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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.
| 1151. |
Find the total number of diprotic acids among the following: `H_(3)PO_(4), H_(2)SO_(4), H_(3)PO_(3), H_(2)CO_(3), H_(2)S_(2)O_(7), H_(3)BO_(3), H_(3)PO_(2), H_(2)CrO_(4), H_(2)SO_(3)` |
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Answer» Diprotic acids are those which furnish two protons per molecule. `H_(2)SO_(4), H_(3)PO_(3), H_(2)CO_(3), H_(2)S_(2)O_(7), H_(3)CrO_(4), H_(2)SO_(3)` `H_(3)BO_(3) +H_(2)O rarr B(OH)_(4)^(Theta) +H^(o+)` (Monobasic) `H-O-underset((H_(2)S_(2)O_(7)))(overset(O)overset(||)underset(O)underset(||)(O)-O)-overset(O)overset(||)underset(O)underset(||)(S)-O-H("Diabasic")` Only those hydrogensa re replaceable form oxy-acids of phoshorous and sulphur which are attached to oxygen. `H_(3)PO_(4)` (Tribasic) `H_(3)PO_(3)` (Dibasic) `H_(3)PO_(2)` (Monobasic) |
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| 1152. |
The `pK_(a)` for acid A is greater than `pK_(a)` for acid B, the strong acid is :-A. Acid BB. Acid AC. Both A and BD. Neither A nor B |
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Answer» Correct Answer - A Because those acid have higher `pK_(a)` value are weak acid. |
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| 1153. |
Accepting the definition that an acid a proton donor, the acid in the following reaction `NH_(3)+H_(2)O rarr NH_(4)^(+) + OH^(-)` isA. `NH_(3)`B. `H^(+)`C. `NH_(4)^(+)`D. `H_(2)O` |
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Answer» Correct Answer - D In this reaction `H_(2)O` acts as a acid. |
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| 1154. |
The solubility of `AgCl(s)` with solubility product `1.6 xx 10^(-10)` is in 0.1 M NaCl solution would beA. ZeroB. `1.26 xx 10^(-5)M`C. `1.6 xx 10^(-9) M`D. `1.6 xx 10^(-11) M` |
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Answer» Correct Answer - C `AgCl hArr Ag^(+) + Cl^(-)` `S hArr S + 0.1` `K_(sp) = [Ag^(+)][Cl^(-)]` `1.6 xx 10^(-10) = S xx (S + 0.1) = S xx 0.1` `1.6 xx 10^(-9) = S` |
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| 1155. |
Accepting the definition that an acid is a proton donor, the acid in the following reaction `NH_(3)+H_(2)O rarr NH_(4)^(+)+OH^(-)` isA. `NH_(3)`B. `H^(+)`C. `NH_(4)^(+)`D. `H_(2)O` |
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Answer» Correct Answer - d In the reaction `NH_(3) +H_(2) O rarr NH_(4)^(+) +OH^(+)` water , `H_(2)O` acts as an acid as it gaives a `H^(+)` to `NH_(3)`. |
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| 1156. |
Which of the following mixture solution has `pH ~~ 1.0`?A. `100 mL M//10 HCI + 100 mL M//10 NaOH`B. `55 mL M//10 HCI +45 mL M//10 NaOH`C. `10 mL M//10 HCI + 90 mL M//10 NaOH`D. `75 mL M//5 HCI +25 mL M//5 NaOH` |
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Answer» Correct Answer - D a. `HCI = 100 xx (1)/(10) = 10 mEq` `NaOH = 100 xx (1)/(10 = 10 mEq` Salt of `S_(A)//S_(B)` is formed, `pH = 7` b. `HCI = 55 xx (1)/(10) = 5.5 mEq` `NaOH = 45 xx (1)/(10) = 4.5 mEq` `1.0 mEq` of `HCI` is left. So `pH` will be in acidic range but not equal to one, since concentration fo `HCI` is very low. c. `HCI =10 xx (1)/(10) = 1mEq` `NaOH = 90 xx (1)/(10) =9 mEq` `8 mEq of NaOH` is left and `pH gt 7`. d. `HCI = 75 xx (1)/(5) = 15 mEq` `NaOH = 25 xx (1)/(5) = 5 mEq` `15 - 5 = 10 mEq of HCI` is left. `:. [HCI] = [H_(3)O^(o+)] = (10 mEq)/((75+25)mL) = (10)/(100) = 0.1M` `pH =- log (10^(-1)) = 1`. |
