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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.
| 1001. |
The pH of `H_(2)O` isA. 7B. `gt 7`C. `lt 7`D. 0 |
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Answer» Correct Answer - A Because pure water has a 7 pH. |
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| 1002. |
Which is not a buffer solutionA. `NH_(4)Cl + NH_(4)OH`B. `CH_(3)COOH + CH_(3)COONa`C. `CH_(3)COONa`D. Borax + Boric acid |
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Answer» Correct Answer - C It is not a mixture of weak acid or base and their strong salt. |
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| 1003. |
In a solution of pH = 5, more acid is added in order to reduce the pH = 2. The increase in hydrogen ion concentration isA. 100 timesB. 1000 timesC. 3 timesD. 5 times |
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Answer» Correct Answer - B `[H^(+)]_(I) = 10^(-5) = [H^(+)]_(II) = 10^(-2)`. Thus increase in `[H^(+)] = (10^(-2))/(10^(-5)) = 1000` times. |
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| 1004. |
The work done in an isothermal reversible expansion of n moles of a gas isA. `int n PdV`B. `int (RT)/(np)dV`C. `int(nRT)/(V) dV`D. `int n RTdV` |
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Answer» Correct Answer - C The question are base on facts. |
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| 1005. |
Precipitation takes place when the ionic productA. equals the solubility productB. exceeds the solubility productC. is less than the solubility productD. is almost zero |
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Answer» Correct Answer - B Factual question. |
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| 1006. |
Explain why `CuS` is more soluble than predicted by the `K_(sp)`. |
| Answer» d. The `S^(2-)` hydrolyses extensively. The amount which dissolves and the amount which exists as `S^(2-)` in solutions are very different. | |
| 1007. |
Calculate the solubility of `CoS` in `0.1M H_(2)S` and `0.15M H_(3)O^(oplus) (K_(sp)` of `CoS = 3 xx 10^(-26))`. `(K_(1) xx K_(2) (H_(2)S) = 10^(-21))` |
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Answer» a. `H_(2)O +H_(2)S overset(K_(1))hArrH_(3)O^(oplus) +HS^(Θ)` `H_(2)O +HS^(Θ) overset(K_(2))hArrH_(3)O^(oplus) +S^(2-)` `K_(1)K_(2) = ([H_(3)O^(oplus)]^(2)[S^(2-)])/([H_(2)S])=10^(-21)` In `0.1M H_(2)S` and `0.15M H_(3)O^(oplus)` `[S^(2-)] = ((0.15)^(2)[S^(2-)])/(0.1) = 10^(-21)` `:. [S^(2-)] = (10^(-22))/(0.0225) = 4 xx 10^(-21)` `K_(sp) = [Co^(2+)] [S^(2-)] = 3 xx 10^(-26)` `[Co^(2+)] = (3xx10^(-26))/(4xx10^(-21)) = 7 xx 10^(-6)M` |
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| 1008. |
The pH of 0.001 M `Ba(OH)_(2)` solution will be :-A. 2B. 8.4C. 11.3D. 13 |
| Answer» Correct Answer - 3 | |
| 1009. |
The dissociation constant of a substitude benzoic acid at `25^(@)C` is `1.0xx10^(-4)`. The pH of a `0.01M` solution of its sodium salt is |
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Answer» Given `K_(a)=10^(-4)` `pK_(a)=4` `C=0.01M` `pH=7+(1)/(2)pK_(a)+(1)/(2)logC=7+(1)/(2)(4)+(1)/(2)(-2)=8` |
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| 1010. |
`100.0mL` of a saturated solution of `Ag_(2)SO_(4)` is added to `250.0mL` of saturated solution of `PbCrO_(4)`. Will may precipitate form and if so what? Given `K_(sp)` for `Ag_(2)SO_(4), Ag_(2)CrO_(4), PbCrO_(4)`,and `PbSO_(4)` are `1.4 xx 10^(-5), 2.4 xx 10^(-12), 2.8 xx 10^(-13)`, and `1.6 xx 10^(-8)`, respectively. |
