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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1201. |
`H_(2)S` gas is passed into one `dm^(3)` of a solution containing 0.1 mole of `Zn^(2+)` and 0.01 mole of `Cu^(2+)` till the sulphide ion concentration reaches `8.1xx10^(-19)` moles. Which one of the following statements is true ? (`K_(sp)` of ZnS and CuS are `3xx10^(-22) and 8xx10^(-36)` respectively)A. Only ZnS precipitatesB. Both CuS and ZnS precipitateC. Only CuS precipitatesD. No precipitation occurs |
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Answer» Correct Answer - B Precipitation occurs if ionic product `gt K_(sp)` Ionic product of `ZnS=[Zn^(2+)][S^(2-)]` `=0.1xx8.1xx10^(-19)=8.1xx10^(-20) gt K_(sp)` Ionic product of CuS `=[cu^(2+)][S^(2-)]` `=0.01xx8.1xx10^(-19)=8.1xx10^(-21) gt K_(sp)` Hence, both will precipitate. |
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| 1202. |
Which of the following is correct regarding the Lewis concept of acids and bases ?A. It cannot explain the cases when a species is donating as well as accepting electron pairs.B. It cannot explain the acidic chaacter of protonic acids.C. It cannot explain quantiatively the strength of acids as well as bases.D. All of these |
| Answer» Carbon monoxide in `[Ni(CO)_(4)]` donates as well as etc., cannot pairs. The acidic character of `HCI, H_(2)SO_(4)`, etc., cannot be explained by the Lewis definition. | |
| 1203. |
The concentration of `OH^(-)` ions in a `0.050 M HNO_(3)` solution isA. `2.0xx10^(-13)`B. `1.0xx10^(-13)`C. `0.5xx10^(-13)`D. `1.5xx10^(13)` |
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Answer» Correct Answer - A Since `HNO_(3)` is a strong acid, it ionizes almost completely. Thus, the concentration of `H^(+)` ions form `HNO_(3)` is equal to the concentration of `H^(+)` ions form `HNO_(3)` is equal to the concentration of `HNO_(3)`, i.e., `5xx10^(-2)` . The contribution of `H^(+)` ions form water will be negligible. `K_(w)=C_(H^(+))C_(OH^(-))` `C_(OH^(-))=(K_(w))/(C_(H^(+)))=(1.0xx10^(-14))/(5xx10^(-2))` `=2.0xx10^(-13)` |
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| 1204. |
If the salt `M_(2)X, QY_(2) and PZ_(3)` have the same solubilities, their `K_(sp)` values are related asA. `K_(sp) (M_(2)X) = K_(sp) (Q Y_(2)) lt K_(sp) (PZ_(3))`B. `K_(sp) (M_(2)X) gt K_(sp) (Q Y_(2)) = K_(sp) (PZ_(3))`C. `K_(sp) (M_(2)X) lt K_(sp) (Q Y_(2)) = K_(sp) (PZ_(3))`D. `K_(sp) (M_(2)X) gt K_(sp) (Q Y_(2)) gt K_(sp) (PZ_(3))` |
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Answer» Correct Answer - A `M_(2)X and QY_(2)` are salts of the type `A_(2)B` for which `K_(sp) = 4 S^(3)`. If S is same, `K_(sp)` will be equal. For `PZ_(3) , K_(sp) = S(3S)^(3)=27S^(4), ` i.e., much higher. Hence, (a) is the correct option. |
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| 1205. |
If for heterogeneous equilibrium, `CaCO_3(s)hArr CaO(s)+CO_2(g), K_(eq)=1` at `1 atm` pressure, the corresponding temperature is given byA. `T=(DeltaG^(Θ))/(DeltaH^(Θ))`B. `T=(DeltaG^(Θ))/(R)`C. `T=(DeltaS^(Θ))/(DeltaH)`D. `T=(DeltaH^(Θ))/(DeltaS^(Θ))` |
