InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 1051. |
To `1L` of a `1.6xx10^(-3)M` aqueous solution of ethylene diamine `(K_(b_(1))=8xx10^(-5),K_(b_(2))=2.7xx10^(-8)),5xx10^(-4)` mole of `Ba(OH)_(2)` is added.Then:A. `pH_(f)-pH_(i)=0.5`B. `alpha_(i)-alpha_(f)=0.12`C. `([C_(2)N_(2)H_(10)^(2+)])/([C_(2)N_(2)H_(10)^(2+)]_(f))=0.128`D. `([C_(2)N_(2)H_(9)^(+)]_(i))/([C_(2)N_(2)H_(9)^(2+)]_(f))=0.128` |
|
Answer» initial :`8xx10^(-5)=(1.6xx10^(-5)alpha^(2))/(1-alpha)" "rArr alpha=(1)/(5)` `:.[OH^(-)]=3.2xx10^(-4)" "pH=10.5` `8xx10^(-5)xx2.7xx10^(-8)=((3.2xx10^(-4))^(2)(C_(2)N_(2)H_(10)^(2+))_(i))/(1.28xx10^(-3))` `rArr2.7xx10^(-8)=[C_(2)N_(2)H_(10)^(2+)]`_(i) Final `pH=11," " 8xx10^(-5)=10^(-3)xxalpha rArr alpha=0.08` `[C_(2)N_(2)H_(9)^(+)]=1.28xx10^(-4)` `2.7xx8xx10^(-13)=((10^(-6))[C_(2)N_(2)H_(10)^(2+)])_(f)/(1.6xx10^(-3))rArr [C_(2)N_(2)H_(10)^(2+)]=1.28xx2.7xx10^(-9)M` So A,B and D are correct but `([C_(2)N_(2)H_(10)^(2+)])/([C_(2)N_(2)H_(10)^(2+)])=(1)/(0.128)` |
|
| 1052. |
Solubilty product of `Mg(OH)_(2),Cd(OH)_(2),Al(OH)_(3)` and `Zn(OH)_(2)` are `4xx10^(-11),8xx10^(-6),8.5xx10^(-23)` and `1.8xx10^(-14)` resectively. The cation, that will precipitate first as hydroxide, on adding limited quantity of `NH_(4)OH` in a solution containing equimolar amount of metal cation, is :A. `Al^(3+)`B. `Zn^(2+)`C. `Mg^(2+)`D. `Cd^(2+)` |
|
Answer» Correct Answer - a |
|
| 1053. |
Which of the following increases with dilute at a given temperature?A. `pH` of `10^(-3)`M acetic acid solutionB. Degree of dissociation of `10^(-3)`C. Degree of dissociation of `10^(-3)` M acetic acidD. Degree of dissociation of `10^(-3)` M aniline |
|
Answer» Correct Answer - a,c,d |
|
| 1054. |
Which one of the following statements is not trueA. The conjugate base of `H_(2)PO_(4)^(-)` is `HPO_(4)^(2-)`B. `pH + pOH = 14` for all aqueous solutionsC. The pH of `1 xx 10^(-8)` M HCl is 8D. 96, 500 coulombs of electricity when passed througha `CuSO_(4)` solution deposits 1 gram equivalent of copper at the cathode |
|
Answer» Correct Answer - C Because pH = 8 is basic nature but HCl is a strong acid. |
|
| 1055. |
An organic dye, cosine used to detect end point of precipitation titration by adsorption is calledA. Absorption indicatorB. Adsorption indicatorC. Chemical indicatorD. None of these |
| Answer» Correct Answer - B | |
| 1056. |
1 M NaCl and 1 M HCl are present in an queous solution. The solution isA. Not a buffer solution with `pH lt 7`B. Not a buffer solution with `pH gt 7`C. A buffer solution with `pH lt 7`D. A buffer solution with `pH gt 7` |
|
Answer» Correct Answer - A It is a neutral solution and its pH = 7. |
|
| 1057. |
Phenolphthalein does not act as an indicator for the titration betweenA. NaOH and `CH_(3)COOH`B. `H_(2)C_(2)O_(4)` and `KMnO_(4)`C. `Ba(OH)_(2)` and HClD. KOH and `H_(2)SO_(4)` |
| Answer» Correct Answer - B | |
| 1058. |
pH of completely dissociated 0.005 M `H_(2)SO_(4)` isA. 3B. 4C. 2D. 5 |