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Answer» For `{:(Ag_(2)SO_(4) hArr,2Ag^(o+)+,SO_(4)^(2-),,),(,2S,S,,):}` `K_(sp) = 4S^(3)` or `S = 3sqrt((K_(sp))/(4)) = 3sqrt((1.4 xx 10^(-5))/(4)) = 1.52 xx 10^(-2)M` For `PbCrOhArr Pb^(2+) + CrO_(4)^(2-)` `K_(sp) = S_(1)^(2)` or `S_(1) = sqrt(K_(sp)) = sqrt(2.8 xx 10^(-13)) = 5.29 xx 10^(-7)M` In solution, conce. of each ion can be given as : Thus, `[Ag^(o+)] = (2S xx 100)/(350) = (2xx1.52xx10^(-2)xx100)/(350)` `= 0.869 xx 10^(-12)M` `[SO_(4)^(2-)] = (S xx 100)/(350) = (1.52 xx 10^(-2) xx 100)/(350) = 0.43 xx 10^(-2)` `[Pb^(2+)] = (S_(1)xx250)/(350) = (5.29 xx 10^(-7)xx250)/(350) = 3.78 xx 10^(-7)` `[CrO_(4)^(2-)] = (S_(1)xx250)/(350) = (5.29xx10^(-7)xx250)/(350) = 3.78 xx 10^(-7)` It is thus evident that `[Ag^(o+)] = [CrO_(4)^(2-)] = (0.869 xx 10^(-2))^(2) xx (3.78 xx 10^(-7))` `=2.85 xx 10^(-11) gt K_(sp) Ag_(2)CrO_(4)` The product is greater than `K_(sp)` of `Ag_(2)CrO_(4)` and thus `Ag_(2)CrO_(4)` will ppt. |
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| 1011. |
Determine the number of mole of AgI which may be dissolved in `1.0` litre of `1M CN^(-)` solution. `K_(SP)` for AgI and `K_(C)` for `Ag(CN)_(2)^(-)` are `1.2xx10^(-17)M^(2)` and `7.1xx10^(19)M^(-2)` respectively. |
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Answer» Given, `AgI(s) hArr Ag^(o+) (aq) +I^(Theta) (aq)`, `K_(sp) = [Ag^(o+)] [I^(Theta)] = 1.2 xx 10^(-17) ..(1)` `Ag^(o+) (aq) +2CN^(Theta) (aq) hArr [Ag(CN)_(2)]^(Theta) (aq),` `K_(f) = ([Ag(CN)_(2)^(Theta)])/([Ag^(o+)][CN^(Theta)]^(2)) = 7.1 xx 10^(19)....(2)` Let `x` mole of `AgI` be dissolved in `CN^(Theta)` solution then, Now `{:(,AgI(s)+,2CN^(Theta)hArr,[Ag(CN)_(2)^(Theta)]+,I^(Theta)),("Mole before",,1,0,0),("reaction",,,,),("Mole after",,(1-2x),x,x),("reaction,,,,):}` By equaiton (1) and (2), `K_(ep) = K_(sp) xx K_(f)` `K_(ep) = ([Ag(CN)_(2)^(Theta)][I^(Theta)])/([CN^(Theta)]^(2)) =1.2 xx 10^(-17 xx 7.1 xx 10^(19)` `K_(ep) = 8.52 xx 10^(2) ....(3)` `:. K_(ep) = 8.52 xx 10^(2) = (x)/((1-2x)^(2)) = (x^(2))/((1-2x)^(2))` or `(x)/(1-2x) = 29.2` Thus, `x = 29.2 - 58.4x` or `x = 0.49` mol |
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| 1012. |
The pH of basic buffer mixtures is given by pH `= pK_(a) + log.(["Base"])/(["Salt"])`, whereas pH of acidic buffer mixtures is given by : pH `= pK_(1) + log.(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio `(["Base"])/(["Salt"])` for `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. A solution containing 0.2 mole of dichloroacetic acid litre solution has `[H^(+)]`A. 0.05 MB. 0.025 MC. 0.10 MD. 0.005 M |
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Answer» Correct Answer - A Since `CHCl_(2).COOH` is relatively strong acid having more `K_(a)`. `{:(CHCl_(2).COONa,rarr,CHCl_(2)COO^(-),+,Na^(+)),(,,0.1,,0.1),(CHCl_(2).COOH ,hArr,CHCl_(2)COO^(-) ,+,H^(+)),(0.2,,0,,0),((0.2 - x),,(x + 0.1),,x):}` `:. K_(a) = ([CHCl_(2)COO^(-)][H^(+)])/([CHCl_(2)COOH])` or `5 xx 10^(-2) = ([0.1 + x][x])/([0.2 - x])` `:. x = 0.05`. |
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| 1013. |