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| 1157. |
The addition of NaCl to AgCl decreases the solubility of AgCl becauseA. Solubility product decreasesB. Solubility product remains constantC. Solution becomes unsaturatedD. Solution becomes supersaturated |
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Answer» Correct Answer - D Because the solution becomes supersaturated. |
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| 1158. |
Solution of X is being titrated against a solution of Y. If phenolphathalen is found to be a suitabe indicator. X and Y could be respectively.A. `NaOH` and `HCl`B. `HCl` and `NaOH`C. `CH_(3)COOH` and `NaOH`D. `HCl` and `NH_(3)(aq)` |
| Answer» Correct Answer - A::B::C | |
| 1159. |
In the following reaction. `AgCl +KI hArr KCl +AgI` as KI is added, the equilibrium is shifted towards right giving more AgI precipitate , becauseA. both AgCl and AgI are sparingly solubleB. the `K_(sp)` of AgI is lower than `K_(sp)` of AgClC. the `K_(sp)` of AgI is higher than `K_(sp)` of AgClD. both AgCl and AgI have same solubility product |
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Answer» Correct Answer - B On adding KI, the equilibrium `AgCl+KI hArr KCl +AgI` is shifted towards right. This means that AgI is less soluble than AgCl. In other words `K_(sp)` of AgI is less than `K_(sp)` of AgCl. |
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| 1160. |
The pH of 10 M HCI solution is:A. less than zeroB. zeroC. 2D. 1 |
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Answer» Correct Answer - B `[H^(+)]=10^(1)`, Thus `pH= -1` which is not true. For highly concentration solutions `pH= -log a_(H)^(+)` and not, `pH= -log[H^(+)]` . For such solutions `pH=0`. |
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| 1161. |
In the titration of a solution of a weak acid `HX` with `NaOH`, the `pH` is `5.8` after `10mL` of `NaOH` solution has been added and `6.40` after `20mL` of the `NaOH` has been affed. What is the ionisation constant of the acid `HX`? |
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Answer» Let the mormality of `HX` and `NaOH = 1`. Let `x mEq` or mmol of `HX` is present initially. When `pH = 5.8`. `["Salt"] = 10 xx 1 = 10 mmol` `["Acid"] = (x - 10) mmol` When `pH = 6.4` `["Salt"] = 20 xx 1 = 20 mmol` `["Acid"] = (x - 20) mmol` `5.8 = pK_(a) +log (10//x - 10) ..(i)` `6.4 = pK_(a) +log (20//x - 20)...(ii)` SOlving equaitons (i) and (ii), calculate the value of `x` and `pK_(a)`. `:. pK_(a) = 6.1` |
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| 1162. |
The addition of `KCl` to `AgCl` decreases the solubility of `AgCl`, becauseA. `K_(sp)` of `AgCl` decreasesB. `K_(sp)` of `AgCl` increasesC. Solution becomes unsaturatedD. Ionic product exceeds the `K_(sp)` value |
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Answer» Correct Answer - D Common ions effect |
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| 1163. |
A solution contains `10mL` of `0.1N NaOH` and `10mL` of `0.05Na_(2)SO_(4), pH` of this solution isA. `7`B. Less than `7`C. Greater than `7`D. Zero |
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Answer» Correct Answer - C `NaOH = 10 xx 0.1 = 1 mEq` `Na_(2)SO_(4) = 10 xx 0.05 = 0.5 mEq` But `pH` of solution id due to only strong base, `NaOH` and will be greater than `7`. Salt of `S_(A)//S_(B)`, has no effect on `pH`, since its `pH` is `7`. |