|
Answer» Correct Answer - C `H_(2)SO_(4)` ionized in two step. |
|
| 1059. |
`1.0L` of solution which was in equilibrium with solid mixture of `AgCI` and `Ag_(2)CrO_(4)` was found to contain `1 xx 10^(-4)` moles of `Ag^(+)` ions, `1.0 xx 10^(-6)`moles of `CI^(-)` ions and `8.0 xx 10^(-4)` moles of `CrO_(4)^(2-)` ions. At constant volume, `Ag^(+)` ions are added slowly to the above mixture till `8.0 xx 10^(-7)` moles of `AgCI` got precipitated. How many moles of `Ag_(2)CrO_(4)` were also precipitated ? Given your answer after multiplying with `10^(6)`. |
|
Answer» Correct Answer - `0768` |
|
| 1060. |
The pH of a `0.005 M H_(2)SO_(4)` solution isA. `2.5`B. `4.5`C. `2.0`D. `1.0` |
|
Answer» Correct Answer - C `[H^(+)] = 2 xx 0.005 = 0.01M` `P^( H) = -log_(10)[H^(+)]` |
|
| 1061. |
What is `[Ag^(+)]` in a solution made by dissolving both `Ag_(2)CrO_(4)` and `Ag_(2)C_(2)O_(4)` until saturation is reached with respect to both salt ? `[K_(sp) = (Ag_(2)C_(2)O_(4)) = 2 xx 10^(-11), K_(sp) (Ag_(2)CrO_(4)) = 2 xx 10^(-12)]`A. `2.80 xx 10^(-4)`B. `7.6 xx 10^(-5)`C. `6.63 xx 10^(-6)`D. `3.52 xx 10^(-4)` |
|
Answer» Correct Answer - D `Ag_(2)CrO_(4)(s) hArr underset(2x+2y)((2Ag)^(+)(aq.)) + underset(x)(CrO_(4)^(2-)(aq.))` `Ag_(2)C_(2)underset(2y+2x)(O_(4)(s))hArr2Ag^(+)underset(y)((aq.))+C_(2)O_(4)^(2-)(aq.)` `(K_(sp_(1)))/(K_(sp_(1))) = x/y = (2xx10^(-12))/(2 xx 10^(-11)) rArr = 0.1 ` `2 xx 10^(-11) = (2 x + 2y)^(2).y` `2 xx 10^(-11) = 4.84y^(3)` `y = 1.6 xx 10^(-4)` `:. x = 0.16 xx 10^(-4)` Total `[Ag^(+)] = 2 x + 2y` `= (2 xx 0.16 + 2 xx 1.6) xx 10^(-4) = 3.52 xx 10^(-4)` |
|
| 1062. |
Calculate `[Ag^(o+)]` in a solution made by dissolving both `AgCrO_(4)` and `AgC_(2)O_(4)` untill saturation is reached with respect to both salts. Given `K_(sp) of Ag_(2)CrO_(4)` and `Ag_(2)C_(2)O_(4)` are `9.0 xx 10^(-12)` and `6.0 xx 10^(-12)`, respectively. |
|
Answer» Assume no hydrolysis of `C_(2)O_(4)^(2-)`. The solubilities of the two salts are too similar to make simplifying assumptions about which provides `Ag^(+)` to the solution. `Ag_(2)CrO_(4)hArr 2Ag^(o+) +CrO_(4)^(2-)` `K_(sp) = 9.0 xx10^(-12) =[Ag^(o+)]^(2) [CrO_(4)^(2-)]` `Ag_(2)C_(2)O_(4)hArr 2Ag^(o+) +C_(2)O_(4)^(2-)` `K_(sp) = 6.0 xx 10^(-12) = [Ag^(o+)]^(2) [C_(2)O_(4)^(2-)]` By electroneutality, `[{:("Total +ve charge Total - ve charge"),("Total charge = charge" xx "conc"):}]` `[Ag^(o+)] = 2[CrO_(4)^(2-)] +2 [C_(2)O_(4)^(2-)]` or `0.5 [Ag^(o+)] = [CrO_(4)^(2-)] +[C_(2)O_(4)^(2-)]` `= (9.0xx10^(-12))/([Ag^(o+)](2)) +(6.0xx10^(-12))/([Ag^(o+)]^(2))` `0.5[Ag^(o+)] = 9.0 xx 10^(-12) + 6.0 xx10^(-12)` `= 1.5 xx10^(-11)` `:. [Ag^(o+)] = 3.1 xx10^(-4)M` |
|
| 1063. |
A solution constains `Zn^(2+)` ions and `Cu^(2+)` ions each of `0.02M`. If the solution is made `1M` in `H^(o+)`, and `H_(2)S` is passed untill the solution is satured, should a precipitate be formed? Given: `K_(sp) ZnS = 10^(-22)`, `K_(sp) Cus = 8 xx 10^(-37)`. In satured solution, `K_(sp) (H_(2)S) = 10^(-22)` |
