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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.
| 951. |
A solution is prepared by dissoving 2.8 g of lime (CaO) in enough water to make 1.00 L of lime water `(Ca(OH)_(2)(aq))`. If solubility of `Ca(OH)_(2)` in water is 1.48 gm/lt. The pH of the solution obtained will be : [log 2=0.3, Ca =40, O =16, H=1]A. 12.3B. 12.6C. 1.3D. 13 |
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Answer» Correct Answer - b |
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| 952. |
The solubility products of `MA, MB, MC` and MD are `1.8xx10^(-10), 4xx10^(-3), 4xx10^(-8)` and `6xx10^(-5)` respectively. If a `0.01M` solution of MX is added dropwise to a mixture containing `A^(-), B^(-), C^(-)` and `D^(-)` ions, then the one to be precipitated first will be:A. `MA`B. `MB`C. `MC`D. `MD` |
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Answer» Correct Answer - A Since all compound `(MA, MB, MC`, and `MD)` are uni-univalent type. All will have the same formula of solubility in `H_(2)O` (i.e., `sqrt(K_(sp)))`. Same concentration of common ion `(M^(o+))` is added, so solubilities of all will be supressed. Hence, the compound with least `K_(sp)` value will be precipitated, i.e., `MA` with `K_(sp) = 1.8 xx 10^(-10)`. |
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| 953. |
At `25^(@)`, the solubility product values of AgCl and AgCNS are `1.8xx10^(-10)` and `1.6xx10^(-11)` respectively. When a solution is saturated with both solids, calculate the ratio `([Cl^(-)])/([CNS^(-)])` and also `[Ag^(+)]` in the solution.A. `1.125, 4xx10^(-6)`MB. `11.25, 1.4xx10^(-5)`MC. `1.25, 4xx10^(-5)`MD. `1.25, 4xx10^(-6)`M |
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Answer» Correct Answer - b |
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| 954. |
Calculate similtaneous solubility of silver thiocyanate and silver bromids in water .Given that `K_(sp)` of silver thiocyanate=`10^(-12)` and `K_(sp)` of silver bromids `=5xx10^(-13)` respcectively. |
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Answer» Let the simultaneous solubility of `AgSCN` be `x` and that of `AgBr` is `y`.Then: `AgSCNhArr underset(x+y)(Ag^(+))+underset(x)(SCN^(-))"",,,,,,,,,,,""AgBrhArr underset(x+y)(Ag^(+))+underset(y)(Br^(-))` `10^(-12)=xx(x+y)`....(i) `5xx10^(-13)=y(x+y)`....(ii) on solving we get `x=2y` so `y=4.08xx10^(-7)M` and `x=8.16xx10^(-7)M` |
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| 955. |
Calcuate solubility of silver oxalate in `10^(-2)M` potassium oxalate solution.Given that `K_(sp)` of silver oxalate `=10^(-10)`. |
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Answer» Let the solubility be `x` mol//L `{:(,AgC_(2)O_(4)hArr,2Ag^(+),,C_(2)O_(4)^(2-),),(t=t_(eq),-,2x,x+10^(-2),~~10^(-2)):}` `K_(sp)=10^(-10)=10^(-2)xx(2x)^(2)rArr (10^(-8))/(2xx2)=x^(2)=rArr x=5xx10^(-5)M` |
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| 956. |
Which of the following is not expected to be correctA. `DeltaHf^(@)(CO_(2))=`negativeB. `DeltaH_("comb")(NO)=`positiveC. `DeltaH_("neut")=`negativeD. `DeltaH_("hyd")(BaCl_(2))=`negative |
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Answer» Correct Answer - B Heat of combustion is exothermic process. |