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Answer» Correct Answer - D According to thermodynamics `DeltaG^(Θ)=DeltaH^(Θ)-TDeltaS^(Θ)` At equilibrium, `DeltaG^(Θ)=0`. Thus, `0=DeltaH^(Θ)-TDeltaS^(Θ)` `DeltaH^(Θ)=TDeltaS^(Θ)` `T=DeltaH^(Θ)//DeltaS^(Θ)` |
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| 1206. |
What is the value of the equilibrium constant for the reaction 2 NO2(g) ------> N2O4(g) at 100o C? |
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Answer» The desired reaction is the reverse of the reaction for which the Kc is known. The equilibrium expression is the reciprocal of that given. K'c = 1/Kc = 1/0.212 = 4.72 |
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| 1207. |
Given : `Ag(NH_(3))_(2)^(+)hArr Ag^(+)+2NH_(3),K_(c)=6.2xx10^(-8) and K_(sp) "of" AgCl=1.8xx10^(-10)` at 298 K. Calculate the concentration of the complex in 1.0 M aqueous ammonia. |
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Answer» `K_(c) = ([Ag^(+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)^(+)])` ...(i) `[NH_(3)]=1.0 M ` (as `NH_(3)` obtained from dissociation of the complex is negligible) To know `[Ag^(+)]`, we have `K_(sp) = [Ag^(+)][Cl^(-)]` But `[Cl^(-)]= "Total" [Ag^(+)]` in the free and combined state `= [Ag^(+)]+[Ag(NH_(3))_(2)^(+)]` `:. (K_(sp))/([Ag^(+)])=[Ag^(+)]+[Ag(NH_(3))_(2)^(+)] or [Ag(NH_(3))_(2)^(+)]=(K_(sp))/([Ag^(+)])-[Ag^(+)] ...(ii)` Substituting the values in eqn. (i), we get `6.2xx10^(-8)=([Ag^(+)][1.0]^(2))/((1.8xx10^(-10))/([Ag^(+)])-[Ag^(+)])` Takiing `[Ag^(+)]=C "mol" L^(-1)`, we have `6.2xx10^(-8)=(C^(2))/(1.8xx10^(-10)-C^(2))~=(C^(2))/(1.8xx10^(-10))` or `C^(2)=11.16xx10^(-18)` ltbr. or `C=3.34 xx 10^(-9) M, i.e., [Ag^(+)]=3.3410^(-9)M` Substituting this value in eqn. (ii), we get `[Ag(NH_(3))_(2)^(+)]=(1.8xx10^(-10))/(3.34xx10^(-9))-3.34xx10^(-9)~=(1.8xx10^(-10))/(3.34xx10^(-9))=0.054 M` |
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| 1208. |
A liquid is in equilibrium with its vapur in a sealed container at a fixed temperature. The volume of the container is suddently increases. (a) What is the intial effect of the change on vapour pressure ? (b) How do rates of evaporation and consideration change intially ? (c) What happens when equilibrium is restored finally and what will be the final vapour pressure ? |
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Answer» (a) Intially, the vapour pressure will decrease because the same account of vapour are now distributed in larger space. (b) The rate of evaporation remains constant at constant temperature in a closed vessel 9discussed in unit 5 under vapur pressure. ). However, the rate of condensation will be low initially because there are fever molecules per unit volume in the vapour phase and hence the number ofcollisions per unit time with the liquid surface decreases. (c ) When equilibriumis restored , rate of evaporation = rate of consensation . The final valour pressure will be same as it was originally because vapour pressure of a liquid depends only on temperature and not volume. |
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| 1209. |