|
Answer» Correct Answer - A::C `H_(2)S hArr 2H^(o+) + S^(2-)` `K_(sp) = [H^(o+)]^(2) [S^(2-)]` `10^(-22) = (1M)^(2) (S^(2-):. [S^(2-)] = 10^(-22)M` i. `Q_(sp) (or IP) of ZnS = [Zn^(2+)] [S^(2-)]` `= 0.02 xx 10^(-22) = 2xx 10^(-24)M` `Q_(sp) lt K_(sp) of ZnS (2 xx 10^(-24) lt 10^(-22))`. Does not precipitate. ii. `Q_(sp) (orIP) of CuS = [Cu^(2+)] [S^(2-)]` `= 0.22 xx 10^(-22) = 2 xx 10^(-24)M` `Q_(sp) of CuS gt K_(sp) of CuS (2 xx 10^(-24) gt 8 xx 10^(-37))` So `CuS` will precipitate. |
|
| 1064. |
Statement-1: NaCl cannot be hydrolysed. Statement-2: It is a salt of strong acid and strong base.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-12B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 and Statement-2 is True |
|
Answer» Correct Answer - a |
|
| 1065. |
Which of the following cannot be hydrolysedA. A salt of weak acid and strong baseB. A salt of strong acid and weak baseC. A salt of weak acid and weak baseD. A salt of strong acid and strong base |
|
Answer» Correct Answer - D A salt of strong acid and strong base cannot be hydrolysed. In this case the equilibrium cannot shifted towards the backward. |
|
| 1066. |
The pH at the equivalence point of a titration may differ from 7.0 because ofA. The self ionisation of waterB. Hydrolysis of the salt formedC. The indicator usedD. The concentration of the standard solution |
| Answer» Correct Answer - B | |
| 1067. |
The solubility of `CaCO_(3)` in water is `3.05 xx 10^(-4)` moles/litre. Its solubility product will beA. `3.05 xx 10^(-4)`B. 10C. `6.1 xx 10^(-4)`D. `9.3 xx 10^(-8)` |
|
Answer» Correct Answer - D `{:(CaCO_(3) ,hArr,Ca^(++)+,CO_(3)^(--)),(" "S,,S,S):}` Solbulity product of `CaCO_(3)` It is a binary electrolyte. `S^(2) = K_(sp) , (3.05 xx 10^(-4)) = K_(sp), K_(sp) = 9.3 xx 10^(-8)`. |
|
| 1068. |
In the titration of `NH_(4)OH` versus `HC1`, the `pH` of the solution at equivalence point is about:A. `5.5`B. `7`C. `8.5`D. `9.5` |
|
Answer» Correct Answer - A Titration of `W_(B)//S_(A). pH` at equivalence point is `~~5.5`. |
|
| 1069. |
pH of water is 7. When a substance Y is dissolved in water, the pH becomes 13. The substance Y is a salt byA. Strong acid and strong baseB. Weak acid and weak baseC. Strong acid and weak baseD. Weak acid and strong base |
|
Answer» Correct Answer - D If we mixed any substance into the solution. Then the value of pH is increased these substance is a salt of weak acid and strong base. |
|
| 1070. |
`pH` of water is `7`. When a substance `Y` is dissolved in water, the `pH` becomes `13`. The substance `Y` is a salt ofA. Weak acid and weak baseB. Strong acid and strong baseC. Strong acid and weak baseD. Weak acid and strong base |
| Answer» Correct Answer - D | |
| 1071. |
