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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.
| 651. |
The `p^(H)` of an aqueous solution of a salt is 10. the salt isA. `KCl`B. `NH_(4)NO_(3)`C. `NaCN`D. `(NH_(4))_(2)SO_(4)` |
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Answer» Correct Answer - C NaCN under to go anionic hydrolysis |
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| 652. |
The following equilibrium is established when hydrogen chloride is dissolved in acetic acid `HCl(aq)+CH_3COOH(aq)hArrCl^(-)(aq)+CH_3COOH_2^(+)(aq)` The set that characterises the conjugate acid-base pairs is :A. `(HCl, CH_(3)COOH)(CH_(3)COOH_(2)^(+),Cl^(-))`B. `(HCl,CH_(3)COOH_(2)^(+))` and `(CH_(3)COOH,Cl^(-))`C. `(CH_(3)COOH_(2)^(+),HCl)` and `(Cl^(-),CH_(3)COOH)`D. `(HCl,Cl^(-)) ` and `(CH_(3)COOH_(2)^(+),CH_(3)COOH)` |
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Answer» Correct Answer - D `underset("Acid"1)(HCl)+underset("Base" 2)(CH_(3)COOH ) hArr underset("Base 1")(Cl^(-))+underset("Acid 2")(CH_(3)COOH_(2)^(+))` |
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| 653. |
The following equilibrium is established when hydrogen chloride is dissolved in acetic acid `HCl(aq)+CH_3COOH(aq)hArrCl^(-)(aq)+CH_3COOH_2^(+)(aq)` The set that characterises the conjugate acid-base pairs is :A. `(HCl, CH_3COOH) and (CH_3COOH_2^(+), Cl^- )`B. `(HCl, CH_3COOH_2^+) and (CH_3COOH, Cl^-)`C. `(CH_2COOH_2^+,HCl) and (Cl^-,CH_3COOH)`D. `(HCl, Cl^-) and (CH_3COOH_2^(+), CH_3COOH)` |
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Answer» Correct Answer - D |
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| 654. |
A solution which is `10^(-3) M` each in `Mn^(2+), Fe^(2+), Zn^(3+)` and `Hg^(3)` is is treated with `10^(-6) M` sulphide ion. If `K_(sp)` of `MnS`, ZnS and `HgS` are `10^(-15)`, `10^(-23), 10^(-20)` and `10^(-54)` respectively. Which of the following will be precipitated first?A. `ZnS`B. `FeS`C. `MnS`D. `HgS` |
| Answer» Correct Answer - D | |
| 655. |
Which of the following acids has the smallest dissociation constant?A. `CH_(3)CHFCOOH`B. `FCH_(2)CH_(2)COOH`C. `B_(1)CH_(2)CH_(2)COOH`D. `CH_(3)CHBrCOOH` |
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Answer» Correct Answer - C Electorn-withdrawing groups increase the acidic strength. Fluorine is more electorn withdrawing than bromine. Also, the acid in which the halogen is attached farther from the acid group is weaker because -I effect decreases with distance. On this basic of this, `Br(CH_(2))_(2)COOH` is the weakest acid of the given choices. |
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| 656. |
The solublity product of iron (III) hydroxide is `1.6 xx 10^(-39)`. If X is the solublity of iron (III) hydroxide, then which one of the following expressions can be used to calculate XA. `K_(sp) = X^(4)`B. `K_(sp) = 9 X^(4)`C. `K_(sp) = 27 X^(3)`D. `K_(sp) = 27 X^(4)` |
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Answer» Correct Answer - D The solubility product `(K_(sp))` of a sparingly soluble salt at a given temperature may be defined as the product of the molar concentrations of its each raised to the power equal to its number of ions present in the equation representing the dissociation of one molecule of the salt. `{:(Fe(OH)_(3), hArr ,Fe^(3+)+,3OH^(-)),(,,X,3X):}` `K_(sp) = [Fe^(3+)][OH^(-)]^(3)` `:. K_(sp) = X(3X)^(3) = 27 X^(4)`. |
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| 657. |