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| 1164. |
An acid solution with `pH=6at 25^(@)C` is diluted by `10^(2)` times. The pH of solution will:A. decrease by 2B. increase by 2C. decrease by `0.95` approximatelyD. increase by `0.95` appromately |
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Answer» Correct Answer - D `[H^(+)]=10^(-6)at pH=6` If `[H^(+)]` are diluted to `10^(2)` times, the new `[H^(+)]=10^(-8)`. Thus pH of solution will be in between 6 to 7 approximately `6.95` |
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| 1165. |
The pH value of a 10 M solution of HCl isA. less than 0B. equal to 0C. equal to 1D. equal to 2 |
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Answer» Correct Answer - A `HCl(aq)rarr H^(+)(aq)+Cl^(-)(aq) [S = sqrt(K_(sp))]` `[HCl]=10M` `rArr [H^(+)]=10 mol//L` `pH=-loh[H^(+)]=-log 10` `=-1`, so the pH is less than zero. |
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| 1166. |
Which one of the following is most solubleA. `CuS (K_(sp) = 8 xx 10^(-37))`B. `MnS(K_(sp) = 7 xx 10^(-16))`C. `Bi_(2)S_(3)(K_(sp) = 1 xx 10^(-70))`D. `Ag_(2)S(K_(sp) = 6 xx 10^(-51))` |
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Answer» Correct Answer - B Solubility is directly proportional to the `K_(sp)`. |
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| 1167. |
The ionic product of water at `25^(@)C` is `10^(-4)`. The ionic product at `90^(@)C` will beA. `1 xx 10^(-20)`B. `1 xx 10^(-12)`C. `1 xx 10^(-14)`D. `1 xx 10^(-16)` |
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Answer» Correct Answer - B When increasing the temperature the value of ionic product also increases. |
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| 1168. |
In which of the following the solubility of AgCl will be minimum ?A. `0.1 M NaNO_(3)`B. WaterC. `0.1 M NaCl`D. `0.1 M NaBr` |
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Answer» Correct Answer - C In 0.1 M NaCl, the solubility of AgCl is minimum due to the phenomenon of common ion effect. |
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| 1169. |
Which of the following is most soluble?A. `Bi_(2)S_(3)(K_(sp)=1xx10^(-70))`B. `MnS(K_(sp)=7x10^(-16))`C. `CuS(K_(sp)=8xx10^(-37))`D. `Ag_(2)S(K_(sp)=6xx10^(-51))` |
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Answer» Correct Answer - B Higher the value of solubility product, higher is its solubility. In all these compounds the MnS is most soluble because its solubility product is maximum. |
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| 1170. |
Solubility of AgCl will be minimum inA. 0.001 M `AgNO_(3)`B. Pure waterC. 0.01 M `CaCl_(2)`D. 0.01 M NaCl |
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Answer» Correct Answer - C 0.01 M `CaCl_(2)` gives maximum `Cl^(-)` ions to keep `K_(sp)` of AgCl constant, decrease in `[Ag^(+)]` will be maximum. |
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| 1171. |
Which of the following mixture solution has `pH ~~ 1.0`?A. `100 mL(M)/(10)HCl +100 mL (M)/(10)NaOH`B. `55 mL(M)/(10)HCl +45 mL (M)/(10)NaOH`C. `10 mL(M)/(10)HCl +90 mL (M)/(10)NaOH`D. `75 mL(M)/(5)HCl +25 mL (M)/(5)NaOH` |
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Answer» Correct Answer - D In choice (D),millimoles of HCl` =75 xx (1)/(5) =15` Millimoles of NaOH `=25 xx (1)?(5)=5` Millimoles of HCl left `=15-5 =10` Volume of solution `=100mL` `[H_(3)O^(+)]=(10 xx 10^(-3) xx 1000)/(100)M` `:. pH =-log 10^(-1)=1` |
