or pH will increase.

Discussion

50.	State Le Chatelier‘s principle.
Answer» It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change.

Discussion

Explore topic-wise InterviewSolutions in .

Which of the following equimolar solutions can act as a buffer solution?(a) NH4Cl and NH4OH(b) HCl and NaCl(c) HCOOH and HCOONa(d) HNO3 and NH4NO3

The range of pH for acidic and basic buffer is where `K_(a)` and `K_(b)` are the acid base dissociation constants, respectively.A. form `pH=pK_(a)+-2 to pH=pK_(b)+-2`B. from `pH=pK_(a)+1to pH=pK_(b)+1`C. from `pH=pK_(a)+-1to pH=pK_(b)+-1`D. from `pH=pK_(a)+1to pH=pK_(b)-1`

Which of the following expression tepresents the Henderson equation for an acidic buffer ?A. `pH=(1)/(2)pK_(a)-(1)/(2)logC`B. `pH=pK_(a)-l"og"(["Conjugate base"])/(["Acid"])`C. `pH=pK_(a)+"log"(["Conjugate base"])/(["Acid"])`D. `pH=pK_(a)`

An acidic buffer solution can be prepared by mixing equimolar amounts ofA. `B(OH)_(3)` and `Na_(2)B_(4)O_(7).10H_(2)O`B. `NH_(3)` and `NH_(4)CI`C. `HCI` and `NaCI`D. `CH_(3)COOH` and `CH_(3)COONa`

Through a solution containing `Cu^(2+) and Ni^(2+),H_(2)S` gas is passed after adding dil HCl, which will precipitate out and why ?

Give an example of acidic buffer?

How does dilution with water affect the pH of a buffer solution?

Through a solution containing Cu2+and Ni2+,H2S gas is passed after adding dill HCl, which will precipitate out and why?

Write the expression for Ksp for AB3 type salt With solubility s.

Write the expression for Ksp for AB type salt With solubility s

What would be the expression for solubility product of calcium phosphate in terms of its molar solubility, S.

The solubility product of `AgI` at `25^(@)C` is `1.0xx10^(-16) mol^(2) L^(-2)`. The solubility of `AgI` in `10^(-4) N` solution of `KI` at `25^(@)C` is approximately ( in `mol L^(-1)`)A. `1.0xx10^(-16)`B. `1.0xx10^(-12)`C. `1.0xx10^(-10)`D. `1.0xx10^(-8)`

40 ml of 0.1 M ammonia is mixed with 20 ml of 0.1 M HCl. What is the pH of the mixture ? (` pK_(b)` of ammonia solution is 4.74)A. 4.74B. 2.26C. 9.26D. `5.00`

Calculate the pH of `0.033` M ammonia solution if `0.033 M NH_(4)Cl` is introduced in this solution at the same temperature `(K_(b)` for `NH_(3)=1.77xx10^(-5)`)

What is the relationship between `pK_(a) and pK_(b)` values where `K_(a) and K_(b)` represent constants of the acid and its conjugate base respectively ?

The `pH` of milk, black coffee, tomato juice, lemon juice and egg white are `6.8, 5.0, 4.2, 2.2` and `7.8` respectively. Calculate corresponding hydrogen ion concentration in each.

Lemon juice has a pH = 2.1 . If all the acid in lemon is citric acid `(H "Cit." hArr H^(+)+"Cit"^(-1))` and `K_(a)` for citric acid is `8.4xx10^(-4)` mole/litre, what is the concentration of citric acid in lemon juice ?

At 298 K, the pH of a lemon juice is 2.32. Calculate its `[H_(3)O^(+)]" and "[OH^(-)].`

`{:(,"Column I",,"Column II (pH)"),((A),"MilK",(p),2.2),((B),"Human saliva",(q),6.4),((C),"Human blood",(r),6.8),((D),"Lemon juice",(s),7.4):}`A. A-p, B-q, C-s, D-rB. A-r, B-q, C-s, D-pC. A-r, B-s, C-q, D-pD. A-q, B-r, C-s, D-p

The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.

The ionization constant of phenol is `1.0xx10^(-10)`. What is the concentration of phenate ion in 0.05 M solution of phenol ? What will be its degree of ionization if the solution is also 0.01 M in sodium phenate ?

An aqueous solution contains 0.10 M `H_(2)S " is "1.0 xx 10^(-7)" and that of " S^(2-) "ion from " HS^(-) "ions is " 1.2 xx 10^(-3) ," then the concentration of " S^(2-) " ions in aqueous solution is "`A. ` 5 xx 10^(-8)`B. `3 xx 10^(-20)`C. `6xx 10^(-21)`D. `5xx 10^(-19)`

Calculate the pH at which `Mg(OH)_(2)` begins to precipitate from a solution containing 0.10 M `Mg^(2+)` ions. `K_(sp)` of Mg`(OH)_(2)=1xx10^(-11)`.