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| 957. |
When ammonium chloride is added to ammonia solution the pH of the resulting solution will beA. IncreasedB. SevenC. DecreasedD. Not changed |
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Answer» Correct Answer - C `NH_(4)Cl` is slat of `S.A + W.B` so it is acidic salt `p^(H)` is decreased |
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| 958. |
Which of the following statements are correct at `25^(@)C` ?A. `pK_(a)` for `H_(3)O^(+)` is `15.74`B. `pK_(b)` for `OH^(-)` is `-1.74`C. `pK_(a)+pK_(b)=pK_(w)` for HCl and ClOHD. Degree of dissociation of water is `1.8xx10^(-7)%` |
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Answer» Correct Answer - b,d |
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| 959. |
Assertion : The `H_(3)O^(+)` has additional water molecules closely associated with it. Reason : In solid state the species `H_(5)O_(2)^(+)` and `H_(9)O_(4)^(+)` have been found to exist.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. 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 |
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Answer» Correct Answer - b |
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| 960. |
Which of the following can be classified as a Bronsted base ?A. `NO_(3)^(-)`B. `H_(3)O^(+)`C. `CH_(3)COOH`D. `NH_(4)^(+)` |
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Answer» Correct Answer - A It can act as donor of electron pair. |
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| 961. |
The concentration of acetic acid `(K_(a)=1.8 xx 10^(-5))` required to give `3.5 xx 10^(-4) ` moles `//` litres of `H_(3)O^(+)` ions isA. `6.8 xx 10^(-3) mol L^(-1)`B. `6.8 mol L^(-1)`C. `1.94 mol L^(-1)`D. `0.194 mol L^(-1)` |
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Answer» Correct Answer - A `K_(a)=([CH_(3)COO^(-)][H_(3)O^(+)])/([CH_(3)COOH])` `1.8 xx 10^(-5)[CH_(3)COOH]=[H_(3)O^(+)]^(2)` `[CH_(3)COOH]=((3.5xx10^(-4))^(2))/(1.8 xx10^(-5))` `=(12.25 xx 10^(-8))/(1.8 xx 10^(-5))=6.80 xx 10^(-3) mol L^(-1)` |
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| 962. |
Given that , the solubility product `K_(sp)` , of AgCl is `1.8 xx 10^(-10)` , the concentration of `Cl^(-)` ions that must just be exceeded before AgCl will precipitate from a solution containing `4 xx 10^(-3)` M `Ag^(+)` ions isA. `4.5 xx 10^(-8) M`B. `4 xx 10^(-8) M`C. `1.8 xx 10^(-8) M`D. `1 xx 10^(-8) M` |
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Answer» Correct Answer - A `[Ag^(+)][Cl^(-)]=1.8 xx 10^(-10)` `[Cl^(-)]=(1. 8 xx 10^(-10))/(4 xx 10^(-3))=4.5 xx 10^(-8)` Thus, if conc. Of `Cl^(-)` ions exceeds `4.5 xx 10^(-8)` M precipitation will occur because ionic product will exceed `K_(sp)`. |
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| 963. |
At certain temperature the `H^(+)` ions concentration of water is `4 xx 10^(-7) M` then the value of `K_(w)` at the same temperature isA. `10^(-14) M^(2)`B. `2.5 xx 10^(-13) M^(2)`C. `1.6 xx 10^(-13) M^(2)`D. `4 xx 10^(-7) M^(2)` |
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Answer» Correct Answer - C `H_(2)O hArr H^(+) + OH^(-)` `4 xx 10^(-7) , 4 xx 10^(7)` `K_(w) = 4 xx 10^(-7) xx 4 xx 10^(-7)` `= 16 xx 10^(-14)` `= 1.6 xx 10^(-13)` |