The following acid base titration graphs are given: (I) Which of the following graph represents titration of i. `NH_(4)OH//HC1(W_(A)//S_(A))` ii. `HNO_(3)//KOH(S_(A)//S_(B))` iii. `C_(6)H_(5)COOH//KOH(W_(A)//S_(B))`A. `{:(Graph,,Titration,,),(I.II.III,,i.ii.iii,,):}`B. `{:(Graph,,Titration,,),(I.II.III,,ii.iii.i,,):}`C. `{:(Graph,,Titration,,),(I.II.III,,iii.ii.i,,):}`D. `{:(Graph,,Titration,,),(I.II.III,,iii.i.ii,,):}` |
| Answer» Correct Answer - B | |
| 1072. |
In complex compounds the central metal atom can act asA. A Lewis acidB. A Lewis baseC. A Bronsted acidD. A Arrhenius acid |
|
Answer» Correct Answer - A The complex compound central metal atom act as `L.A` due to lonepair aceptor aceptro from ligands |
|
| 1073. |
The following `pH` range where the indicator shows change in colour are given i. `4 - 9.7` ii. `7.46-10.0` iii. `6.5-4` Which of the above `pH` range represent titration of I. Strong acid/strong base `(S_(A)//S_(B))`, II. Weak acid/strong base `(W_(A)//S_(B))`, III. Weak base/strong acid `(W_(B)//S_(A))`A. `(i) rarr I, (ii) rarr II, (iii) rarr III`B. `(iii) rarr I, (ii) rarr II, (i) rarr III`C. `(i) rarr I, (iii) rarr II, (i) rarr III`D. `(i) rarr I, (iii) rarr II, (ii) rarr III` |
| Answer» Correct Answer - A | |
| 1074. |
The following `pH` range where the indicator shows change in colour are given i. `4 - 9.7` ii. `7.46-10.0` iii. `6.5-4` Which of the above `pH` range represent titration of I. Strong acid/strong base `(S_(A)//S_(B))`, II. Weak acid/strong base `(W_(A)//S_(B))`, III. Weak base/strong acid `(W_(B)//S_(A))`A. `(i) rarr I, (ii)rarrII, (iii) rarr III`B. `(iii) rarr I, (ii) rarr II, (i) rarr III`C. `(ii) rarr I, (iii) rarr II, (i) rarr III`D. `(i) rarr I, (III) rarr II, (ii) rarr III` |
| Answer» Correct Answer - A | |
| 1075. |
Which of the following is Lewis baseA. `BCl_(3)`B. `CH_(4)`C. `NH_(3)`D. `HNO_(3)` |
|
Answer» Correct Answer - C In `NH_(3)`, central atom nitrogen having lonepair hence it is lewis base |
|
| 1076. |
Which of thef following is not Lewis acidA. `AlCl_(3)`B. `BaCl_(2)`C. `BCl_(3)`D. `SnCl_(4)` |
|
Answer» Correct Answer - B `BaCl_(2)` is not lewis acid |
|
| 1077. |
What will be the pH of an aqueous solution of 1.0 M ammonium formate? Given :`pK_(a)=3.8 and pK_(b)=4.8`A. `7.5`B. `3.4`C. `6.5`D. `10.2` |
|
Answer» Correct Answer - C `pH = 1/2(pK_(w) + pK_(a) - pK_(b))` |
|
| 1078. |
For any weak acid indicator (Hln) colour change is observed whenA. `([ln^(-)])/([Hln])=(1)/(10)`B. `pH=pK_(a)-1`C. Both (A ) and (B)D. None of these |
|
Answer» Correct Answer - C `Hln hArr H^(+)+ln^(-)` `pH =pK_(a)+log.([ln^(-)])/([Hln])` When colour of the unionized form is seen `([ln^(-)])/([Hln])=(1)/(10):. pH =pK_(a)-1` |
|
| 1079. |
Dissociation of an indicator can be considered as `Hln hArr H^(+)+ln^(-)`. Colour of Hln is P and `ln^(-)` is Q. Given the ratio of concentration of Hln to `ln^(-)` ranges from 10 to `1//10` then which of the following statements are correct.A. Solution assumes P colour, when `pH lepK_(ln-1)`B. Solution assumes P colour, when `pH ge pH _(ln-1)`C. Solution assumes Q colour , when `pH gepK_(ln-1)`D. Solution assumes Q colour , when `pH lepK_(ln-1)` |