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| 1172. |
Which one of the following can be classified as a Bronsted baseA. `NO_(3)^(-)`B. `H_(3)O^(+)`C. `NH_(4)^(+)`D. `CH_(3)COOH` |
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Answer» Correct Answer - A Those compound which accept `H^(+)` is called Bronsted base `NO_(3)^(-)` accept `H^(+)` and form `HNO_(3)`. So it is a base. |
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| 1173. |
Which are strong acid ? (P)`HCIO_3` , (Q)`H_2SeO_3` , (R)`H_3AsO_4`A. P onlyB. R onlyC. P and R onlyD. Q and R only |
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Answer» Correct Answer - A |
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| 1174. |
When the acids, `HCIO_3, H_3BO_3, H_3PO_4` , are arranged in order of increasing strength , which order is correct ?A. `H_3BO_3 lt H_3PO_4 lt HCIO_3`B. `HCIO_3 lt H_3BO_3 lt H_3PO_4 `C. `H_3PO_4 lt HCIO_3 lt H_3BO_3`D. `H_3BO_3 lt HCIO_3 lt H_3PO_4 ` |
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Answer» Correct Answer - A |
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| 1175. |
When 0.1 mole solid NaOH is added to 1 L of 0.01 M `NH_3(aq)` , then which statement is wrong ? (`K_b=2xx10^(-5)`, log 2 =0.3)A. Degree of dissociation of `NH_3` approaches to zeroB. Change in pH by adding NaOH would be 1.85C. In solution , `[Na^+]` =0.1M , `[NH_3]`=0.1 M , `[OH^-]` =0.2 MD. on adding of `OH^-, K_b` of `NH_3` does not change . |
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Answer» Correct Answer - c |
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| 1176. |
When the compounds below are arranged in order of increasing solubility in water, which order is correct? A. `BaCO_(3), BaSO_(4), CaCO_(3), CaSO_(4)`B. `BaSO_(4), CaCO_(3), CaSO_(4), BaCO_(3)`C. `CaSO_(4), CaCO_(3), BaCO_(3), BaSO_(4)`D. `BaSO_(4), BaCO_(3), CaCO_(3), CaSO_(4)` |
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Answer» Correct Answer - d |
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| 1177. |
The correct relationship between the `pH` of isomolar solutions of sodium oxide `(pH_(1))`, sodium sulphide `(pH_(2))`, sodium selenide `(pH_(3))` and sodium telluride `(pH_(4))` isA. `pH_1 gt pH_2 gt pH_3 gt pH_4`B. `pH_1 lt pH_2 lt pH_3 lt pH_4`C. `pH_1 lt pH_2 lt pH_3 = pH_4`D. `pH_1 gt pH_2 = pH_3 gt pH_4` |
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Answer» Correct Answer - a |
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| 1178. |
What is the correct relationship between the pH of isomolar solutions of sodium oxide `(pH_(1))`, sodium sulphide `(pH_(2))`, sodium selenide `(pH_(3))` , and sodium telluride `(pH_(4)) ` ?A. `pH_(1)gt pH_(2)~~pH_(3)gt pH_(4)`B. `pH_(1)lt pH_(2)lt pH_(3) lt pH_(4)`C. `pH_(1)lt pH_(2)lt pH_(3)~~ pH_(4)`D. `pH_(1)gt pH_(2)gt pH_(3)gtpH_(4)` |
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Answer» Correct Answer - D The correct order of pH of isomolar solution of sodium oxide `(pH_(1))`, sodium sulphide `(pH_(2))`, sodium selenide `(pH_(3))` and sodium telluride `(pH_(4))` is `pH_(1)gt pH_(2)gt pH_(3)gt pH_(4)` because in aqueous solution, they are hydrolysed as follows. `Na_(2)O+2H_(2)O rarr underset("Base")(2NaOH)+H_(2)O` `Na_(2)S+2H_(2)O rarr underset("Strong base")(2NaOH)+underset("Weak acid")(H_(2)Se)` `Na_(2)Te+2H_(2)O rarr underset("strong base")(2NaOH)+underset("Weak acid")(H_(2)Te)` On moving down the group acidic character of oxides increases. Order of acidic strength Order of acidic strength `H_(2)Te gt H_(2)Se gt H_(2)S gt H_(2)O` Order of neutralisation of NaOH `H_(2)Te gt H_(2)Se gt H_(2)S gt H_(2)O` Hence, their aqueous solution have the following order of basic character due to neutralisation of NaOH with `H_(2)O,H_(2)S,H_(2)Se` and `H_(2)Te`. (`because pH` of basic solution is higher than acidic or least basic solution) |