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| 964. |
Calcualte the `pH` at which an indicator with `pK_(b) = 4` changes colour. |
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Answer» `"Ind"^(Theta) +H_(2)O hArr "Hin" + overset(Theta)(O)H` At the colour change, `["Ind"^(Θ)] = ["Hin"]` `K_(b) = ([overset(Θ)OH]["Hin"])/(["Ind"^(Θ)]) = 1.0 xx 10^(-4) = [overset(Θ)OH]` and `pOH = 4.0` `:. pH = 10.00` |
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| 965. |
Bromophenol blue is an indicator with a `K_(a)` value of `5.84 xx 10^(-5)`. What is the percentage of this indicator in its basic form at a `pH` of `4.84`? |
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Answer» `K_(a) = 5.84 xx 10^(-5), pK_(a) = 4.2336` `pH = pK_(a) + log ["Ind"^(Θ)//"Hin"]` `4.84 = 4.2336 + log ["Ind"^(Θ)//"Hin"]` `:. Log ["Ind"^(Θ)//"Hin"] = 0.6064`, `["Ind"^(Θ)//"Hin"] = 4.04` `%` of basic form `= (["Ind"^(Θ)])/(["Ind"^(Θ)]+["HIn"]) xx 100` `["HIn"//"Ind"^(Θ)] = 1.4//04...(i)` Add 1 to both sides of equaiton (i) `(["HIn"])/(["Ind"^(Θ)])+1 = (1)/(4.04) +1:. (["HIn"]+["Ind"^(Θ)])/(["Ind"^(Θ)]) =(5.04)/(4.04)` `(["Ind"^(Θ)])/(["HIn"]+["Ind"^(Θ)]) xx100 = (4.04)/(5.04) xx 100 = 80%` |
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| 966. |
Solubility of `Al(OH)_(3) = S, K_(sp)` will beA. `108 S^(3)`B. `27 S^(3)`C. `4 S^(2)`D. `27 S^(4)` |
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Answer» Correct Answer - D `{:(Al(OH)_(3)(s),hArr,Al^(3+)(aq),+,3OH^(-)(aq)),(,,s,,3s):}` `k_(sp) = [Al^(3+)][OH^(-)]^(3) = (S)(3S)^(3)` `K_(sp) = 27 S^(4)`. |
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| 967. |
The molar solubility of `PbI_(2)` in `0.2 M Pb(NO_(3))_(2)` solution in terms of solubility product, `K_(sp)`A. `(K_(sp)//0.2)^(1//2)`B. `(K_(sp)//0.8)^(1//3)`C. `(K_(sp)//0.4)^(1//2)`D. `(K_(sp)//0.8)^(1//2)` |
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Answer» Correct Answer - D `= (0.2 + s) (2s)^(2)` |
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| 968. |
When solid lead iodide is added to water, the equilibrium concentration of `I^(-)` becomes `2.6 xx 10^(-3)M` . What is the `K_(sp)` for `PbI_(2)` ?A. `2.2 xx 10^(-9)`B. `8.8 xx 10^(-9)`C. `1.8 xx 10^(-8)`D. `3.5 xx 10^(-8)` |
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Answer» Correct Answer - B `PbI_(2) hArr Pb^(2+)+2I^(-)` On the basis of this equation, the concentration `Pb^(2+)` ions will be half of the concentration of `I^(-)` ions . Thus, `[I^(-)]=2.6 xx 10^(-3)M` and `[Pb^(2+0]=1.3 xx 10^(-3)M` `:. K_(sp)=[Pb^(2+)][I^(-)]^(2)` `=(1.3 xx 10^(-3))xx (2.6 xx 10^(-3))^(2) =8.8 xx 10^(-9)` |
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| 969. |
The pH of the solution containing 10 mL of 0.1 NaOH and 10 mL of 0.05 N `H_(2)SO_(4)` would beA. 1B. `0`C. 7D. `gt7` |
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Answer» Correct Answer - D Milliequivalent of NaOH `=10 xx 0.1 =1` Milliequivalent of `H_(2)SO_(4)=10 xx 0.05 =0.5` Therefore, on mixing the two solutions milliequivalents of alkali are left `,. pH gt 7` |
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| 970. |