Calculate pH at which `Mg(OH)_(2)` begins to precipitate from a solution containing `0.10M Mg^(2+)` ions. `(K_(SP)of Mg(OH)_(2)=1xx10^(-11))`

Calculate pH at which `Mg(OH)_(2)` begins to precipitate from a solution containing `0.10M Mg^(2+)` ions. `(K_(SP)of Mg(OH)_(2)=1xx10^(-11))`A. 4B. 6C. 9D. 7

The `K_(sp)` of `Mg(OH)_(2)` is `1xx10^(-12). 0.01M Mg^(2+)` will precipitate tate at the limiting pH ofA. `1.3`B. `9`C. `3.5`D. `8`

In the reaction `A+B hArrC+D` what will happen to the equlibrium if concentration of A is increased?

An aqueous solution of `CuSO_(4)` is acidic while that of `Na_(2)SO_(4)` is neutral. Explain.

The aqeous solutions of `HCOONa, C_(6)H_(5)NH_(3)CI`, and KCN are, respectively,A. acidic, acidic, basicB. acidic, basic, neutralC. basic, neutral, neutralD. basic, acidic, basic

One "mole" of `N_(2)O_(4)(g)` at `300 K` is kept in a closed container under `1` atm. It is heated to `600 K`, when `20%` by mass of `N_(2)O_(4)(g)` decomposes to `NO_(2)(g)`. The resultant pressure is

One "mole" of `N_(2)O_(4)(g)` at `300 K` is kept in a closed container under `1` atm. It is heated to `600 K`, when `20%` by mass of `N_(2)O_(4)(g)` decomposes to `NO_(2)(g)`. The resultant pressure isA. `1.2` atmB. `2.4` atmC. `2.0` atmD. `1.0` atm

When `alpha-D` glucose is dissolved in water , it undergoes mutarotation to form an equilibrium mixture of `alpha - D` glucose and `beta - D` glucose containing `63*6` % of the latter. Calculate `K_(c)` for the mutarotation.

The equilibrium constant for the reaction `H_(2) (g) + Br_(2) (g) hArr 2 HBr (g) " at " 1024 K " is " 1*6 xx 10^(5)` Find the equilibrium pressure of all gases if `10*0` bar of HBr is introduced into a sealed container at 1024 K.

For the following, Kc = 6.3 x 1014 at 100 KNO(g) + O3(g) ⇌ NO2(g) + O2(g)Both the forward and reverse reactions in the equilibrium are elementary bi-molecular reactions. What is Kc, for the reverse reaction?

`1*5` moles of `PCl_(5)` are heated at constant temperature in a closed vessel of 4 litre capacity. At the equilibrium point, `PCl_(5)" is "35 % " dissociated into " PCl_(3) and Cl_(2).` Calculate the equilibrium constant.

Calculate the pH at the equivalence point during the titration of `0.1M, 25 mL CH_(3)COOH` with `0.05M NaOH` solution. `[K_(a)(CH_(3)COOH) = 1.8 xx 10^(-5)]`

15 g sample of `BaO_(2)` is heated to `794^(@)C` in a closed evacuated vessel of 5 litre capacity. How many g of peroxide are converted to `BoO(s)`? `2BaO_(2)(s)hArr2BaO(s)+O_(2)(g), K_(rho)=0.5` atm

The equilibrium constant for the reaction `CaSO_(4).H_(2)O(s)hArr CaSO_(4).3H_(2)O(s)+2H_(2)O(g)` is equal toA. `([CuSO_(4).5H_(2)O][H_(2)O]^(2))/([CuSO_(4).5H_(2)O])`B. `([CuSO_(4).3H_(2)O])/([CuSO_(4).5H_(2)O])`C. `[H_(2)O]^(2)`D. `[H_(2)O]`

Identify the correct order of solubility of `Na_(2)S,CuS` and ZnS in aqueous solutionA. `CuS gt ZnS gt Na_(2)S`B. `ZnS gt Na_(2)S gt CuS`C. `Na_(2)S gt CuS gt ZnS`D. `Na_(2)S gt ZnS gt CuS`

If pH of a saturated solution of `Ba(OH)_(2)` is 12, the value of its `K_((sp))` isA. `5.00 xx 10^(-7) M^(3)`B. `4.00 xx 10^(-6) M^(3)`C. `4.00 xx 10^(-7) M^(3)`D. `5.00 xx 10^(-6) M^(3)`

`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` is

The pH an aqueous solution of `Ba(OH)_(2)` is `10.0`. If the `K_(sp)` of `Ba(OH)_(2)` is `1.0xx10^(-9)`, the concetration of `Ba^(2+)` ions in the solution isA. `1.0xx10^(-5)M`B. `1.0xx10^(-1)M`C. `1.0xx10^(-4)M`D. `1.0xx10^(-2)M`

What is meant by the conjugate acid-base pair ? Find the conjugate acid/base for the secies : `HNO_(2), CN^(-), HClO_(4),F^(-),OH^(-),CO_(3)^(2-) and S^(2-)`

If `CO_(2)` is allowed to escape from the following reaction at equilibrium `CO_(2)+H_(2)O+H_(2)CO_(3)hArr 2H^(+) + 2HCO_(3)^(-)`A. pH will decreaseB. pH will remain constantC. pH will increaseD. forward reaction will be favoured

State Le Chatelier‘s principle.

The equilibrium constant for the reaction `H_(2) (g) + Br_(2) (g) hArr 2 HBr (g) " at " 1024 K " is " 16 xx 10^(5)` Find the equilibrium pressure of all gases if `100` bar of HBr is introduced into a sealed container at 1024 K.