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| 1179. |
The correct relationship between the `pH` of isomolar solutions of sodium oxide `(pH_(1))`, sodium sulphide `(pH_(2))`, sodium selenide `(pH_(3))` and sodium telluride `(pH_(4))` isA. `pH_(1) gt pH_(2)~~pH_(3) gtPH_(4)`B. `pH_(1) lt pH_(2) lt pH_(3) lt PH_(4)`C. `pH_(1) lt pH_(2) lt pH_(3) ~~PH_(4)`D. `pH_(1) gtpH_(2) gt pH_(3) gt PH_(3)` |
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Answer» Correct Answer - D Acidic character increases in the order `H_(2)O lt H_(2)S lt H_(2)Se lt H_(2)Te` Basic character of conjugate bases will be in the order `O^(2-)gt S^(2-) gt Se^(2-) gt Te^(2-)` More the basic character, more the pH `:.` pHs are in the order `pH_(1)gt pH_(2)gt pH _(3) gt pH _(4)` |
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| 1180. |
pH of 0.1 M `Na_2HPO_4` and 0.2 M `NaH_2PO_4` are respectively : (`pK_a` for `H_3PO_4` are 2.2,7.2,12.0)A. 4.7,9.6B. 9.6,4.7C. 4.7,5.6D. 5.6,4.7 |
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Answer» Correct Answer - b |
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| 1181. |
Which acid -base indicator will give the best results for the titration of an aqueous ammonia solution with 0.10 M HCI ? `{:("Indicator","Colour Change","pH range"),("Methyl violet ", red to blue,0 to 2),("Methyl red", red to yellow, 4 to 6),("Cresol red", yellow to "purple", 7 to 9),("Phenolphthalein", "colourless" to "pink", 8 to 10):}`A. Methyl violetB. Methyl redC. Cresol redD. Phenolphthalein |
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Answer» Correct Answer - b |
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| 1182. |
Which one of the followinf is acid salt?A. `Na_(2)S`B. `Na_(2)SO_(3)`C. `NaHSO_(3)`D. `Na_(2)SO_(4)` |
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Answer» Correct Answer - C Salt of `S_(B)//W_(A)` are called acid salt. `Na_(2),Na_(2)SO_(3),Na_(2)SO_(4)` are salts of `S_(a)//S_(B)`. But `NaHSO_(4)` is salt of `W_(A)//S_(B)`. |
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| 1183. |
Solution of aniline hydrochloride is X due to hydrolysis of Y.X and Y are:A. Basic , `C_6H_5NH_3^+`B. Acidic , `C_6H_5NH_3^+`C. Basic , `Cl^-`D. Acidic , `Cl^-` |
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Answer» Correct Answer - b |
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| 1184. |
Solution of aniline hydrochloride is X due to hydrolysis of Y.X and Y are:A. basic, `C_(6)H_(5)NH_(3)^(+)`B. acidic, `C_(6)H_(5)NH_(3)^(+)`C. basic, `CI^(-)`D. acidic, `CI^(-)` |
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Answer» Correct Answer - B `C_(6)H_(5)NH_(3)^(+)CI^(-)`, `underset(Y)(C_(6)H_(5)NH_(3)^(+))+H_(2)OhArr C_(6)H_(5)NH_(3)OH+underset(X)(H^(+))` Thus Y is `C_(6)H_(5)NH_(3)^(+)` and X is acidic. |
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| 1185. |
Which one is not an acid salt?A. `NaH_(2)PO_(4)`B. `NaH_(2)PO_(2)`C. `NaH_(2)PO_(3)`D. All of the above are acid salts |