The solublity of AgCl is formed when equal volumes of the following are mixed. [`K_(sp)` for `AgCl = 10^(-10)`]A. `2.0 xx 10^(-5) M`B. `1.0 xx 10^(-4) M`C. `5.0 xx 10^(-9) M`D. `2.2 xx 10^(-4) M` |
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Answer» Correct Answer - C `{:(,CaCl_(2),hArr,Ca^(2+)+,2Cl^(-)),("Concen.",0.04,,0.04,0.08):}` For AgCl `AgCl hArr Ag^(+) + Cl^(-)` `K_(sp) = [Ag^(+)][Cl^(-)]` `4.0 xx 10^(-10) = [S][S+0.08]` `4.0 xx 10^(-10) = S^(2) + S xx 0.08 [S^(2) lt lt lt 1]` `4.0 xx 10^(-10) = S xx 0.08` `S = (4.0 xx 10^(-10))/(0.08) = 0.5 xx 10^(-8) = 5 xx 10^(-9)`. |
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| 658. |
Which is incorrect for buffer solutionA. It contains weak acid and its conjugate baseB. It contains weak weak base and its conjugate acidC. In this there is very less change is pH value when very less amount of acid and base is mixedD. None of the above |
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Answer» Correct Answer - D All are true - (a) is true acid buffer, (b) for basic buffer, (c) is called buffer solution. |
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| 659. |
A precipitate of AgCl is formed when equal volumes of the following are mixed. [`K_(sp)` for `AgCl = 10^(-10)`]A. `10^(-4) M AgNO_(3)` and `10^(-7) M HCl`B. `10^(-5) M AgNO_(3)` and `10^(-6) M HCl`C. `10^(-5) M AgNO_(3)` and `10^(-4) M HCl`D. `10^(-6) M AgNO_(3)` and `10^(-6) M HCl` |
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Answer» Correct Answer - C For the precipitation of an electrolyte, it is necessary that the ionic product must exceed its solubility product. |
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| 660. |
Which of the following salt is acidicA. `Na_(2)SO_(4)`B. `NaHSO_(3)`C. `Na_(2)SO_(3)`D. `Na_(2)S` |
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Answer» Correct Answer - B It contains replacable H atom. |
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| 661. |
`Fe(OH)_(3)` can be separated from `Al(OH)_(3)` by the addition ofA. NaCl solutionB. Dil. HCl solutionC. NaOH solutionD. `NH_(4)Cl` and `NH_(4)OH` |
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Answer» Correct Answer - D Due to common ion effect of `NH_(4)^(+)`, concentration of `OH^(-)` decreases, as `K_(sp)Fe(OH)_(3) lt K_(sp)Al(OH)_(3)` thus `Fe(OH)_(3)` get precipitated first. |
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| 662. |
In a neutral solutionA. `H_(3)O^(+)` ions are not presentB. `OH^(-)` ions are not presentC. Both `H_(3)O^(+)` ions and `OH^(-)` ions are not presentD. Both `H_(3)O^(+)` ions and `OH^(-)` ions are present in small but equal concentration. |
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Answer» Correct Answer - D In neutral solution `[H^(+)]=[OH^(+)]` `=1xx10^(-7)` mole `L^(-1)` at `298 K` |
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| 663. |
Which is least soluble in waterA. AgClB. AgFC. AgID. `Ag_(2)S` |
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Answer» Correct Answer - D It is a less ionic, so that least soluble in water. |
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| 664. |
Given that the dissociation constant for `H_(2)O` is `K_(w) = 1 xx 10^(-14) "mole"^(2)//"litre"^(2)`. What is the pH of a 0.001 molar KOH solutionA. `10^(-11)`B. `10^(-3)`C. 3D. 11 |
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Answer» Correct Answer - D pH = 14 - pOH = 14 - 3 = 11. |
