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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.
| 1251. |
The aqueous solution of `CuSO_(4)` isA. AcidicB. BasicC. NeutralD. Amphoteric |
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Answer» Correct Answer - A `CuSO_(4)` is a salt of weak base, `(Cu(OH)_(2))` and strong acid `(H_(2)SO_(4))`. |
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| 1252. |
Dissociation of `H_(3)PO_(4)` takes place in following stepsA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `H_(3)PO_(4) overset("I step")hArr H^(+) + H_(2)PO_(4)^(-)` `H_(2)PO_(4)^(-) overset("II step")hArr + HPO_(4)^(--)` `HPO_(4)^(--) overset("III step")hArr H^(+) + PO_(4)^(---)` |
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| 1253. |
Correct statement isA. `NH_(4)Cl` gives alkaline solution in waterB. `CH_(3)COONa` gives acidic solution in waterC. `CH_(3)COOH` is a weak acidD. `NH_(4)OH` is a strong base |
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Answer» Correct Answer - C `CH_(3)COOH` is weak acid shows dissociation equilibrium as `CH_(3)COOH hArr CH_(3)COO^(-) + H^(+)`. |
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| 1254. |
Neutralization of an acid with base invariably results in the production ofA. `H_(3)O^(+)`B. `OH^(-)`C. `H_(2)O`D. `H^(+)` and `OH^(-)` |
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Answer» Correct Answer - C `NaOH + HCl underset("Reaction")overset("Neutralization")hArr underset("Salt")(NaCl) + H_(2)O` |
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| 1255. |
Formation of detive bond is neutralization according to the following acid base theoryA. ArrheniusB. LewisC. Bronsted-LowryD. Lowry theory |
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Answer» Correct Answer - B `overset(..)NH_(3) + squareBF_(3) rarr NH_(3) rarr BF_(3)` |
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| 1256. |
A solution contains `0.1 M` is `Cl^(-)` and `10^(-4) M CrO_(4)^(2-)`. If solid `AgNO_(3)` is gradually added to this solution, what will be the concentration of `Cl^(-)` when `Ag_(2)CrO_(4)` begins to precipitate? `(Ksp (AgCl) = 10^(-10) M^(2)`, `K_(sp) (Ag_(2)CrO_(4)) = 10^(-12)M^(3))`A. `10^(-6)M`B. `10^(-4) M`C. `10^(-5) M`D. `10^(-9) M` |
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Answer» Correct Answer - A `[Ag^(+)] ` required for precipitation of `AgCl` `[Ag^(+)] = (ksp[AgCl])/([Cl^(-)]) = 10^(-9)` `[Ag^(+)]` required for precipitation of `Ag_(2)CrO_(2)` `[Ag^(+)] =sqrt((ksp[Ag_(2)CrO_(4)])/([CrO_(4)^(2-)])) = sqrt((10^(-12))/(10^(-4)) ) = 10^(-4)` At this point, the concentration of `Cl^(-)` ion in the solution can be calculated from `Ksp(AgCl)` `[Cl^(-)] = (ksp(AgCl))/([Ag^(+)]) = (10^(-10))/(10^(-4)) = 10^(-6) M` |
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| 1257. |
The correct order of increasing basicity of the given conjugate bases `(R=CH_(3))` isA. `RCObarO lt HC -= barC lt barR lt bar(N)H_(2)`B. `barR lt HC -= barC lt RCObarO lt bar(N)H_(2)`C. `RCObarO lt bar(N)H_(2) lt HC -= bar(C) lt bar(R)`D. `RCObar(O) lt HC -= bar(C) lt bar(N)H_(2) lt bar(R)` |
| Answer» Correct Answer - D | |
| 1258. |
If `P^(@)` and `P_(s)`, the V.P. of solvent and solution respectively and `N_(1)` and `N_(2)` are the mole fraction of solvent then :A. `p=p^(@)n_(1)`B. `p^(@)-p=n`C. `p=p^(@)n_(1)`D. `p=p^(@)(n_(1)//n_(2))` |
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Answer» Correct Answer - A Factual question. |
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| 1259. |
20 mL of `C_(2)H_(5)OH` when mixed with 20ml of water,the volume of the resulting solution isA. 40 mLB. `lt40 mL`C. ` gt 40 mL`D. none of these |
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Answer» Correct Answer - C Becasuse of positive deviation. |