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Answer» Correct Answer - D Salt of `S_(B)//W_(A)` are called acid salts. a. `NaH_(2)PO_(4)` (Salt of `NaOH +H_(3)PO_(4)) (S_(B)//W_(A))` b. `NaH_(2)PO_(2)` (Salt of `NaOH +H_(3)PO_(2)) (S_(B)//W_(A))` c. `NaH_(2)PO_(3)` (Salt of `NaOH +H_(3)PO_(3)) (S_(B)//W_(A))`All are acid salts. |
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| 1186. |
Slaked lime, `Ca(OH)_(2)` is used extensively in sewage treatment. What is themaximum pH that can be estabilished in `Ca(OH)_(2)(aq)` ? `Ca(OH)_(2(s))hArr Ca_((aq.))+2OH_((aq.))^(-), (K_(SP)=5.5xx10^(-6))`A. `1.66`B. `12.3471`C. `7.0`D. `14.0` |
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Answer» Correct Answer - B `K_(SP)of Ca(OH)_(2)=4S^(3)=5.5xx10^(-6)` `:. S= 3sqrt((5.5xx10^(-6))/(4))=1.11xx10^(-2)` `:. [OH^(-)]=2S=2xx1.11xx10^(-2)` `:. pOH=1.6528 :. pH =12.3471` |
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| 1187. |
Slaked lime, `Ca(OH)_(2)` is used extensively in sewage treatment. What is themaximum pH that can be estabilished in `Ca(OH)_(2)(aq)` ? `Ca(OH)_(2(s))hArr Ca_((aq.))+2OH_((aq.))^(-), (K_(SP)=5.5xx10^(-6))`A. `1.66`B. `12.35`C. `7.0`D. `14.0` |
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Answer» Correct Answer - b |
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| 1188. |
If ammonia is added to pure water, the concentration of a chemical species already present will decrease. The species isA. `O^(2-)`B. `OH^(-)`C. `H_(3)O^(+)`D. `NH_(4)^(+)` |
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Answer» Correct Answer - C In pure water, there is an equilibrium `H_(2)O +H_(2)O hArr H_(3)O^(+) +OH^(-)` Addition of ammonia disturbs this equilibrium because `NH_(3)` reacts with `H_(3)O^(+)` ions as `NH_(3)+H_(3)O^(+) hArr NH_(4)^(+)+H_(2)O` |
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| 1189. |
An aqeous of acid is characterised by the presence ofA. `H^(+)` ionsB. `H_(2)^(+)` ionsC. `H_(4)O^(+)` ionsD. `H_(3)O^(+)` ions. |
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Answer» Correct Answer - D As per definition, acid gives `H_(2)O^(+)` ions in aqueous solution |
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| 1190. |
The difference between `DeltaH` and `Delta E` at constant voluem is equal toA. `V Delta P`B. RC. P Delta VD. (3)/(2)R |
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Answer» Correct Answer - A `DeltaH=DeltaE+Delta(PV)`or `DeltaH=DeltaE+P DeltaV +V DeltaP` At constant volume `Delta V=0` `:. Delta H -Delta E =V Delta P` |
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| 1191. |
A chemical reaction with `Delta H gt0` carried out at constant temperature and pressure will necessarily be spontaneous ifA. `Delta G gt0`B. `Delta S lt0`C. `Delta H gt T Detla S`D. `Delta H lt T Delta S` |
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Answer» Correct Answer - D For a reaction to be spontaneous `Delta G` should be negative. It is possible only if `Delta H lt T Delta S ` . (Here `Delta S =+ve)` |
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| 1192. |
Which of the following reactions is `Delta H` less than `Delta E` ?A. `C_(6)H_(12)O_(6)(s) +6O_(2)(g) rarr 6CO_(2)(g) +6H_(2)O(l)`B. `N_(2)O_(4)(g) rarr 2NO_(2)(g)`C. `N_(2)(g)+O_(2)(g) rarr 2NO(g)`D. `2SO_(2)(g)+O_(2)(g) rarr 2SO_(3)(g)` |
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Answer» Correct Answer - D For reaction where `Delta n=-ve, Delta H ltDelta E`. |