The solubility product of AgCI in water is`1.5xx10^(-10)`. Calculate its solubility in `0.01M NaCI`. |
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Answer» Let the solubility of AgCI in water be S mol `L^(-1)`. `AgCI(s) hArr underset(S)(Ag^(+)(aq)) + underset(S)(CI^(-)(aq))` `NaCI overset((aq))(to) underset(0.01M)(Na^(+) (aq)) +underset(0.01M)(CI^(-) (aq))` `"Now"" "[Ag^(+)]=S , [CI^(-)] =[CI^(-)] " from " AgCI + [CI^(-)] " from " NaCI` `=S + 0.01 ~~ 0.01 ("as" S lt lt 0.01)` `"Now"" "K_(sp) =[Ag^(+)][CI^(-)] " or " 1.5 xx 10^(-10) =S xx 0.01` `S=(1.5 xx 10^(-10))/(0.01) =1.5 xx 10^(-8) "mol "L^(-1)` |
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| 1210. |
5 moles of `SO_(2) and 5 " moles of " O_(2) ` are allowed to react . At equilibrium , it was found that 60% `SO_(2)` is used up . If the pressure of the mixture is one atmosphere , the partial pressure of `O_(2)` isA. `0*52` atmB. `0*21` atmC. ` 0*41`atmD. `0*82` atm |
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Answer» Correct Answer - C ` {:(,2 SO_(2)(g),+,O_(2)(g),hArr,2 SO_(3)(g)),("Intial",5,,5,,0):}` As 60 % `SO_(2)` is used up, no of moles of `SO_(2) "used up" = 60/100 xx 5 =3` ` :. " No. of moles of " SO_(2) " at equilibrium " = 5-3=2` As 2 moles of `SO_(2) " react with 1 mole of " O_(2)` `=1/2 xx 3 = 1.5 " moles" ` i.e. No. of moles of `O_(2)` at equilibrium `=5-1*5 = 3.5 ` As 2 moles of `SO_(2) " produce 2 moles of " SO_(3) ` `:. " No . of moles of "SO_(3) " equilibrium = 3 moles"` `= 2 + 3*5 + 3 = 8*5 ` As 2 moles of `SO_(2)"produce 2 moles of"SO_(3)` ` :. " No. of moles of " SO_(3) " at equilibrium" = 3 moles ` `:. "Total no. of moles at equilibrium "` |
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| 1211. |
Which of the following is the strongest base ?A. `C_(6)H_(5)NH_(2)(pK_(b)=9.42)`B. `C_(6)H_(5)NHCH_(3)(pK_(b)=9.15)`C. `C_(6)H_(5)N(CH_(3))_(2)(pK_(b)=8.94)`D. `C_(6)H_(5)NHC_(2)H_(5)(pK_(b)=8.89)` |
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Answer» Correct Answer - D The smaller the value of `pK_(b)`, the larger is the value of `K_(b)` and the stronger is the base. |
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| 1212. |
What will be the value of pH of 0.01 mol `dm^(-3) cH_(3)CO OH (K_(a) = 1.74 xx 10^(-5))`?A. 3.4B. 3.6C. 3.9D. `3.0` |
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Answer» Correct Answer - A `pH = (1)/(2) (pK_(a)-log c)=(1)/(2) [-log (1.74xx10^(-5))-log 10^(-2)]` `=(1)/(2) [ (5-0.2405) + 2] = (1)/(2) (6.76) = 3.38 ~= 3.4`. |
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| 1213. |
What is `[H^(+)]` in mol/L of a solution that is 0.20 M in `CH_(3)CO ON a ` and 0.10M in `CH_(3)CO OH` ? `K_(a)` for `CH_(3)CO OH = 1.8 xx 10^(-5)`.A. `9.0xx10^(-6)`B. `3.5xx10^(-4)`C. `1.1xx 10^(-5)`D. `1.8xx10^(-5)` |
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Answer» Correct Answer - A The given solution is an acidic buffer. Hence, its pH will be `pH = pK_(a) +log. (["Salt"])/(["Acid"])` `=-log.(1.8xx10^(-5))+log.(0.20)/(0.10)` or `-log(h^(+)]=log.(2)/(1.8xx10^(-5))` or `log.(1)/([H^(+)])=log.(2)/(1.8xx10^(-5))` or `[H^(+)]=(1.8xx10^(-5))/(2) = 0.9xx10^(-5)M` `=9xx10^(-6)M` |