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| 1260. |
The pH of normal KOH isA. 1B. 0C. 14D. 7 |
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Answer» Correct Answer - C It is a strong base. |
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| 1261. |
The concentration of hydrogen ion in water isA. 8B. `1 xx 10^(-7)`C. 7D. 43472 |
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Answer» Correct Answer - B `[H^(+)][OH^(-)] = 10^(-14), (10^(-7))(10^(-7)) = 10^(-14)`. |
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| 1262. |
Given reason for the statement that the `pH` of an aqueous solution of sodium acetate is more than `7`. |
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Answer» Sodium acetate is a salt of strong base and weak-acid. It undergoes anionic hydrolysis and forms an alkaline solution in water, therefore `pH gt 7`. `CH_(3)COONa +H_(2)O hArr CH_(3)COOH +NaOH` `CH_(3)COO^(Theta) +H_(2)O hArr CH_(3)COOH +overset(Theta)OH` |
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| 1263. |
The aqueous solution of sodium acetate with minimum pH isA. `0.01 M`B. `0.001 M`C. `0.0001 M`D. `0.1 M` |
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Answer» Correct Answer - C `CH_(3)COONa` is a salt of weak acid `(CH_(3)COOH)` and strong base `(NaOH)`. So `CH_(3)COONa` is a strong base. |
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| 1264. |
Methy`1` red has a `K_(a) = 10^(-5)`. The acid form `"Hin"` is red and its conjugate base, `Ind^(Θ)` is yellow. Complete the following table: `{:("pH",3,5,7),(["Ind"^(Theta)]//["HIn"],-,-,-),("Colour",-,-,-):}` |
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Answer» `10^(-2), 1, 10^(2)`, Red, Orange, Yellow. i. `pH = 3, [H^(o+)] = 10^(-3)M` `[("Ind"^(Θ))/("HIn")] = (K_(a))/([H^(o+)]) = (10^(-5))/(10^(-3)) = 10^(-2)` (Colour of acid form predominates, i.e., red) ii. `pH = 5, [H^(o+)] = 10^(-5)M` `:. [("Ind"^(Θ))/("HIn")] = (10^(-5))/(10^(-5)) = 1` (Colour is mixture of red and yellow, i.e. orange) iii. `pH = 7, [H^(o+)] = 10^(-7)M` `[("Ind"^(Θ))/("HIn")] = (10^(-5))/(10^(-7)) = 10^(2)` (Colour of basic form predominates, i.e., yellow colour) |
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| 1265. |
The solubility of `TI_(2)S` in pure `CO_(2)` -free water is `6.3 xx 10^(-6)M`. Assume that the dissolved `S^(2-)` ion hydrolyses almost completely to `HS^(Θ)` and that the further hydrolysis to `H_(2)S` is neglected. What is the `K_(sp).(K_(2)(H_(2)S) = 10^(-14))`. |
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Answer» `TI_(2)S hArr 2TI^(oplus) +S^(2-), K_(sp) = [TI^(oplus)]^(2)][S^(2-)]` `S^(2-) +H_(2)O hArr HS^(Θ) + overset(Θ)OH` `K_(h) = (K_(w))/(K_(2)) = (10^(-14))/(10^(-14)) = 1.0` `[TI^(oplus)] = 2 (6.3 xx 10^(-6)), [overset(Θ)OH] = [HS^(Θ)] = 6.3 xx 10^(-6)` `K_(h) = ((6.3 xx 10^(-6))^(2))/([S^(2-)]) = 1.0, :. [S^(2-)] = (6.3 xx 10^(-6))^(2)` `K_(sp) = (6.3 xx 10^(-6))^(2) [2(6.3 xx 10^(-6))]^(2) = 6.3 xx 10^(-21)` |
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| 1266. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)`A buffer of acetic acid `(pK_(a) = 4.8)` with sodium acetate will be, when `CH_(3)COOH` and `CH_(3)COONa` are present in equivalent amounts has pH limits equal toA. `0` to `4.8`B. `3.8 ` to `5.8`C. `4.3` to `5.3`D. `4.8` |
| Answer» Correct Answer - B | |
| 1267. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)` Which among the following solution will the most efficient bufferA. `0.1M CH_(32)COONa + 0.01M CH_(3)COOH`B. `0.1M NH_(4)Cl + 0.1MNH_(4)OH`C. `0.001M HCOOH+ 0.002M HCOONa`D. All the above |
| Answer» Correct Answer - B | |
| 1268. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)` The buffer capacity is equal toA. `(Deltan)/(DeltapH)`B. `(pH)/(Deltan)`C. `+-1pK_(a)`D. None of these |