The pH of basic buffer mixtures is given by pH `= pK_(a) + log.(["Base"])/(["Salt"])`, whereas pH of acidic buffer mixtures is given by : pH `= pK_(1) + log.(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio `(["Base"])/(["Salt"])` for `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. The amount of `(NH_(4))_(2)SO_(4)` to be added to `500 mL` of 0.01 M `NH_(4)OH` solution (`pK_(a)` for `NH_(4)^(+)` is 9.26) to prepare a buffer of pH 8.26 isA. 0.05 moleB. 0.025 moleC. 0.10 moleD. 0.005 mole |
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Answer» Correct Answer - B `pH = pK_(a) + log.(["Base"])/(["Salt"]) rArr ["Base"] = (0.01 xx 500)/(500) = 0.01` `[NH_(4)^(+)] = (a xx 2)/(500)`, Let a millimole of `(NH_(4))_(2)SO_(4)` are added. `:. [Salt] = [NH_(4)^(+)]`. `pH = 9.26 + log [(0.01)/(2a//500)]` `8.26 = 9.26 + log.(0.01 xx 500)/(2a) :. a = 25` `:.` Mole of `(NH_(4))_(2)SO_(4)` added `= 0.025`. |
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| 1014. |
The ionization constant of ammonium hydroxide is `1.77xx 10^(-5)` at 298 K. Hydrolysis constant of ammonium chloride isA. `5.65 xx 10^(-10)`B. `6.50 xx 10^(-12)`C. `5.65 xx 10^(-13)`D. `5.65 xx 10^(-12)` |
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Answer» Correct Answer - A `K_(h) = (K_(w))/(K_(b)) = (1 xx 10^(-14))/(1.77 xx 10^(-5)) = 5.65 xx 10^(-10)` |
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| 1015. |
The ionization constant of ammonium hydroxide is `1.77xx10^(-5)` at `298 K`. Hydrolysis constant of ammonium chloride isA. `5.65xx10^(-10)`B. `6.50xx10^(-12)`C. `5.65xx10^(-13)`D. `5.65xx10^(-12)` |
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Answer» Correct Answer - A Given, `K_(a)(NH_(4)OH)=1.77xx10^(-6)` `NH_(4)OH hArr NH_(4)^(+)+OH^(-)` `K_(a)=([NH_(4)^(+)][OH^(-)])/([NH_(4)OH])=1.77xx10^(-5)` ..(i) Hydrolysis of `NH_(4)Cl` takes place as, `NH_(4)Cl+H_(2)O rarr NH_(4)OH+HCl` or `NH_(4)^(+)+H_(2)O rarr NH_(4)OH+H^(+)` Hydrolysis constant, `K_(h)=([NH_(4)OH][H^(+)])/([NH_(4)^(+)])` ....(ii) or `K_(h)=([NH_(4)OH][H^(+)][OH^(-)])/([NH_(4)^(+)][OH^(-)])` ....(iii) From Eqs. (i), (ii) and (iii) `K_(h)=(K_(w))/(K_(a)) " " [because [H^(+)][OH^(-)]=K_(w)]` `=(10^(-14))/(1.77xx10^(-5))=5.65xx10^(-10)` |
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| 1016. |
The ionization constant of ammonium hydroxide is `1.77xx10^(-5)` at `298 K`. Hydrolysis constant of ammonium chloride isA. `5.65 xx 10^(-12)`B. `5.65 xx 10^(-10)`C. `5. 65 xx 10^(-12)`D. `5.65 xx 10^(-13)` |
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Answer» Correct Answer - B Hydrolysis constant, `K_(h)=(K_(w))/(K_(b))=(1 xx 10^(-14))/(1.77 xx 10^(-5))` `=(10)/(1.77)xx10^(-10)=5.65 xx 10^(-10)` |
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| 1017. |
The pH of gastric juice in our stomach isA. 0.01B. 2C. 7D. 14 |
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Answer» Correct Answer - B 2 |
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| 1018. |
Number of equivalents of HCl present in 100 mL of its solution whose pH is 4:A. `10^(-4)`B. `10^(-3)`C. `10^(-2)`D. `10^(-5)` |
| Answer» Correct Answer - 4 | |
| 1019. |
In decinormal solution, `CH_(3)COOH` acid is ionised to the extent of `1.3%`. If `log 1.3 = 0.11,` what is the `pH` of the solution?A. `3.89`B. `2.89`C. `4.89`D. Unpredictable |
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Answer» Correct Answer - B `{:(,CH_(3)COOH+,H_(2)OhArr,CH_(3)COO^(Theta)+,H_(3)O^(o+)),("Initial",1,,0,0),("Final",C(1-alpha),,Calpha,Calpha):}` C = Decinormal `= 0.1N` `alpha = 1.3% = (1.3)/(100) = 1.3 xx 10^(-2)` `[H_(3)O^(o+)] =C alpha = 0.1 xx 1.3 xx 10^(-2) = 1.3 xx 10^(-3)` `pH =- log (1.3 xx 10^(-3)) =- log 1.3 - log 10^(-3)` `=- 0.11 +3 = 2.89` |