Which of the following has a higher value for `K_(h)` at `27^(0)c`A. `NaF`B. `NaCl`C. `NaBr`D. `NaI` |
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Answer» Correct Answer - A Extent of hydrolysis decreases with increases of size of anion |
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| 971. |
Calculate the percentage hydrolysis in `0.003M` aqueous solution of NaOCN. `(K_(a)for HOCN=3.33xx10^(-4))`A. `10^(-3)`B. `10^(-2)`C. `10^(-4)`D. `10^(-5)` |
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Answer» Correct Answer - B `NaOCN + H_(2)O hArr NaOH + HCN` `h = sqrt((K_(H))/(C)) = sqrt((K_(W))/(K_(a)xxc))` |
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| 972. |
Calculate the percentage hydrolysis in `0.003M` aqueous solution of NaOCN. `(K_(a)for HOCN=3.33xx10^(-4))` |
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Answer» `NaOCN+H_(2)OhArrNaOH+HCN` `h= sqrt([(K_(H))/(C)])=sqrt([(K_(w))/(K_(a).C)])=sqrt((10^(-14))/(3.33xx10^(-4)xx0.003))` `h= 10^(-4)` `:. %` hydrolysis `= (10^(-4))/(100)=10^(-2)%` |
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| 973. |
If the pH value is 4.5 for a solution then what is the value of `H^(+)` concentrationA. `3.162 xx 10^(-5)` mol/LB. `31.62 xx 10^(-5)` mol/LC. `0.3162 xx 10^(-5)` mol/LD. `3.162 xx 10^(-8)` mol/L |
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Answer» Correct Answer - A `[H^(+)] = 10^(-pH)` `= 10^(-4.5) mol//K, = 10^(0.5) xx 10^(-5)` `= sqrt(10) xx 10^(-5)` `3.162 xx 10^(-5) mol//L`. |
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| 974. |
What is the `pH` of the following solutions: a. `10^(-7)M NaOH` b. `10^(-8)M NaOH` c. `10^(2) M NaOH` |
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Answer» a. `pH` of `10^(-7) M NaOh` First method: The concentration of `overset(Θ)OH` ions from pure water is `10^(-7)M`. So add `[overset(Θ)OH]` from water and `[overset(Θ)OH]` from `NaOH` to get `[overset(Θ)OH]_("Total")` `:. [overset(Θ)OH]_("Total") = 10^(7)M +2xx10^(-7)M` `pOH =- "log" [overset(Θ)OH]_("Total")` `=- log (2xx10^(-7))` `=- log2 - log 10^(-7)` `=- 0.3011 +7 = 6.6989` `:. pH = 14 - 6.6989 = 7.3011` Second method: Due to common ion `(overset(Θ)OH)`, the supression of ionsiation of `H_(@)O` takes place in the presence of `NaOH`. Let `[H_(3)O^(o+)] = [overset(Θ)OH] = xM` `2H_(2)O(l)hArr underset(x M)(H_(3)O^(oplus))(aq)+ underset(x M)(overset(Theta)(O)H)(aq)...(i)` `NaOH rarr underset(10^(-7)M)(Na^(o+)(aq)) + underset(10^(-7)M)(overset(Θ)OH (aq)) ...(ii)` `[H_(3)O^(o+)]_("Total")` from equaitons (i) and (ii) `= (x +10^(-7))M`. At equilibrium `[H_(3)O^(o+)] [overset(Θ)OH] = 10^(-14) = Kw` `(x) (x +10^(-7)) = 10^(-4)` or `x^(2) +10^(-7) x - 10^(-14) = 0` `:. x = (-10^(-7)+sqrt((10^(-7))^(2)+(4xx10^(-14))))/(2)` `:. [H_(3)O^(o+)] = x = 0.616 xx 10^(-7)M` Therefore, `[overset(Θ)OH]_("total") = (10^(-7) +0.6.18 xx 10^(-7))M` `:. pOH =- log (1.618 xx 10^(-7)) = 6.7910` `pH = 14 - 6.7910 = 6.209` b. `pH` of `10^(-8)M NaOH` First method : `[overset(Θ)OH]_("total") = 10^(-8) ("from NaOH") +10^(-7) ("from" H_(2)O)` `= 10^(-7) (10^(-1) +1) = 1.1 xx 10^(-7)` `pOH =- log (1.1 xx 10^(-7)) =- log 1.1 - log 10^(-7)` `=- 0.0414 +7= 6.9586` `pH = 14 - 6.9586 = 7.0414`. Second method : Ket `x = [overset(Θ)OH] = [H_(3)O^(o+)]` from `H_(2)O` The `[overset(Θ)OH]_("Total")` is generated from the ionisation of NaOH dissociated and from ionisation of `H_(2)O` `:. K_(w) = (10^(-8)+x) (x) = 10^(-4)` or `x^(2) +10^(-8)x - 10^(-14) = 0` `:. x =(-10^(-8) +sqrt((10^(-8))^(2)+(4xx10^(-14))))/(2)` `[H_(3)O^(o+)] = x = 9.5 xx 10^(-8)` So, `pH = 7.02` c. pH of `10^(-2) M NaOh` `{:(,NaOHrarr,Na^(o+)+,overset(Θ)OH,),("Initial concentration",10^(2),0,0,),("Concentration after dissociation",0,10^(2),10^(2),):}` `pOH =- log (10^(2)) =-2` `pH = 14 - (-2) = 16` `pH gt 12`, only means that the `[overset(Θ)OH] gt 1M`. So pH scale becomes in between 2 and 16. |
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| 975. |
pH of a solution having 0.00001 gm ions of ions of hydrogen per litre isA. 5B. 4C. `10^(-5)`D. `10^(-4)` |
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Answer» Correct Answer - A `[H^(+)] = 1 xx 10^(-5) M` `pH = - log 10^(-5)` pH = 5 log 10 = 5. |
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| 976. |
a. Calculate the percentage hydrolysis of `0.003M` aqueous solution of `NaOH. K_(a)` for `HOCN = 3.3 xx 10^(-4)`. b. What is the `pH` and `[overset(Theta)OH]` of `0.02M` aqueous solution of sodium butyrate. `(K_(a) = 2.0 xx 10^(-5))`. |
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Answer» a. `NaOCN + H_(2)O hArr NaOH + HCN` `h = sqrt(((K_(h))/(c))) = sqrt(((K_(w))/(K_(a).C)))` `=sqrt((10^(-14))/(3.33xx10^(-4)xx0.003))` `h = 10^(-4)` `:. %` hydrolysis `= 10^(-4) xx 100 = 10^(-2)`. b. `{:(,NaBu+,H_(2)OhArr,NaOH+,BuH),("Conc before hydrolysis",1,,0,0),("Conce after hydrolysis",1-h,,h,h):}` `:. [OH^(Theta)] = C.h = sqrt(((K_(h))/(c)))` `= sqrt((K_(h).C)) = sqrt(((K_(w))/(K_(a)).C ))` `[OH^(Theta)] = (10^(-14)xx0.2)/(2xx106(-5)) = sqrt(10^(-10)) = 10^(-5)`, `pOH = 5, pH = 9` |
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| 977. |
Assuming methanol to undergo self dissociation to give `H^+` and `CH_3O^(-)` what will be its percentage dissociation if a 0.5 M solution has `[H^+]` equal to `2.5xx10^(-4) M` ?A. `5%`B. `10%`C. `0.05%`D. `0.01%` |
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Answer» Correct Answer - c |
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| 978. |
The pH of water at `25^(@)C` is nearlyA. 2B. 7C. 10D. 12 |
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Answer» Correct Answer - B `[H^(+)] = [OH^(-)]` `K_(w) = [H^(+)][OH^(-)] = 10^(-14)` `:. [H^(+)] = 10^(-7), pH = -log[H^(+)] = 7`. |
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| 979. |
For a strong acidA. `alpha` is very highB. `K_(a)` is very highC. `P^(K_(a))`, is a very lowD. All are correct |
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Answer» Correct Answer - D For a stron acid Ka is very high `P^(ka)` is very low |
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| 980. |
If solubility of AgCl in 0.2 M solution of `AgNO_(3)` is represented as `yxx10^(-10)` then find the value of y. `("Given": K_(sp(AgCl))=10^(-10))` |
| Answer» Correct Answer - 5 | |
| 981. |
At `80^(@)C` distilled water has `[H_(3)O^(+)]` concentration equal `[OH^(-)] 1xx10^(-6) "mole"//litre`. The value of `K_(w)` at this temperature will beA. `1 xx 10^(-8)`B. `1 xx 10^(-6)`C. `1 xx 10^(-9)`D. `1 xx 10^(-12)` |