| Answer» Correct Answer - A | |
| 1269. |
Which of the following statement in incorrect ?A. Acidic buffer solution must be acidic in nature.B. Basic buffer solution can be of acidic, basic or neutral in nature.C. A solution of salt of weak acid and weak base, acts as a buffer solution.D. A solution in which `NaHCO_(3)` and `Na_(2)CO_(3)` are present, acts as a buffer solution. |
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Answer» Correct Answer - a |
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| 1270. |
What volume of 0.2 M `RNH_(3)CI` solution should be added to 100 mL of 0.1 M `RNH_(2)` solution to produce a buffer solution of pH=8.7 ? [Given : `pK_(b)` of `RNH_(2)=5`, log 2=0.3]A. 50 mLB. 100 mLC. 200 mLD. None of these |
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Answer» Correct Answer - b |
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| 1271. |
Which one is amphoteric oxide ?A. `SO_(2)`B. `B_(2)O_(3)`C. `ZnO`D. `Na_(2)O` |
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Answer» Correct Answer - C ZxnO is soluble in acid and alkalies both. |
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| 1272. |
Out of the following acid base-reactions, reaction (b) and (c) are possible. a. `PH_(3) + overset(o+)NH_(4) rarr overset(o+)PH_(4) +NH_(3)` b. `NH_(3) + Poverset(o+)H_(4) rarr Noverset(o+)H_(4) + PH_(3)` c. `(CH_(3))_(3) P + overset(o+)NH_(4) rarr (CH_(3))_(3) overset(o+)PH + NH_(3)` d. `(CH_(3))_(3) N + PH_(4)^(o+) rarr (CH_(3))_(2) overset(o+)NH + PH_(3)` |
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Answer» Correct Answer - A Proton affinity of `NH_(3)`is greater than `PH_(3)`. Proton affinity `(CH_(3))_(3)P` is grater than `NH_(3)`. |
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| 1273. |
The solution which consumes `[H^(+)]` or `[OH^(-)]` or both simultaneously from externally added base in order to give negligible change in `pH`, is known as buffer solution. In general, the solution resists the change in pH. Buffer solution does not mean that there does not occur a pH change in pH. Buffer solution des not mean that there does not occur a pH change at all. It implies the pH change occurs but in neglibible amount. There are two types of buffer (i) Acidic buffer: it is a mixture of weak acid and its salt acid strong base. (ii) Basic buffer : It is a mixture of weak base and its salt with strong acid. Which of the following mixture will be a buffer solution when dissolved in `500 mL` of water:A. `0.200` mol of aniline and `0.200` mol of HClB. `0.200` mol of aniline and `0.400` mol of `NaOH`C. `0.200` mol of NaCl and `0.100` mol of `HCl`D. `0.200` mol of aniline and `0.100` mol of `HCl` |
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Answer» Correct Answer - D For the formation of Buffer solution meq of weak base `gt` meq of strong acid. |
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| 1274. |
The aqueous solution `P^(H)` is zero. What is the nature of the solution.A. Slightly acidicB. Strongly acidicC. NeutralD. Basic |
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Answer» Correct Answer - B `P^(H) = 0` the solution is strongly acidic |
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| 1275. |
If `pH_(a)` is more than `pH_(b)`, the `p_(H)` of the aqueous solution of the salt formed by the above acidic and base isA. `7`B. `gt 7`C. `lt 7`D. `0` |
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Answer» Correct Answer - B `P^(Ka) gt P^(Kb)`, the `P^(H)` of the aqeous solution of the salt is `gt 7` |
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| 1276. |
Which of the following is an acidic salt/A. `Na_(2)SO_(4)`B. `NaHSO_(3)`C. `Na_(2)SO_(3)`D. `K_(2)SO_(4)` |