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| 1214. |
Calculate the concentration of `H_3O^(+)` of a mixture (solution) that is `0.010M` in `CH_3COOH` and `0.20M` in `NaCH_(3-)COO`. `(K_a=1.8xx10^(-5))` |
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Answer» Strategy: Write the suitable equations for both `NaCH_3COO` and `CH_3COOH` and the ionization constant expression for `CH_3COOH`. Then, represent the equilibrium concentrations algebraically and substitute into the `K_a` expression and solve for the unknown. Solution: The suitable equations and the ionization constant expression are `Na^(+)CH_3COO^(-)rarr Na^(+)+CH_3COO^(-)` (to completion) `CH_3COOH+H_2O=H_3O^(+)+CH_3COO^(-)` (reversible) `K_a=([H_3O^(+)][H_3COO^(-)])/([CH_3COOH])=1.8xx10^(-5)` This expression of `K_a` is valid for all solutions that contain `CH_3COOH`. In solutions that contain both `CH_3COOH` and `NaCH_3COO`, acetate ions `(CH_3COO^(-))` come from two sources. The ionization constant is satisfied by the total `CH_3COO^(-)` concentration. Because `NaCH_3COO` is completely dissociated, the `[CH_3COO^(-)]` from `NaCH_3COO` will be `0.20 mol L^(-1)`: `{:(NaCH_3COO(aq.),rarr,Na^(+)(aq.),+,CH_3COO^(-)(aq.)),(0.20M,,0.0,,0.0),(0.0,,0.20M,,0.20M):}` Let `x=[CH_3COOH]` that ionizes. Then x is also equal to `[H_3O^(+)]` and `[CH_3COO^(-)]` from `CH_3COOH`. The total concentration of `CH_3COO^(-)` is `(0.20+x)M` and the concentration of unionized `CH_3COOH` is `(0.10-x)M`. `{:(,CH_(3)COOH(aq.)+H_(2)O(l)hArrH_(3)O^(+)(aq.)+CH_(3)COO^(-)(aq.)),("Initial (M)"," 0.10 0 0.20"),("Change (M)"," -x +x +x"),("Equilibrium (M)",bar(" (0.10-x) x (0.20+x) ")):}` The substitution of equilibrium concentrations into the ionization constant expression for acetic acid gives `K_a=(C_(H_3O^+)+C_(CH_3COO^-))/(C_(CH_3COOH))` `1.8xx10^(-5)=((x)(0.20+x))/((0.10-x))` The small value of `K_a` suggets that x is very small. This leads to two assumptions: (i) `x lt lt 0.10`. So `(0.10-x)~~0.10`. This implies that very little `CH_3COOH` ionizes. (ii) `x lt lt 0.20`. So `(0.20+x)~~0.20`. This implies that most of the `CH_3COO^(-)` comes from `NaCH_3COO` and very little `CH_3COO^(-)` comes from the ionization of `CH_3COOH`. It is reasonable to assume that `x` (from the ionization of `CH_3COOH`) is small, because `CH_3COOH` is a weak acid, and its ionization is further suppressed by the high concentration of `CH_3COO^(-)` formed by the soluble salt, `NaCH_3COO`. Applying these assumptions gives `(0.20x)/(0.10)=1.8xx10^(-5)` `x=9.0xx10^(-6)` `:. C_(H_3O^(+))=x=9.0xx10^(-6)` |
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| 1215. |
If the concentration of the weak monoprotic acid HA is C mmol `L^(-1)` and its ionization constant is `K_(a)`, thenA. `C_(H)^(+)=C//2`B. `C_(H)^(+)=sqrt(C )`C. `C_(H)^(+)=sqrt(K_(a)C)`D. `C_(H)^(+)=C//C_(a)` |