|
Answer» Correct Answer - A,B,C Maximum pH to see distinct colour is `pK_(ln-1)` and minimum pH to see a colour is `pK_(ln+1)` |
|
| 1080. |
Arrange the following bases in order of decreasing basicity: `S^(2-),CH_(3)COO^(Θ), CH^(Θ), NH_(3),F^(Θ)` |
|
Answer» Stronger the acid, weaker is conjugate base and vice versa. First arrange them in decreasing acidic character. `HF gt CH_(3) COOH gt NH_(4)^(o+) gt HCN gt HS^(Θ)` Therefore, the decreasing order of basic character. `S^(2-) gt CN^(Θ) gt NH_(3) gt CHCOO^(Θ) gt F^(Θ)` |
|
| 1081. |
Determine `[overset(Θ)OH]` of a `0.050M` solution of ammonia to which sufficient `NH_(4) CI` has been added to make the total `[NH_(4)^(o+)]` equal to `0.1M`. |
|
Answer» `NH_(3) + H_(2)O hArr NH_(4)^(o+) + overset(Θ)OH` `K_(b) = ([NH_(4)^(o+)][overset(Θ)OH])/([NH_(3)])=((0.1)x)/((0.050)) = 1.8 xx 10^(-5)` thus `x = [overset(Θ)OH] = 9.0 xx 10^(-6)` |
|
| 1082. |
Assertion (A): On addition of `NH_(4)CI` to `NH_(4)OH, pH` decreases but remains grater than `7`. Reason (R) : Addition of `overset(o+)NH_(4)` ion decreases ionication of `NH_(4)OH`, thus `[overset(Theta)OH]` decreases and also `pH` decreases.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
|
Answer» Correct Answer - A Both `(A)` and `(R)` are true. A basic buffer is formed whose `pH` is always `gt7`. |
|
| 1083. |
`H_(2)S` is passed through `(CH_(3)COO)_(2)` Zn and `ZnCI_(2)` solutions separately. White ppt. will be noticed in :A. `(CH_(3)COO)_(2)Zn` solutionB. `ZnCI_(2)` solutionC. both (a) and (b)D. none of these |
|
Answer» Correct Answer - A `Zn(CH_(3)COO)_(2)` will furnich `CH_(3)COOH`, a weak acid which will not influence dissociation of `H_(2)S` to apprecible amount and thus ZnS will be prercipitated. |
|
| 1084. |
Calculate the minimum mass of `NaCI` necessary to dissolve `0.01mol AgC1` in `100L` solution. (Assume no change in volume) `(K_(f)AgC1_(2)^(Θ) = 3 xx 10^(5)) (K_(sp)AGC1 = 10^(-10))` |
|
Answer» `AgC1 + C1^(Θ) hArr AgC1_(2)^(Θ)` `[AgC1_(2)^(Θ)] = (0.01mol)/(100L) = 10^(-4)M` `K_(f) = ([AgC1_(2)^(Θ)])/([AgC1][C1^(Θ)])` `K_(f) = ([AgC1_(2)^(Θ)])/([Ag^(o+)][C1^(Θ)]^(2)) = 3 xx 10^(5)` `[Ag^(o+)] = (K_(sp))/([C1^(Θ)])` `K_(f) = ([AgC1_(2)^(Θ)])/(K_(sp)[C1^(Θ)]), [C1^(Θ)] = ([AgC1_(2)^(Θ)])/(K_(sp)K_(f))` `= (10^(-4))/((10^(-10))(3xx10^(5)))` `= 3.3 M` `= (3.3 mol L^(-1)) xx (100L) xx 58.5 g mol^(-1)` `= 19305 g` `= 19.3 Kg` |
|
| 1085. |
If `K_(a)` for formic acid is `2.0 xx 10^(-4)` mol `L^(-1)` then `K_(h)` for `HCOO^(-)` isA. `2.0 xx 10^(-10)`B. `1.414 xx 10^(-12)`C. `10^(-13)`D. `5 xx 10^(-11)` |
|
Answer» Correct Answer - D `K_(h)=(K_(w))/(K_(a))=(10^(-14))/(2.0 xx 10^(-4))` |
|
| 1086. |