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Answer» Correct Answer - B `NaHSO_(3)` have acidic hydrogen so it is acidic salt |
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| 1277. |
What is the `pH` of `m//20 KNC_(8)H_(4)O_(4)` solution .Given :`H_(2)C_(8)O_(4)` is a diabasic acid with `pK_(a_(1))& pK_(a_(2))` as `2.94 & 5.44` respectively. |
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Answer» `NaNO_(3)" "[C]=0.1M` `pH=(pK_(a_(1))+pK_(a_(2)))/(2)=(2.94+5.44)/(2)=4.19` |
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| 1278. |
pOH of 0.1 molar aqueous NaCN solution found to be 2, then calculate value of dissociation constant of HCN in its aqueous solution.A. `1.11xx10^(-3)`B. `10^(-2)`C. `10^(-11)`D. `9xx10^(-12)` |
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Answer» Correct Answer - d |
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| 1279. |
Which buffer solution out of the following will have `pH gt 7`?A. `CH_(3)COOH + CH_(3)COONa`B. `HCOOH + HCOOK`C. `CH_(3)COONH_(4)`D. `NH_(4)OH +NH_(4)CI` |
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Answer» Correct Answer - D a. Acidic buffer, and have `pH lt 7` b. Also acidic buffer, `pH lt 7` c. Salt of `W_(A)//W_(B), pH = 7` .d Basic buffer, and `pH gt`. |
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| 1280. |
Which of the following is most soluble?A. `Bi_(2)S_(3) (K_(sp) = 1 xx 10^(-70))`B. `MnS (K_(sp) = 7xx10^(-16))`C. `CuS(K_(sp) = 8 xx 10^(-37))`D. `Ag_(2)S (K_(sp) = 6 xx 10^(-51))` |
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Answer» Correct Answer - B Most soluble compound is that which have highest `K_(sp)` value. `K_(sp)` of `MnS (7 xx 10^(-6))` is highest. |
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| 1281. |
`10^(-6)M HCl` is diluted to `100` times. Its `pH` is:A. `6`B. `8`C. `6.98`D. `7.02` |
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Answer» New Concentration of `HCl=(10^(-6))/(100)=10^(-8)M` `[H^(+)]=10^(-7)+10^(-8)`(approximately) (Little less than `10^(-7)` form water). |
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| 1282. |
Which of the following solutions will have pH close to 1.0 ?A. `100mL of M//10 HCl+100mL of M//10NaOH`B. `55mL of M//10 HCl +45mL of M//10 NaOH`C. `10mL of M//10 HCl+90mL of M//10 NaOH`D. `75mL of M//5 HCl+25mL of M//5NaOH` |
| Answer» `{:(a,HCl,NaOH),(,"No of milli eq."=(1)/(100)xx10=10,(1)/(10)xx100=10),(,"so solution is Neutral",),(b,(1)/(10)xx55=5.5,(1)/(10)xx45=4.5),(,[H^(+)]=(1)/(100)=10^(-2)M pH=2,),(c,(1)/(10)xx10=1,(1)/(10)xx90=9"Basic"),(d,(1)/(5)xx75=15,(1)/(5)xx25=5),(,[H^(+)]=0.1MpH=1,):}` | |
| 1283. |
`Ph` of `3xx10^(-3)M` solution of `H_(3)X` will be Assuming `alpha_(1)=1,alpha_(2)=1//3,alpha_(3)=` negligibleA. `2.40`B. `3.0`C. `3.4771`D. `4.0` |
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Answer» Correct Answer - 1 `H_(3)X rarr H^(+)+H_(2)X^(-)` `3xx10^(-3)" "3xx10^(-3)` `{:(H_(2)X,hArr,H^(+),+,HX^(-)),(3xx10^(-3),,3xx10^(-3),,0),(3xx10^(-3)-Calpha_(2),,3xx10^(-3)+Calpha_(2),,):}` `(H^(T))=3xx10^(-3)+3xx10^(-3)xx(1)/(3)` `=4xx10^(-3)` `pH=3-log4=-0.602~~2.4` |
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| 1284. |
Degree of hydrolysis of `0.25 M CH_(3)COOHNa` is `10%` what will be the degree of hydrolysis if concentration of `CH_(3)COONa` is made `0.01M`A. `10%`B. `1%`C. `50%`D. `75%` |
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Answer» Correct Answer - 3 `(h_(1))/(h_(2))=sqrt((c_(2))/(c_(1)))rArr(0.1)/(h_(2))=sqrt((0.01)/(0.25))rArrh=0.5rArr50%(3)` |
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| 1285. |
If 20 mL of 0.4 N NaOH solution completely neutralises 40 mL of a dibasic acid, the molarity of the acid solution is `:`A. 0.1 MB. 0.2 MC. 0.3 MD. 0.4 M |