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| 665. |
By adding 20 ml 0.1 N HCL to 20 ml 0.001 N KOH, the pH of the obtained solution will beA. 2B. 1.3C. 0D. 7 |
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Answer» Correct Answer - B 20 ml. of 0.1N HCl `= (0.1)/(1000) xx 20 eq. = 2 xx 10^(-3) g eq`. 20 ml. of 0.001 KOH `= (0.001)/(1000) xx 20` gm eq. `= 2 xx 10^(-5)` g eq. `:.` HCl left unneutralised `= 2(10^(-3) - 10^(-5))` `= 2 xx 10^(-3) (1- 0.01) = 2 xx 0.99 xx 10^(-3) = 1.98 xx 10^(-3)` g eq. Volume of solution = 40 ml. `:. [HCl] = (1.98 xx 10^(-3))/(40) xx 1000 M = 4.95 xx 10^(-2)`. `:. pH = 2 - log 4.95 = 2 - 0.7 = 1.3`. |
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| 666. |
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)Cl` |
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Answer» Correct Answer - D `NH_(4)OH//NH_(4)Cl` is basic buffer and has pH `gt7` |
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| 667. |
Initially, 0.8 mole of `PCl_(5)` and 0.2 mol of `PCl_(30` are mixed in one litre vessel. At equilibrium, 0.4 mol of `PCl_(3)` is present. The value of `K_(c )` for the reaction `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` would beA. `0.1 mol L^(-1)`B. `0.05 mol L^(-1)`C. `0.013 molL^(-1)`D. ` 0.66 mol L^(-1)` |
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Answer» Correct Answer - A `K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.4 xx 0.2 )/(0.6)=0.13 mol L^(-1)` |
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| 668. |
The pH of a 0.001 M NaOH will beA. 3B. 2C. 11D. 12 |
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Answer» Correct Answer - C 0.001 M of NaOH means `[OH^(-)] = 0.001` `= 10^(-3) M rArr pOH = 3` `pH + pOH = 14 rArr pH = 14-3 = 11`. |
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| 669. |
Amongst the following solutions, the buffer solution is Or A basic buffer is made by mixing the solutionA. `NH_(4)Cl+NH_(4)OH` solutionB. `NH_(4)Cl + NaOH` solutionC. `NH_(4)OH + HCl` solutionD. `NaOH + HCl` solution |
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Answer» Correct Answer - a `NH_(4)Cl` and `NH_(4)OH` is a buffer solution (weak base and salt of strong acid). |
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| 670. |
If the dissociation constants of two weak acids `HA_(1)` and `HA_(2)` are `K_(1)` and `K_(2)`, then the relative strengths of `HA_(1)` and `HA_(2)` are given byA. `(K_(1))/(K_(2))`B. `(K_(1)+sqrt(K_(2)))/(K_(2)+sqrt(K_(1)))`C. `((K_(1))/(K_(2)))^(1//2)`D. `((K_(1))/(K_(2)))^(3//2)` |
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Answer» Correct Answer - C The dissociation of a weak acid is related to its `K_(a)` value as `{:(,HA,hArr ,H^(+),+,A^(-)),("Initial conc. (M)",1.0,,0,,0),("Eqm. conc.",C(1-alpha),,Calpha,,Calpha):}` or `K_(a)=(C alpha.C alpha)/(C(1-alpha))=(C alpha^(2))/((1-alpha))` If `a lt lt 1` and `C=1` mol then `K_(a)=alpha^(2)` or `alpha=(K_(a))^(1//2)` Ratio of relative strengths of two acids depend on depends of dissociation i.e., `alpha` `:. (alpha_(1))/(alpha_(2))=((K_(1))/(K_(2)))^(1//2)` |
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| 671. |
At any temperature , the proton concentration of water isA. `10^(-14) M`B. `Kw`C. `gt 10^(-7) M`D. `sqrt(K_(w))` |
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Answer» Correct Answer - D `K_(w) = [H^(+)] [OH^(-)]` in water `[H^(+)] = [OH^(-)]` `:. [H^(+)] = sqrt(K_(q))` |