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Answer» Correct Answer - D `K_(w)=[H_(3)^(+)O][OH^(-)]` Concentration of `[H_(3)^(+)O]` in distilled water `=1 xx 10^(-6) mol L^(-1)` Now `[H_(3)^(+)]=[OH^(-)]` `:. K_(w)=(1 xx 10^(-6))xx (1xx 10^(-6))=1 xx 10^(-12)` |
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| 982. |
At `80^(@)C` distilled water has `[H_(3)O^(+)]` concentration equal `[OH^(-)] 1xx10^(-6) "mole"//litre`. The value of `K_(w)` at this temperature will beA. `1 xx 10^(-8)`B. `1 xx 10^(-12)`C. `1 xx 10^(-14)`D. `1 xx 10^(-6)` |
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Answer» Correct Answer - B `[H_(3)O^(+)]=[OH^(-)]=1xx10^(-6) mol L^(-1)` Thus `K_(w)=[H_(3)O^(+)][OH^(-)]` `=1 xx 10^(-6)xx1xx10^(-6)` `K_(w)=1xx10^(-12)` |
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| 983. |
The molar solubility of AgCl in `1.8 M AgNO_(3)` solution is (`K_(sp)` of `AgCl = 1.8 xx 10^(-10)`)A. `10^(-5)`B. `10^(-10)`C. `1.8 xx 10^(-5)`D. `1.8 xx 10^(-10)` |
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Answer» Correct Answer - B `1.8 M AgNO_(3) "……" 1.8 xx 10^(-10)` `1M AgNO_(3) "…….." ? ` Molar sububility `= (1.8 xx 10^(-10))/(1.8) = 10^(-10)` |
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| 984. |
At `80^(@)C`, distilled water has `[H_(3)O^(+)]` concentration equal to `1 xx 10^(-6)` mole/litre. The value of `K_(w)` at this temperature will beA. `1 xx 10^(-6)`B. `1 xx 10^(-9)`C. `1 xx 10^(-12)`D. `1 xx 10^(-15)` |
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Answer» Correct Answer - C `K_(w) = [H_(3)O^(+)][OH^(-)]` Concentration of `H_(3)O^(+)` in distilled water `= 1 xx 10^(-6)` mol/l. Now `[H_(3)O^(+)] = [OH^(-)]` `K_(w) = [1 xx 10^(-6)] xx [1 xx 10^(-6)] = 1 xx 10^(-12)`. |
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| 985. |
What is the `pH` of the resulting solution when equal volumes of `0.1 M NaOH` and `0.01 M HCl` are mixed?A. `12.65`B. `2.0`C. `7.0`D. `1.04` |
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Answer» Correct Answer - A Key Concept - When equal volumes of acid and base are mixed, then resulting solution become alkaline if concentration of base is taken high. Let normality of the solution after mixing 0.1 M NaOH and 0.01 M HCl is N. `therefore N_(1)V_(1)-N_(2)V_(2)=NV` or `0.1xx1-0.01xx1=Nxx2` Since, normality of NaOH is more than that of HCl. Hence, the resulting solution is alkaline. or `[overset(-)(O)H]=N=(0.09)/(2)=0.045 N` or `pOH=-log(0.045)=1.35` `therefore pH=14-pOH=14-1.35=12.65` |
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| 986. |
The `K_(b)` for `AgCl` is `2.8 xx 10^(-10)` at a given temperature. The solubility of `AgCl` in `0.01` molar `HCl` solution at this temperature will be :A. `2.5 xx 10^(-8) molL^(-1)`B. `2.8 xx 10^(-12) molL^(-1)`C. `5.6 xx 10^(-8) molL^(-1)`D. `2.8 xx 10^(-4) molL^(-1)` |
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Answer» Correct Answer - A `K_(sp) = [Ag^(+)][Cl^(-)]` `2.8 xx 10^(-10) = [Ag^(+)]xx 0.01` `[Ag^(+)] = (2.8 xx 10^(-10))/(10^(-2)) = 2.8 xx 10^(-8)"mol" L^(-1)` |
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| 987. |