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Answer» Correct Answer - A 0.4 NaOH = 0.4 M NaOH `2 NaOH +H_(2)A rarr 2NaA+2H_(2)O` `underset((NaOH))((M_(1)V_(1))/(n_(1)))=underset((H_(2)A))((M_(2)V_(2))/(n_(2)))` `(0.4 M xx 20 mL)/(2 mol )=(M_(2) xx 40 mL)/(1 mol)` `:. M_(2)=(0.4 xx 20 M)/(40 xx2)=0.1 M` |
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| 1286. |
An element which can exist as a positive ion in acidic solution and also as a negative ion in basic solution is said to be….. |
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Answer» Correct Answer - A::C An element which can exist as a positive ion in acidic solution and also as a negative ion in basic solution is said to be amphoteric. |
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| 1287. |
If the equilibrium constant for the reaction of weak acid HA with strong base is `10^(9)`, then pH of `0.1M` Na A is:A. 5B. 9C. 7D. 8 |
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Answer» Correct Answer - B `HA +NaOH hArrNaA +H_(2)O` `K=10^(9)` `:.` For `NaA +H_(2)O hArr NaOH +HA,K=10^(-9)` Initial 0.1 M At eqm `(0.1 -alpha)M " "alpha M " "alphaM` `([NaOH][HA])/([NaA])=10^(-9)` `(alpha^(2))/(0.1-alpha)=10^(-9)` As `alpha` is very small , `0.1 -alpha ~~0.1` `alpha^(2)=10^(-9)xx0.1` `alpha=10^(-5)` `[OH^(-)]=[NaOH]=10^(-5)M` ltbr. `pH =14 0pOH =14-5=9` |
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| 1288. |
If the equilibrium constant for the reaction of weak acid HA with strong base is `10^(9)`, then pH of `0.1M` Na A is: |
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Answer» Correct Answer - 9 `HA+NaOH rarr NaA+H_(2)O` or `HA+OH^(-)rarr A^(-)+H_(2)O` `K_(eq.)=10^(9)=([A^(-)][H_(2)O])/([HA][OH^(-)])` Also `HAhArrH^(+)+A^(-)` `K_(a)= ([H^(+)][A^(-)])/([HA])` `:. (K_(eq.))/(K_(a))=(1)/(K_(w))` or `K_(a)=10^(9)xx10^(-14)=10^(-5)` Thus, for `A^(-)H_(2)OhArrHA+OH` `[OH^(-)]= c.h=c. sqrt((K_(H))/(c ))= sqrt((K_(w).c)/(K_(a)))` `=sqrt((10^(-14)xx-0.1)/(10^(-5)))=10^(-5)` `:. [H^(+)]=10^(-9) implies pH =9` |
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| 1289. |
The pH value of `N//10` NaOH isA. 7B. 10C. 12D. 13 |
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Answer» Correct Answer - D For NaOH, `N//10 `NaOH `=0.1 ` M NaOH `[NaOH ]=0.1 M =10^(-1) M` `[OH^(-)] =[NaOH] =10^(-1)` `[H^(+)][OH^(-)]=10^(-14)` `:. [H^(+)]=(10^(-14))/([OH^(-)])=(10^(-14))/(1 xx 10^(-1))=10^(-13)` `pH =-log log[H^(+)]=-(log 10^(-13))=13` |
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| 1290. |
For the following equilibrium `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` `K_(p)` is formed to be equal to `K_(x )`. This is attained atA. 1 atmB. 0.5 atmC. 2 atmD. 4 atm |
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Answer» Correct Answer - A `K_(p)=K_(x )(P)^(Deltan)` `K_(x)` is equilibrium constant when concentration are take as mole fraction. `K_(p)=K_(x)(P)` `K_(p)=K_(x)` when `P=1` atm |
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| 1291. |
For the following equilibrium `N_(2)O_(4)(g)hArr 2NO_(2)(g)` `K_(p)` is found to be equal to `K_(c)`. This is attained when `:`A. `T=1K`B. `T=12.18K`C. `T=27.3K`D. `=273K` |
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Answer» Correct Answer - B `K_(p)=K_(c ) (RT)^(Deltan)` `Delta n =2-1=1` `:. K_(p)=K_(c)(RT)` `K_(p)=K_(c)` when `RT =1` `:. T=(1)/(R)` `=(1)/(0.0821)=12.18K` |
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| 1292. |
If NaOH is titrated HCI, variation of conductance (y-axis) with addition of HCI (x-axis) will be:A. B. C. D. |
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Answer» Correct Answer - B On addition of `NaOH,H^(+)` combines with `OH^(-)` to form undissociated water and thus faster moving `H^(+)` are replaced by slow moving `Na^(+)` to show decrease in conductance. Addition of `NaOH` at equivalence point will increase conductance abruptly due to free `H^(+)` and `CI^(-)` ions. |