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| 1020. |
A physician wishes to prepare a buffer solution at pH = 3.58 that efficiently resists changes in pH yet contains only small concentration of the buffering agents. Which of the following weak acids together with its sodium salt would be best to useA. m - chlorobenzoic acid `(pK_(a) = 3.98)`B. p - chlorocinnamic acid `(pK_(a) = 4.41)`C. 2, 5-dihydroxy benzoic acid `(pK_(a) = 2.97)`D. Acetoacetic acid `(pK_(a) = 3.58)` |
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Answer» Correct Answer - D `pH = pK_(a) + log.(["Salt"])/(["Acid"])`. For highest buffer capacity `(["Salt"])/(["Acid"]) ~~ 1` `pH = pK_(a) + log_(10)1` `pH = pK_(a)` So, `pK_(a)` should be 3.58. |
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| 1021. |
The pH value of decinormal solution of `NH_(4)OH` which is 20% ionised, isA. `13.30`B. `14.70`C. `12.30`D. `12.95` |
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Answer» Correct Answer - C For `NH_(4)OH`. `[OH^(-)] = C.alpha , c = (1)/(10)M, alpha = 0.2` `[OH^(-)] = (1)/(10) xx 0.2 = 2 xx 10^(-2)M` `pOH = - log [OH^(-)] = log[2 xx 10^(-2)], pOH = 1.7` `pH = 14 - pOH = 14-1.7 = 12.30`. |
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| 1022. |
The aqueous solution of which of the following salt has the lowest pHA. NaClOB. `NaClO_(2)`C. `NaClO_(3)`D. `NaClO_(4)` |
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Answer» Correct Answer - D `NaClO_(4)` is a salt of strong acid `HClO_(4)`. So it is a strong acid salt. |
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| 1023. |
The pH of solution(s) is (are) defined as the:A. negative logarithm of the hydrogen ion concentrationB. logarithm of reciprocal of hydrogen ion concentrationC. negative power raised on 10 in order to express `[H^(+)]` ion conncentrationD. none of these |
| Answer» Correct Answer - A::B::C | |
| 1024. |
The correct order of basic strength isA. `H_(2)O lt OH^(-) lt CH_(3)OH lt CH_(3)O^(-)`B. `CH_(3)OH lt H_(2)O lt CH_(3)O^(-) lt OH^(-)`C. `H_(2)O lt CH_(3)OH lt OH^(-) lt CH_(3)O^(-)`D. `OH^(-) lt H_(2)O lt CH_(3)O^(-) lt CH_(3)OH` |
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Answer» Correct Answer - C Strong acids has a weak conjugate base and vice-versa. `underset("(Increasing order of basic strength)")overset(H_(2)O lt CH_(3)OH lt OH^(-) lt CH_(3)O^(-))rarr` |
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| 1025. |
The pH at the equivalence point of a titration may differ from `7.0` because ofA. the self ionization of waterB. hydrolysis of the salt formedC. the indicator usedD. the concentration of the standard solutions. |
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Answer» Correct Answer - B Equivalence point of a titrations may differ from 7 due to salt hydrolysis. |
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| 1026. |
The `pH` of a solution of weak base at neutralisation with strong acid is `8. K_(b)` for the base isA. `1.0 xx 10^(-4)`B. `1.0 xx 10^(-6)`C. `1.0 xx 10^(-8)`D. None of these |
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Answer» Correct Answer - B At half neutralisation, `[B^(o+)] = [BOH] , ["Salt"] = ["base"]` `pOH = pK_(b) + "log" ([B^(o+)])/([BOH])` `pOH = pK_(b), (pH = 8, pOH = 6)` `:. pK_(b) = 6, K_(b) = 1 xx 10^(-6)` |