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| 672. |
Neutralisation constant of `HCOOH` with a strong bse is `10^(8)`. What is the `pH` of `0.01 M HCOOK` solution at `25^(0)C`? |
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Answer» Correct Answer - 9 `HCOOH + KOH hArr HCOOK + H_(2)O` `K_("neu") = (1)/(K_(h)) = (K_(a))/(K_(w))` `K_(a) = 10^(8) xx 10^(-14) = 10^(-6) , pH = 7+3-1 = 9` |
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| 673. |
Three bases `XOH,YOH & ZOH` has `pK_(b)` values `2,3, & 4` reespectively the strongest conjugate acid is `:`A. `XOH_(2)^(+)`B. `YOH_(2)^(+)`C. `ZOH_(2)^(+)`D. All same |
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Answer» Correct Answer - 3 `XOH,YOH & ZOH pK_(b)=2,3 & 4 ` respectively Basic strength will be `XOH gt YOHgtZOH` hence strength of conjugate acid will be. |
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| 674. |
One `dm^(3)` solution containing `10^(-5)` moles each of `Cl^(-)` ions and `CrO_(4)^(2-)` ions is treated with `10^(-4)` moles of silver nitrate. Which one of the following observation is made. `[K_(sp) Ag_(2)CrO_(4) = 4 xx 10^(-12)] [K_(sp)AgCl = 1 xx 10^(-10)]`A. Precipitation does not occurB. Silver chromate gets precipitated firstC. Silver chloride gets precipitated firstD. Both silver chromate and silver chloride start precipitation simultaneously |
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Answer» Correct Answer - C Ionic product of `AgCl = [10^(-4)][10^(-5)] = 10^(-9)` Ionic product of `Ag_(2)CrO_(4) = [10^(-8)][10^(-5)] = 10^(-13)` Ionic product of `AgCl gt K_(sp)` of AgCl So, precipitation will take place. Ionic product of `Ag_(2)CrO_(4) lt K_(sp)` of `Ag_(2)CrO_(4)` So, precipitation will not take place. |
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| 675. |
What name is given to the reaction between hydrogen ion and hydroxyl ionA. HydrogenenationB. HydroxylationC. HydrolysisD. Neutralization |
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Answer» Correct Answer - D `H^(+) + OH^(-) hArr H_(2)O`, value than stronger the acid. |
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| 676. |
Which will not affect the degree of ionizationA. TemperatureB. ConcentrationC. Type of solventD. Current |
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Answer» Correct Answer - D Current does not effect the degree of ionization. |
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| 677. |
Neutralisation constant of `HCOOH` with a strong base is `10^(8)`. What is the pH of `0.01 M` HCOOK solution? At `25^(0)C` |
| Answer» Correct Answer - 9 | |
| 678. |
The values of dissociation constants of some acids (at `25^(@)C`) are as follows. Indicate which is the strongest acid in waterA. `1.4 xx 10^(-2)`B. `1.6 xx 10^(-4)`C. `4.4 xx 10^(-10)`D. `4.3 xx 10^(-7)` |
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Answer» Correct Answer - A More is `K_(a)`, lesser is `pK_(a)(pK_(a) = - log K_(a))` more is acidic strength. |
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| 679. |
Calculate the concentration of hydrogen ion in the acidic solution with `pH` a. `4.3` b. `5.8239` c. `3.155` |
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Answer» a. `pH =- log [H_(3)O^(o+)]` `:. Log [H_(3)O^(o+)] =0 4.3 =- 4 - 0.3 +1 -1` `= bar(5).7` `:. [H_(3)O^(o+)] = "Antilog" (bar(5).7)` `= 5 xx 10^(-5) N = 0.5 xx 10^(-4) N` b. `pH =- log [H_(3)O^(o+)] = 5.8239` `:. log [H_(3)O^(o+)] =- 5.8239 =- 5 - 0.8239 +1 - 1` `= bar(6).1761` `:. [H_(3)O^(o+)] = "Antilog" (bar(6).1761)` `= 1.5 xx 10^(-6)N` c. `pH =- log [H_(3)O^(o+)] = 3.155` `:. log [H_(3)O^(o+)] =- 3.155 =- 3 - 0.155 +1 - 1` `= bar(4).845` `:. [H_(3)O^(o+)] = "Antilog" (bar(4).845)` `= 7 xx 10^(-4) = 0.7 xx 10^(-3) N` |