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| 1193. |
In which of the following neutralisation reaction will the enthalpy of neutralisation be the smallest ?A. NaOH with HClB. HCl with `NH_(4)OH`C. NaOH with `CH_(3)COOH`D. `H_(3)PO_(4)` with NaOH |
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Answer» Correct Answer - B It is neutralisation of strong acid with weak base. |
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| 1194. |
`{:("Column I",,"Column II"),((A) Fe(OH)_(3),,(i)K_(sp)=s^(2)),((B) Ag_(2)CrO_(4) ,, (ii)K_(sp) =27s^(4)),((C) CH_(3)COOAg,,(iii) K_(sp) =108s^(5)),((D)Ca_(3)(PO_(4))_(2),,(iv) K_(sp) =4s^(3)):}`A. (A) `to` (iii) , (B) `to` (ii) , (C )`to` (iv) , (D) `to` (i)B. (A) `to` (ii) , (B) `to` (iv) , (C )`to` (i) , (D) `to` (iii)C. (A) `to` (i) , (B) `to` (iii) , (C )`to` (ii) , (D) `to` (iv)D. (A) `to` (iv) , (B) `to` (i) , (C )`to` (iii) , (D) `to` (ii) |
| Answer» Correct Answer - 2 | |
| 1195. |
The solubility product of `Ag_(2)CrO_(4)` is `32 xx 10^(-12)`. What is the concentration of `CrO_(4)^(-)` ions in that solutionA. `2 xx 10^(-4) M`B. `16 xx 10^(-4) M`C. `8 xx 10^(-4) M`D. `8 xx 10^(-8) M` |
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Answer» Correct Answer - A `underset(S)(Ag_(2)CrO_(4)) rarr underset(2S)(2Ag^(+)) + underset(S)(CrO_(4)^(--))` `K_(sp) = (2S)^(2)S = 4S^(3)` `S = ((K_(sp))/(4))^((1)/(3)) = ((32 xx 10^(-12))/(4))^((1)/(3)) = 2 xx 10^(-4) M`. |
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| 1196. |
During the neutralisation of an acid by a base, the end point refers to the completion of reaction. The detection of end point in acid-base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthalein) or a weak base (Methyl orange). At `50%` ionisation which depends on the medium. the anion furnished by an indicator (acid) cation furnished colour to solution at end point. For example, phenolphthalein dissociation `underset("Colourless")(HIn)toH^(+)+underset("Pink")(In^(-)),K_(HIn)=([H^(+)][In^(-)])/([HIn])` is favoured in presence of alkali and pink colour of phenolphthalein ion is noticed as soon as the medium changes to alkali nature. The end point of acid-base neutralisation not necessarily coincides with eqivalence point but it is closer to equivalence point. Also at equivalence point of acid-base neutralisation, pH is not necessarily equal to 7. Bromophenol blue is an acid having dissociation constant `5.48xx10^(-5)`. The percentage of coloured ion furnished at a ph of 4.84 is :A. `80%`B. `40%`C. `20%`D. `90%` |
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Answer» Correct Answer - a |
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| 1197. |
During the neutralisation of an acid by a base, the end point refers to the completion of reaction. The detection of end point in acid-base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthalein) or a weak base (Methyl orange). At `50%` ionisation which depends on the medium. the anion furnished by an indicator (acid) cation furnished colour to solution at end point. For example, phenolphthalein dissociation `underset("Colourless")(HIn)toH^(+)+underset("Pink")(In^(-)),K_(HIn)=([H^(+)][In^(-)])/([HIn])` is favoured in presence of alkali and pink colour of phenolphthalein ion is noticed as soon as the medium changes to alkali nature. The end point of acid-base neutralisation not necessarily coincides with eqivalence point but it is closer to equivalence point. Also at