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| 1293. |
Which of the following will not produce a precipitate with dilute silver nitrate solution ?A. `CO_(3(aq.))^(2-)`B. `CrO_(4(aq.))^(2-)`C. `I_((aq.))^(-)`D. `F_((aq.))^(-)` |
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Answer» Correct Answer - D AgF is soluble in water. |
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| 1294. |
Of the given anions, the strongest Bronsted base isA. `CIO^(Theta)`B. `CIO_(2)^(Theta)`C. `CIO_(3)^(Theta)`D. `CIO_(4)^(Theta)` |
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Answer» Correct Answer - A Stronger is the acid, weaker is the conjugate base and vice versa. `{:(Base,,Acid),(CIO^(Theta),,HCIO),(CIO_(2)^(Theta),,HCIO_(2)),(CIO_(3)^(Theta),,HCIO_(3)),(CIO_(4)^(Theta),,HCIO_(4)):}` Acidic order: `HCIO_(4) gt HCIO_(3) gt HCIO_(2) gt HCIO` Conjugate base order: `CIO_(4)^(Theta) lt CIO_(3)^(Theta) lt CIO_(2)^(Theta) lt CIO^(Theta)` `:.` Strongest base `=CIO^(Theta)` |
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| 1295. |
Of the given anions, the strongest bronsted base isA. `ClO^(-)`B. `ClO_(2)^(-)`C. `ClO_(3)^(-)`D. `ClO_(4)^(-)` |
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Answer» Correct Answer - A Among the given oxyacids HClO is the weakest as acidity depends on the oxidation state (+1 in HClO, which is lowest). Hence its conjugate base `(ClO^(-))` will be the strongest. |
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| 1296. |
Statement: The acidic nature of some cations is: `B^(3+)gtBe^(2+)gtNa^(+)gtK^(+)` Explanation: More is the effective nuclear charge on cation more is its acidic nature.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. |
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Answer» Correct Answer - C Explanation is correct reason for statement. |
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| 1297. |
Relation between hydrolysis constant and dissociation constant are given. Which is the correct formula for `MgCl_(2)`.A. `K_(h) = (K_(w))/(K_(a))`B. `K_(h) = (K_(w))/(K_(b))`C. `K_(h) = (K_(w))/(K_(a) xx K_(b))`D. `K_(w) = (K_(h))/(K_(b))` |
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Answer» Correct Answer - B For a salt of weak base and strong acid like `MgCl_(2)`, the relation is as `K_(h) = (K_(w))/(K_(b))`. |
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| 1298. |
Which is nucleophileA. `BF_(3)`B. `NH_(3)`C. `BeCl_(2)`D. `H_(2)O` |
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Answer» Correct Answer - B `NH_(3)` presence of lone pair of electrons. |
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| 1299. |
The solubility product of `As_(2)O_(3)` is `10.8 xx 10^(-9)`. It is 50% dissociated in saturated solution. The solubility of salt isA. `10^(-2)`B. `2 xx 10^(-2)`C. `5 xx 10^(-3)`D. `5.4 xx 10^(-9)` |
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Answer» Correct Answer - B `As_(2)O_(3) hArr 2As^(3+) + 3O^(2-)` `K_(sp) = (2s)^(2) (3s)^(3)` `10.8 xx 10^(-9) = 108 S^(5)` `S = 10^(-2)` If dissociation is 50% then solubility is `2 xx 10^(-2)` mol/L. |
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The solubility product of `As_(2)S_(3)` is `2.8 xx 10^(-72)`. What is the solubility of `As_(2)S_(3)`A. `1.09 xx 10^(-15)` mole/litreB. `1.72 xx 10^(-15)` mole/litreC. `2.3 xx 10^(-16)` mole/litreD. `1.65 xx 10^(-36)` mole/litre |
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Answer» Correct Answer - A `K_(sp) = [As^(3+)][S^(2-)]`, `S = 5sqrt((K_(sp))/(108)) = 5sqrt((2.8 xx 10^(-72))/(108)) = 1.09 xx 10^(-15)` |
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