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| 1027. |
Which of the following is (are) correct order(s) for basic strength ?A. `NaOHgtNH_(4)OHgtH_(2)O`B. `NH_(3)gtN_(2)H_(4)gtNH_(2)OH`C. `NH_(3)gtH_(2)OgtHF`D. `OH^(-)ltC_(2)H^(-)ltNH_(2)^(-)ltC_(2)H_(3)^(-)ltC_(2)H_(5)^(-)` |
| Answer» Correct Answer - A::B::C::D | |
| 1028. |
The dissociation constant of monobasic acids `A, B,C` and `D` are `6 xx 10^(-4),5xx10^(-5), 3.6 xx 10^(-6)`, and `7 xx 10^(-10)`, respectively. The `pH` values of their `0.1M` aqueous solutions are in the order.A. `A lt BltC`B. `AgtBgtC`C. `A=B=C`D. `AgtBltC` |
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Answer» Correct Answer - A More is the dissociation constant, more is the concentration of `H^(+)` ions and hence smaller is the pH because `pH =-log [H^(+)]` |
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| 1029. |
The dissociation constant of a weak monobasic acid `K_(a)` is `1xx 10^(-5)` . The pH of 0.1 M of that acid would beA. 5B. 1C. 2D. 3 |
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Answer» Correct Answer - D The dissociation constant of acid `=10^(-5)` or `K=10^(-5)` The concentration of acid `0.1 M =10^(-1)M` `pH=(1)/(2)K_(a)-(1)/(2)log C` `pH=(1)/(2)xx 5-(1)/(2)log 10^(-1)` `pH =2.5 +(1)/(5)=3` |
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| 1030. |
The dissociation constant of monobasic acids `A, B,C` and `D` are `6 xx 10^(-4),5xx10^(-5), 3.6 xx 10^(-6)`, and `7 xx 10^(-10)`, respectively. The `pH` values of their `0.1M` aqueous solutions are in the order.A. `HX gt HX gt HY gt HD`B. `HDgtHYgtHXgtHA`C. All the solutions have same pHD. `HXgtHDgtHYgtHA`. |
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Answer» Correct Answer - B `[H^(+)]=sqrt(KxxC)`. Thus for same value of C, larger the value of K, greater is `[H^(+)]`, hence lower is the value of pH. |
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| 1031. |
The dissociation constant of monobasic acids `A, B,C` and `D` are `6 xx 10^(-4),5xx10^(-5), 3.6 xx 10^(-6)`, and `7 xx 10^(-10)`, respectively. The `pH` values of their `0.1M` aqueous solutions are in the order.A. `D gt C gt B gt A`B. `A gt B gt C gt D`C. `A =B = C = D`D. None |
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Answer» Correct Answer - A Higher the `K_(a)`value, stronger is the acid and high concentration of `H^(o+)`ions and low `pH` value. Thus, the order of `pH` is: `D gt C gt B gt A`. |
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| 1032. |
The sulphide ion concentration `[S^(2-)]` in saturated `H_(2)S` solution is `1 xx 10^(-22)`. Which of the following sulphides should be quantitatively precipitated by `H_(2)S` in the presence of dil. HCl `{:("Sulphide",,"Solubility Product"),("(I)",,1.4xx10^(-16)),("(II)",,1.2xx10^(-22)),("(III)",,8.2xx10^(-46)),("(IV)",,5.0xx10^(-34)):}`A. I, IIB. III, IVC. II, III, IVD. Only I |
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Answer» Correct Answer - B Basic radicals of group II & IV are precipitated by `H_(2)S` in the form of their sulphides. Iind group in acidic medium & IV group in alkaline medium. They precipitate when ionic product increases than solubility product. |
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| 1033. |
Autoprotolysis constant of `NH_(3)` isA. `[NH_(4)^(+)][NH_(3)]`B. `[NH_(2)^(-)][NH_(3)]`C. `[NH_(4)^(+)][NH_(2)^(-)]`D. `([NH_(4)^(+)])/([NH_(2)^(-)])` |