`pH` calculation upon dilute of strong acid solution is generally done by equating `n_(H)` in original solution & diluted solution.However . If strong acid solution is very dilute then `H^(+)` from water are also to be considered take `log3.7=0.568` and answer the following questions. A `1` litres solution of `pH=4`(solution of a strong acid ) is added to the `7//3` litres of water.What is the `pH `of resulting solution?A. `4.52`B. `4.365`C. `4.4`D. `4.432` |
|
Answer» Initial `pH=4` `[H^(+)]=10^(-4)` `N_(1)V_(1)=N_(2)V_(2)rArr 10^(-4)=N_(2)xx[1+(7)/(3)]` `10^(-4)=N_(2)xx(10)/(3)rArr N_(2)=3xx10^(-5)" " gt10^(-6)` so `[H^(+)]` of water is not consider `[H^(+)]=3xx10^(-5)`" " so" "`pH=5-log(3)`" "`=5-0.48=4.52` |
|
| 1087. |
For a `0.072 M NaSO_(4)` solution,select the incorrect option(s):`(K_(a_(1))` and `K_(a_(2))` of `H_(2)SO_(4)=oo` & `1.2xx10^(-2))`A. `pH=1.62`B. `pH=7.39`C. Degree of hydrolysis `h=(1)/(sqrt(6))`D. `h=3.4xx10^(-6)` |
|
Answer» `1.2xx10^(-7)=(7.2xx10^(-2)alpha^(2))/(1-alpha)rArr 6alpha^(2)+alpha-1=0rArr alpha=(1)/(3)` So `pH=-log ((0.072)/(3))=1.62` |
|
| 1088. |
In a saturated solution of electrolyte, the ionic product of their concentration are constant at constant temperature and this constant for electrolyte is known asA. Ionic productB. Solubility productC. Ionization constantD. Dissociation constant |
| Answer» Correct Answer - B | |
| 1089. |
The product of ionic concentration in a saturated solution of an electrolyte at a given temperature is constant an is known asA. Ionic product of the electrolyteB. Solubility productC. Ionization constantD. Dissociation constant |
|
Answer» Correct Answer - B Definition of `K_(sp)` |
|
| 1090. |
Why only `As^(+3)` gets precipitated as `As_(2)S_(3)` and not `Zn^(+2)` as ZnS when `H_(2)S` is passed through an acidic soluiotn containing `As^(+3)` and `Zn^(+2)` ?A. Solubility product of `As_(2)S_(3)` is less than that of ZnSB. Enough `As^(+3)` are present in acidic mediumC. Zinc salt does not ionise in acidic mediumD. Solubility product changes in presence of an acid. |
|
Answer» Correct Answer - A This is because `K_(sp)` of `As_(2)S_(3) lt ZnS`. |
|
| 1091. |
Statement: In acidic medium, `Zn^(2+)` is not precipitated by `H_(2)S`. Explanation: Common ion effect reduces the concentration of `S^(2-)` to a minimum level.A. S is correct but E is wrong.B. S is wrong but E is correc.C. Both S and E are correct and E is correct explanation of S.D. Both S and E are correct but E is correct explanation of S. |
|
Answer» Correct Answer - C `Zn^(2+)` is precipitated by `S^(2-)` ions is alkaline solution. |
|
| 1092. |
Statement: In acidic medium, `Zn^(2+)` is not precipitated by `H_(2)S`. Explanation: Common ion effect reduces the concentration of `S^(2-)` to a minimum level.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
|
Answer» Correct Answer - A Both `(A)` and `(R)` are correct and `(R)` is the correct of explanation of `(A). Zn^(2+)` is precipitated as `ZnS` in basic medium. |
|
| 1093. |