equivalence point of acid-base neutralisation, pH is not necessarily equal to 7. The indicator phenolphthalein is a tautomeric mixture of two forms as given below: Which of the following statements are correct? The form II is referred as quinonoid form and is deeper is colour. (Q)The form I is referred as quinonoid form and is light in colour. (R) The form II is more stable in alkaline medium. (S) The change is pH from acidic to alkaline solution bring in the more and more conversion of I form to II form. (T) The form I is more stable in acidic medium.A. P,Q,R and SB. P,R,S and TC. R,S and TD. Q,R,S and T |
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Answer» Correct Answer - b |
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| 1198. |
During the neutralisation of an acid by a base, the end point refers to the completion of reaction. The detection of end point in acid-base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthalein) or a weak base (Methyl orange). At `50%` ionisation which depends on the medium. the anion furnished by an indicator (acid) cation furnished colour to solution at end point. For example, phenolphthalein dissociation `underset("Colourless")(HIn)toH^(+)+underset("Pink")(In^(-)),K_(HIn)=([H^(+)][In^(-)])/([HIn])` is favoured in presence of alkali and pink colour of phenolphthalein ion is noticed as soon as the medium changes to alkali nature. The end point of acid-base neutralisation not necessarily coincides with eqivalence point but it is closer to equivalence point. Also at equivalence point of acid-base neutralisation, pH is not necessarily equal to 7. The dissociation constant on an acid-base indicator which furnishes coloured anion is `1xx10^(-5)`. The pH solution at which indicator will furnish its colour is :A. `5.2`B. `5.6`C. `8.4`D. `8.48` |
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Answer» Correct Answer - b |
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| 1199. |
During the neutralisation of an acid by a base, the end point refers to the completion of reaction. The detection of end point in acid-base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthalein) or a weak base (Methyl orange). At `50%` ionisation which depends on the medium. the anion furnished by an indicator (acid) cation furnished colour to solution at end point. For example, phenolphthalein dissociation `underset("Colourless")(HIn)toH^(+)+underset("Pink")(In^(-)),K_(HIn)=([H^(+)][In^(-)])/([HIn])` is favoured in presence of alkali and pink colour of phenolphthalein ion is noticed as soon as the medium changes to alkali nature. The end point of acid-base neutralisation not necessarily coincides with eqivalence point but it is closer to equivalence point. Also at equivalence point of acid-base neutralisation, pH is not necessarily equal to 7. The dissociation constant of an acid-base indicator which furnishes colorued cation is `1xx10^(-5)`. The pH of solution at which indicator will furnish its colour is :A. 5B. 9C. 6D. 10 |
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Answer» Correct Answer - b |
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| 1200. |
An aqueous solution whose pH = 0 isA. AlkalineB. AcidicC. NeutralD. Amphoteric |
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Answer» Correct Answer - B pH = 0 means `[H^(+)] = 10^(@) = 1M`. Hence solution is strongly acidic. |
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