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Answer» Correct Answer - C `NH_(3) + NH_(3) NH_(4)^(+) + NH_(2)^(-)`. |
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| 1034. |
Which of the following is (are) the example(s) of autoprotolysis ?A. `NH_(3)+NH_(3)hArrNH_(4)^(+)+NH_(2)^(-)`B. `CH_(3)OH+CH_(3)OHhArrCH_(3)OH_(2)^(+)+CH_(3)O^(-)`C. `HCOOH+HCOOHhArrHCOOH_(2)^(+)+HCOO^(-)`D. `HCI+HFhArrH_(2)C^(-)+F^(-)` |
| Answer» Correct Answer - A::B::C | |
| 1035. |
An aqueous solution of sodium carbonate has a pH greater than 7 becauseA. It contains more carbonate ion than `H_(2)O` moleculesB. Contains more hydroxide ions than carbonate ionsC. `Na^(+)` ions react with waterD. Carbonate ions react with `H_(2)O` |
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Answer» Correct Answer - B `pH gt 7` = Basic It means contain more hydroxide ion than carbonate ions. |
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| 1036. |
The pH of millimolar HCl isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B `pH = log.(1)/(H^(+)), pH = log.(1)/(10^(-3)), pH = 3`. |
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| 1037. |
Highest pH 14 is given byA. `0.1 M H_(2)SO_(4)`B. `0.1 M NaOH`C. 1N NaOHD. 1N HCl |
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Answer» Correct Answer - C 1N NaOH solution have highest pH `[OH^(-)] = 1, pOH = 0, pH + pOH = 14` `pH = 14- 0 =14` |
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| 1038. |
The pH of the solution: 5ml of `(M)/(5)`, HCl + 10 ml of `(M)/(10)` NaOH isA. 5B. 3C. 7D. 8 |
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Answer» Correct Answer - C Milliequivalents of HCl `= 5 xx (1)/(5) = 1`. Millequivalents of `NaOH = 10 xx (1)/(10) = 1` `:. 5ml (M)/(5)HCl = 10 ml(M)/(10) NaOH` Hence the solution will be neutral i.e, pH = 7. |
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| 1039. |
Among the following which one is wrong?A. Degree of dissociation of a weak electrolyte increases with dilution.B. Increase in temperature increases the ionisationC. Strong electrolytes are ionised completely even at moderate concentrations.D. Addition of `NH_(4)Cl` to `NH_(4)OH` increases the ionisation of the latter. |
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Answer» Correct Answer - D Due to commonion effect weakelectrolyte ionisation decreases |
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| 1040. |
The degree of ionisation does not depends on?A. TemperatureB. CurrentC. Nature of solventD. Concentration |
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Answer» Correct Answer - B Degree of ionisation depends on (1) temperature (2) nature of sovlent (3) concententration |
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| 1041. |
Select the correct statements:A. pH of `NaHCO_(3)` solution can be given by `pK_(H_(2)CO_(3))+pK_(HCO_(3)^(-))//2`B. `AI^(3+)` ion is amphotericC. `K_(SP)` values of metal nitrates are very-very highD. Liquid `SO_(2)` is aprotic solvent `Na_((aq.))^(+)` is conjugate base of `NaOH_((aq.))` |
| Answer» Correct Answer - A::B::C::D | |
| 1042. |
100 ml of `0.2 M H_(2)SO_(4)` is added to 100 ml of 0.2 M NaOH. The resulting solution will beA. AcidicB. BasicC. NeutralD. Slightly basic |