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Answer» Correct Answer - C `{:(,HA(aq.),hArrH^(+)(aq.),+A^(-)(aq.)),("Inital (M)",C,0,0),("Change (M)",C-Calpha,Calpha,Calpha),("Equilibrium (M)",C-Calpha,Calpha,Calpha):}` where `alpha` is the degree of ionization. `K_(a)=(C_(H^(+))C_(A^(-)))/(C_(HA))` `=((Calpha)(Calpha))/(C(1-alpha))` `=Calpha^(2)` because `alpha lt lt 1` or `alpha=sqrt((K_(a))/(C ))` `C_(H^(+))=Calpha=Csqrt((K_(a))/(C ))=sqrt(K_(a)C )` |
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| 1216. |
The `pK_(a)` of a weak acid `(HA)` is `4.5`. The `pOH` of an aqueous buffered solution of `HA` in which `50%` of the acid is ionized is:A. `4.5`B. `7.0`C. `9.5`D. `2.5` |
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Answer» Correct Answer - C Ionization of weak acid (HA) is represented as `HA(aq.)hArr H^(+)(aq.)+A^(-)(aq.)` When HA is `50%` ionized. `C_(HA)=C_(A^(-))` According to the Henderson equation, `pH=pK_(a)+log.(C_(A^(-)))/(C_(HA))` `=pK_(a)+log 1` `=pK_(a)=4.5` At `25^(@)C`, `pH+pOH=pK_(w)=14` Thus, `pOH=14-pH` `=14-4.5` `=9.5` |
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| 1217. |
The `pK_(a)` of weak acid (HA) is 4.5 . The pOH of an aqueous buffered solution of HA is which 50% of the acid is ionized isA. `7.0`B. `4.5`C. `2.5`D. `9.5` |
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Answer» Correct Answer - D When HA is 50% ionized , [HA] = `[A^(-)]` `pH = pK_(a)+log .([A^(-)])/([HA])=pK_(a)=4.5 ` (Given) `pOH = 14- pH = 14 - 4.5 = 9.5. |
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| 1218. |
Which of the following acids is not completely ionized in aqueous solution ?A. Hydrochloric acidB. Sulphuric acidC. Acetic acidD. Nitric acid |
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Answer» Correct Answer - C Though all are polar covalent compounds, mineral acids `(HCI, H_(2)SO_(4)`, and `HNO_(3))` are completely ionized into its constituent ions, while acetic acid is only partially ionized `(65%)`. The extent to which ionization occurs depends upon the strength of the bond and the extent of solvation of ions produced. |
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| 1219. |
Equal volumes of the following `Ca^(2+)` and `F^(-)` solutions are mixed. In which of the solutions will precipitation occur ? `(K_(sp) "of" CaF_(2)=1.7xx10^(-10))` (1) `10^(-2) M Ca^(2+) + 10^(-5) M F^(-)` (2) `10^(-3) M Ca^(2+)+10^(-3)M F^(-)` (3) `10^(-4) M Ca^(2+)+ 10^(-2)M F^(-)` (4)`10^(-2) M Ca^(2+) + 10^(-3) M F^(-)` Select the correct answer using the codes given below:A. `10^(-2) M Ca^(2+) + 10^(-5) M F^(-)`B. `10^(-3) M Ca^(2+) + 10^(-3) M F^(-)`C. `10^(-4) M Ca^(2+) + 10^(-2) M F^(-)`D. `10^(-2) M Ca^(2+) + 10^(-3) M F^(-)` |
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Answer» Correct Answer - D Ionic product of `CaF_92)=[Ca^(2+)][F^(-)](2)`. Concentration of ions will be halved after mixing. Thus, ionic products will be (1) `(10^(-2))/(2)xx((10^(-5))/(2))^(2)=(1)/(8)xx10^(-12)` (2) `(10^(-3))/(2)xx((10^(-3))/(2))^(2)=(1)/(8)xx10^(-9)` (3) `(10^(-4))/(2)xx((10^(-2))/(2))^(2)=(1)/(8)xx10^(-8)` (4) `(10^(-2))/(2)xx((10^(-3))/(2))^(2)=(1)/(8)xx10^(-8)` In (2), (3) and (4), ionic product `gt K_(sp)`. Hence, precipitation will occur in (2), (3) and (4). |
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