The pH of 0.01 molar solution of HCl will beA. `0.001`B. 3C. 2D. 6 |
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Answer» Correct Answer - B pH of `0.001 M HCl = 10^(-3) M [H^(+)], pH = 3`. |
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| 988. |
The pH of a `10^(-8)` molar solution of HCl in water isA. 8B. `-8`C. Between 7 and 8D. Between 6 and 7 |
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Answer» Correct Answer - D Molar conc. of HCl `= 10^(-8)`. `:.` pH = 8. But this cannot not be possible as pH of an acidic solution can not be more than 7. So we have to consider `[H^(+)]` coming from `H_(2)O`. Total `[H^(+)] = [H^(+)]_(HCl) + [H^(+)]_(H_(2)O)` IOnisation of `H_(2)O : H_(2)O rarr H^(+) + OH^(-)` `K_(w) = 10^(-14) = [H^(+)][OH^(-)]` Let x be the conc. of `[H^(+)]` from `H_(2)O` or `[H^(+)] = x [OH^(-)]_(H_(2)O)` `:. 10^(-14) = (x + 10^(-8))(x)` or `x = 9.5 xx 10^(-8) M` `:. [H^(+)] = 10^(-8) + 9.5 xx 10^(-8) = 10.5 xx 10^(-8)` or pH `= -log(10.5 xx 10^(-8)) = 6.98` |
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| 989. |
A white salt is readily soluble in water and gives a colourless solution with a pH of a about 9. The salt would beA. `NH_(4)NO_(3)`B. `CH_(3)COONa`C. `CH_(3)COONH_(4)`D. `CaCO_(3)` |
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Answer» Correct Answer - B pH of 9 means the salt solution should be fairly basic. |
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| 990. |
Some salts although containing two different metallic elements give test for only one of them in solution. Such salts areA. Double saltsB. Normal saltsC. Complex saltD. Basic salts |
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Answer» Correct Answer - C Complex salts contain two different metallic elements but give test for only one of them. e.g., `K_(4)[Fe(CN)_(6)]` does not give test `Fe^(3+)` ions. |
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| 991. |
The dissociation constant of a weak acid is `1.0 xx 10^(-5)`, the equilibrium constant for the reaction with strong base isA. `1.0 xx 10^(-5)`B. `1.0 xx 10^(-9)`C. `1.0 xx 10^(9)`D. `1.0 xx 10^(14)` |
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Answer» Correct Answer - C `HA hArr H^(+) + A^(-) , K_(a) = ([H^(+)][A^(-)])/([HA])` ....(i) `K = ([A^(-)])/([HA][OH^(-)])` ....(ii) Dividing (i) by (ii) `(K_(a))/(K) = [H^(+)] [OH^(-)] = K_(w) = 10^(-14)` `K = (K_(a))/(K_(w)) = (10^(-5))/(10^(-14)) = 10^(9)` |
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| 992. |
The pH of a soft drink is 3.82. Its hydrogen ion concentration will beA. `1.96 xx 10^(-2)` mol/lB. `1.96 xx 10^(-3)` mol/lC. `1.5 xx 10^(-4)` mol/lD. `1.96 xx 10^(-1)` mol/l |
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Answer» Correct Answer - C `pH = 3.82 = -log[H^(+)] :. [H^(+)] = 1.5 xx 10^(-4)` mole/litre |
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| 993. |
Components of buffer solution are 0.1 M HCN and 0.2 M NaCN. What is the pH of the solutionA. 9.61B. 6.15C. 2D. 4.2 |
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Answer» Correct Answer - A `pH = pK_(a) + log[("salt")/("acid")]` `= 9.30 + log[(0.2)/(0.1)] = 9.30 + 0.3010 = 9.6`. |
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| 994. |