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Answer» Correct Answer - A M.eq. of 0.2 M `H_(2)SO_(4) = (2 xx 0.2 M)/(1000) xx 100 = 0.04 m//l` M.eq. of 2M NaOH `= (0.2)/(1000) xx 100 = 0.02 m//l` left `[H^(+)] = .04 - .02 = .02` Total volume `= 200 = (.02)/(200) =.0001 = 10^(-4)M` pH =4. |
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| 1043. |
Which is not a conjugate acid/base pair ?A. `H_2CO_3 and CO_3^(2-)`B. `HSO_4^- and SO_4^(2-)`C. `H_2PO_4^(-) and HPO_4^(2-)`D. `H_3O^(+) and H_2O` |
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Answer» Correct Answer - A |
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| 1044. |
Assertion (A): Methy red has `K_(a) = 10^(-5)`and the acid form, "Hin" is red and its conjugate base `Ind^(Theta)` is yellow. Reason (R) : `{:(pH=,3,5,7,),((["Ind"^(Theta)])/(["Hin"])=,10^(-2),1,10^(2),),("Colour" =,"Red","Orange","Yellow",):}`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 | |
| 1045. |
Refer to above, the ratio of solubility of `CH_(3)COOAg` in a buffer solution with `pH = 4` and in `H_(2)O` isA. `1//2`B. `2`C. `1//3`D. `3` |
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Answer» Correct Answer - B `S` in buffer `= 2 xx 10^(-6)M` `S` in `H_(2)O = sqrt(K_(sp)) = (106(-12))^(1//2) = 10^(-6)M` `:. ("S buffer")/(S H_(2)O) = (2xx10^(-6))/(10^(-6)) = 2`. |
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| 1046. |
Which indicator should be used in titrating KOH against oxalic acid ?A. LitmusB. Methyl orangeC. Methyl redD. Phenolphthalein. |
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Answer» Correct Answer - D Phenolphthalein is used in titration of strong base and weak acid. |
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| 1047. |
In the estimation of an oxalate with `KMnO_(40` solution which of the following is used as indicator ?A. Methyl orangeB. PhenolphthaleinC. StarchD. None of the above. |
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Answer» Correct Answer - D `KMnO_(4)` acts as a self indicator. |
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| 1048. |
Write the conjugate acids of the following bases: a. `C_(5)H_(5)N` b. `overset(o+)N_(2)H_(5)` |
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Answer» In each case, the `C_(A)` is formed the base by the addition of a paoton. a. `C_(5)H_(5)overset(o+)NH` b. `N_(2)H_(6)^(2-)` |
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| 1049. |
Given the acid ionization constants when the conjugate bases are arranged in order of increasing base strength , which order is correct ? A. `CIO_2^(-),CIO^-,HPO_4^(2-),CN^-`B. `CIO_2^(-),HPO_4^(2-),CIO^-,CN^-`C. `CN^-,HPO_4^(2-),CIO^(-),CIO_2^-,`D. `CN^-,CIO^(-),HPO_4^(2-),CIO_2^-` |
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Answer» Correct Answer - b |
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| 1050. |
In a `0.2M` aqueous solution of Ethylene diamine `(H_(2)NCH_(2)CH_(2)NH_(2)):` Given `K_(b_(1))=8xx10^(-5)` & `K_(b_(2))=2.7xx10^(-8)`A. `[OH^(-)]=2xx10^(-2)M`B. `[C_(2)N_(2)H_(10)^(2+)]=2.7xx10^(-8)M`C. Both (A) and (B)D. None of these |
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Answer» `[OH^(-)]=sqrt(K_(b_(1))xxC)=sqrt(1.6xx10^(-5))=4xx10^(-3)M` `K_(b_(1))xxK_(b_(2))=([OH^(-)]^(2)[C_(2)H_(10)N_(2)^(2+)])/([C_(2)H_(6)H_(8)])rArr 8xx2.7xx10^(-13)=(16xx10^(-3)[C_(2)H_(10)N_(2)^(2+)])/(0.2)` `:.[C_(2)H_(10)N_(2)^(2)]=2.7xx10^(-8)M` |
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