At `25^(@)C K_(b)` for `BOH=1.0xx10^(-12).0.01M` solution of `BOH` has `[OH^(-)]`:A. `1.0xx10^(-6)M`B. `1.0xx10^(-7)M`C. `1.0xx10^(-5)M`D. `2.0xx10^(-6)M` |
|
Answer» Correct Answer - B `[OH^(-)]=Calpha=Csqrt((K_(b))/(C ))` `= sqrt((K_(b).C))= sqrt(1.0xx10^(-12)xx0.01)` `=1.0xx10^(-7)M` |
|
| 1094. |
At `25^(@)C`, the dissociation constant of a base. BOH is `1.0xx10^(-12)`. The concentration of hydroxyl ions in `0.01`M aqueous solution of the base would beA. `2.0xx10^(-6)mol L^(-1)`B. `1.0xx10^(-5)mol L^(-1)`C. `1.0xx10^(-6)mol L^(-1)`D. `1.0xx10^(-7)mol L^(-1)` |
|
Answer» Correct Answer - D Base, BOH is dissociated as follows `BOH hArr B^(+)+OH^(-)` So, the dissociation constant of BOH base `K_(b)=([B^(+)][OH^(-)])/([BOH])` ….(i) At equilibrium `[B^(+)]=[OH^(-)]` `therefore K_(b)=([OH^(-)]^(2))/([BOH])` Given that `K_(b)=1.0xx10^(-12)` and `[BOH]=0.01 M` Thus, `1.0xx10^(-12)=([OH^(-)]^(2))/(0.01)` `[OH^(-)]^(2)=1xx10^(-14)` `[OH^(-)]=1.0xx10^(-7)"mol L"^(-1)` |
|
| 1095. |
At `25^@C, K_b` for BOH=`1.0xx10^(-12)`.0.01 M solution of BOH has `[OH^-]` :A. `1.0xx10^(-6) M`B. `1.0xx10^(-7) M`C. `1.0xx10^(-5) M`D. `2.0xx10^(-6) M` |
|
Answer» Correct Answer - b |
|
| 1096. |
The acid dissociation constant `K_(a)` of acetic acid is `1.74 xx 10^(-5) ` at 298 K. The pH of a solution of 0.1 M acetic acid isA. 2.88B. 3.6C. `4.0`D. `1.0` |
|
Answer» Correct Answer - A `CH__(3)COOH hArr CH_(3)COO^(-) +H^(+)` `alpha=sqrt((K_(a))/(C))=sqrt((1.74 xx 10^(-5))/(0.1 ))` `=sqrt(1.74 xx 10^(-4)) =1.32 xx 10^(-2)` `[H^(+)]=C alpha =0.1 xx 1.32 xx 10^(-2) =1.32 xx 10^(-3)` `pH =- log 1.32 xx 10^(-3)` `=3- log 1.32 =2.88 ` |
|
| 1097. |
The pH of gestric juice is normallyA. Greter than `1.5` and less than `2`B. Less than `1.5`C. Greater than `2` and less than `3`D. Less than `1` and greater than `0` |
|
Answer» Correct Answer - A `p^(H)` of gasteric juice is normally greater than `1.5` ans less than `2` |
|
| 1098. |
In the titration of a weak acid of known concentratin with a standard solution of a strong base, a ph water was used to follow the progress of the titration. Which of the following is true of this experiment.A. The pH at the equivalence point depends on the indicator usedB. The graph of pH versus volume of base added rises gradually at first and then much more rapidally.C. The graph of pH versus volume of base added shows no sharp rise.D. The `[H^(+)]` at the equivalence point equals the ionization constant of the acid |
| Answer» Correct Answer - B | |
| 1099. |
The `P^(H)` of `HCl` is `5`, its is diluted by `1000` times Its `P^(H)` will beA. `5`B. `8`C. `2`D. `7` |
|
Answer» Correct Answer - D For very `10` times dilution `p^(H)` value increased by one unit |
|
| 1100. |
Consider the titration of `0.1000 M` `NH_(3)(K_(b) = 1.76 xx 10^(-5))` with `0.1000 M HNO_(3)`. The pH at the equivalence point isA. between `2.0` and `4.0`B. between `4.5` and `6.5`C. approximately `7.0`D. between `7.5` of `9.5` |
|
Answer» Correct Answer - B same s Q.No. `14` |
|