Assertion (A): `pH` value of `HCN` solution decreases when `NaCN` is added to it. Reason (R) : `NaCN` provides a common ion `CN^(Theta) to `HCN`.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. |
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Answer» Correct Answer - D `(A)` is wrong because addition of `NaCN` to `HCN`, due to common ion `(CN^(Theta))`. The degree of dissociation of `HCN` is supressed and hence less `[H^(o+)]` and increase in `pH. (R)` is correc. |
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| 995. |
Assertion (A): `pH` of water increases with an increase in temperature. Reason (R) : `K_(w)` or water increases with increase in temperature.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. |
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Answer» Correct Answer - D `(A)` is wrong . Dissociation of water is an endothermic reaction, so increasing `T` will increase `[H^(o+)]` and hence decreases in `pH. (R)` is correct. |
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| 996. |
Which oxychloride has maximum pHA. NaClOB. `NaClO_(3)`C. `NaClO_(3)`D. `NaClO_(4)` |
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Answer» Correct Answer - A Maximum pH HClO is a weak acid all of these. So that the salt of weak acid is also weak. |
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| 997. |
The pH vaue of 0.1 M NaOH solution is (when there is a given reaction `[H^(+)][OH^(-)] = 10^(-14)`A. 13B. 12C. 11D. 2 |
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Answer» Correct Answer - A Given, concentration of `NaOH = 0.1 M` `[OH^(-)] = 0.1` Given `[H^(+)] [OH^(-)] = 10^(-14)` `[H^(+)] = 10^(-13) , pH = 13`. |
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| 998. |
The concentration of NaOH solution is `10^(-8)`M . Find out the `(OH^(-))` concentrationA. `10^(-8)`B. Greater than `10^(-6)`C. `10^(-6)`D. Lies between `10^(-6)` and `10^(-7)` |
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Answer» Correct Answer - A Concentrate of `[OH^(-)]` in `NaOH = 10^(-8)` and in `H_(2)O = 10^(-7)`. Total `[OH^(-)] = 10^(-8) + 10^(-7) = 1.1 xx 10^(-7)` pOH = 7 - log 1.1 pH = 14-7 + log 1.1 `pH = 7+ log 1.1 gt 7`. |
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| 999. |
Assertion (A): `pH` of `HCI` solution is less than that of acetic acid of the some concentartion. Reason (R) : In equimolar solution, the number of titrable protons present in `HCI` is less than that present in acetic acid.A. If both (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. |
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Answer» Correct Answer - C `(A)` is correct because `HCI` is a strong electrolyte resulting in higher `[H^(o+)]` (and less `pH)` then acetic acid. `(R)` is wrong, `HCI` will produce more protons than acetic acid at same concentration. |
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| 1000. |
Assertion (A): `pH` of `HCI` solution is less than that of acetic acid of the some concentartion. Reason (R) : In equimolar solution, the number of titrable protons present in `HCI` is less than that present in acetic acid.A. Both A and R true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - C | |