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| 680. |
Calculate `pH` of a. `0.002 N CH_(3) COOH` having `2.3%` dissociation. b. `0.002N NH_(4) OH` having `2.3%` dissociation. |
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Answer» a. `0.002 N CH_(3)COOH`: Acetic acid is weak electrlyte and partially associated. `{:(,CH_(3)COOHhArr,CH_(3)COO^(Theta)+,H^(o+),),("Conc before dissocitation",1,0,0,),("Conc after dissociation",1-alpha,alpha,alpha,):}` `:. [H^(o+)] = C alpha = 2 xx 10^(-3) xx (2.3)/(100) = 4.6 xx 10^(-5) M` `:. pH = -log[H^(oplus)] = - log(4.6 xx 10^(5))` `pH = 4.3372` b. `0.002 N NH_(4)OH: NH_(4)OH` is weak base and partially dissociated. `{:(,NH_(4)OHhArr,NH_(4)^(Theta)+,overset(Theta)OH,),("Conc before dissocitation",1,0,0,),("Conc after dissociation",1-alpha,alpha,alpha,):}` `:. [overset(Theta)OH] = C alpha = 2 xx 10^(-3) xx (23)/(100) = 4.6 xx 10^(-5) M` `because pOH = log [overset(Theta)OH] =- log (4.6 xx 10^(-5))` `:. pOH =- 4.3372 :. pH = 14 - 4.3372` `pH = 9.6628` |
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| 681. |
What is the `pH` of the following solutions: a. `10^(-8)M HCI` b. `5 xx 10^(-8) M HCI` c. `5xx10^(-10)M HCI` d. `10^(2) M HCI` |
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Answer» `pH` of `10^(-8) M HCI` First method: `[H_(3)O^(o+)] = 10^(-8)` (from HCI) `= 10^(-7)` (from `H_(2)O)` `= 10^(-7) (10^(-1) +1) = 1.1 xx 10^(-7)` `pH =- log (1.1 xx 10^(-7))` `=- log 1.1 - log 10^(-7)` `=- 0.414 +7 = 6.9586` Second method: Let `x = [oversetΘ)OH] = [H_(3)O^(o+)]` from `H_(2)O`. The `[H_(3)O^(o+)]_("total")` is generated from the ionisation of `HCI` dissolved and from ionisation of `H_(2)O`. `:. K_(w) = (10^(-8)+x) (x) = 10^(-14)` or`x^(2) +10^(-8)x - 10^(-14) = 0` `:. x=(-10^(-8)+sqrt((10^(-8))_(2)+(4xx10^(-14))))/(2)` `[overset(Θ)OH] = x = 9.5 xx 10^(-8)M` So, `pOH = 7.02` and `pH = 6.98` b. `pH` of `5xx10^(-8)M HCI` : If the contribution of `[H_(3)O^(o+)]` from `HCI` is considered, i.e., `[H_(3)O^(o+)] = 5xx10^(-8)M`. Then `pH` would be `gt7`. This is not possible because `[H_(3)O^(o+)]` of nay acid solution, no matter how dilute it is, cannot be less than that of pure water alone. in such cases, the concentration of `[H_(3)O^(o+)]` made by water has to be taken. Let `[H_(3)O^(o+)]` from water be `x M` in the presence of `5 xx 10^(-8) M HCI`. `{:(,2H_(2)O,hArr,H_(3)O^(oplus),+,[overset(Theta)(OH)]),(,,,xM,,xM),(HCl+,H_(2)O,hArr,H_(3)O^(oplus),+,Cl^(Theta)),(,,,5xx10^(-8)M,,5 xx10^(-8)M):}` At equilibrium, `[H_(3)O^(o+)] = (x +5 xx 10^(-8)), [overset(Θ)OH] = x`. `K_(w) = [H_(3)O^(o+) [overset(Θ)OH] = (x +5 xx 10^(-5)) (x) = 10^(-14)` `x = 0.78 xx 10^(-7) M` `:. [H_(3)O^(o+)] = (5 xx 10^(-8) +0.78 xx 10^(-7))` `= 1.28 xx 10^(-7)` `pH =- log (1.28 xx 10^(-7)) =- (0.11-7) =6.89` c. `pH` of `5 xx 10^(-10) M HCI`: `HCI` is so diulte that its contribution to `[H_(3)O^(o+)]` is negligible as compared with the ionisation of water. Thus, `[H^(o+)] = 10^(-7)` and therefore, `pH = 7.00` d. `pH` of `10^(-2)M HCI:` `{:("Initial concentration",HCIrarr+H_(2)Orarr,H_(2)Orarr+,CI^(Θ),),("Concentration after",10^(2),0,0,),("dissociation",0,10^(2),10^(2),):}` `pH =- log (10^(-2)) =- 2` A nagative `pH` only means that the `[H^(o+)] gt 1M`. However, in actual practive, a negative `pH` is uncommon. Firstly, even strong acids (say `100% H_(2)SO_(4))` become partially dissociated ay hight concentration. According to Sorenson, `pH` is related to thermodynamic activites rather than `[H^(o+)]`, i.e., on `a_(H)^(o+) = [H^(o+)] f_(H^(o+))`. In dilute solution activity coeffiecient, `f_(H^(o+))` is near enough to unity and thus, `a_(H^(o+)) = [H^(o+)]`. At high concentrations, the activity coefficient is less than unity. Thus, `pH` denfined by `-log [H^(o+)]`, which is not only of littele theoretical significance, but it in fact cannot be measured directly. Therefore, `pH` is redefined as : `pH =- log_(10) a_(H^(o+))` (This is what `apH` meter reading is a meausre of) i.e., `pH` of `10^(2) M HCI` cannot be calculated untiall `f_(H^(o+)` is known. Nevertheless, there is mathematically no basis for not having a negative `pH`. |
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| 682. |
Calculate the `pH` of the following solutions: a `10^(-2) M HCI` b `10^(-3) M H_(2)SO_(4)` c `0.2 xx 10^(-2) M NaOH` d `0.3 xx 10^(-3) M Ca (OH)_(2)` |
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Answer» Since `HCI` is monobasic strong acid, and is completely ionised a. `pH = - log (10^(-2)) = 2` b. Since `H_(2)SO_(4)` is dibasic strong (`n` factor `= 2`). Therefore concentration of `H^(o+)` ions `= 2 xx 10^(-3)N` `{:(,H_(2)SO_(4)rarr,2H^(o+)+,SO_(4)^(2-),),("Initial",10^(-3)M,0,0,),("Final",0,2xx10^(-3)M,10^(-3)M,):}` `pH =- log (2xx 10^(-3)) =- log2 - log (10^(-3))` `=- 0.3 =3 = 2.7` c. First method: Since `NaOH` is monoacidic strong base, so `[overset(Θ)OH] = 0.2 xx 10^(-2)M = 2xx 10^(-3)`. So, first calculate `pOH` of `NaOH` and then calculate `pH` accordingly. `pOH =- log [overset(Θ)OH] =- log(2) - log (10^(-3))` `=- 0.3 +3 = 2.7` `pH = 14 - poH = 14 - 2.7 = 11.3` Second method: First calculate `[H_(3)O^(o+)]` and then calculate `pH` accordingly. `K_(w) = [H_(3)O^(o+)] [overset(Θ)OH]` `:. [H_(3)O^(o+)] = (K_(w))/(overset(Θ)([OH])) = (10^(-14))/(2xx10^(-3))` `= 0.5 xx 10^(-11) = 5 xx 10^(-12)` `:. pH =- log 5 - log (10^(-12)) =- 0.7 +12 = 11.3` d. Since `Ca(OH)_(2)` is diacidic strong base, thus `[overset(Θ)OH] = 2xx 0.3 xx 10^(-3) N (n` factor `= 2`) or `{:(,Ca(OH)_(2)rarr,Ca^(2+)+,2oversetΘOH,),(Initial,0.3xx10^(-3)M,0,0,),("Final",0,0.3xx10^(-3),2xx0.3xx10^(-3)M,):}` `:. pOH =- log (2xx 0.3 xx 10^(-3)N)` `=- log (6xx 10^(-4)N)` `=- log (3xx2) - log (10^(-4))` `=- log 3 - log 2+4` `=- 0.48 - 0.3 +4 = 3.22` Therefore, `pH = 14 - 3.22 = 10.78` |
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| 683. |
`K_(b)` for `NH_(4)OH` is `1.8 xx 10^(-5)`. The `[overset(Theta)OH]` of `0.1 M NH_(4)OH` isA. `5.0 xx 10^(-2)`B. `4.20 xx 10^(-3)`C. `1.34 xx 10^(-3)`D. `1.8 xx 10^(-6)` |
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Answer» Correct Answer - C `K_(b) = 1.8 xx 10^(-5)` `pK_(b) = 4.7447` `pOH_(W_(B)) = (1)/(2) (pK_(b) - log 0.1)` `= (1)/(2) (4.7447 +1) = 2.87`. `[overset(Theta)OH] = "Antilog" (-2.87)` `= "Antilog" (-2-0.87 + 1-1)` `= "Antilog" (bar(3).13)` `= 1.34 xx 10^(-3)` |
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| 684. |
Calculate the `[overset(Theta)OH]` of `[NH_(2)C_(2)H_(4)NH_(3)]^(o+)` and `[H_(3)N-C_(2)H_(4)NH_(3)]^(2+)` in `0.15M` ethylene diamine (aq) if `NH_(2)C_(2)H_(4)NH_(2)+H_(2)O hArr NH_(2)C_(2)H_(4)overset(o+)NH_(3)+overset(Theta)OH (K_(1) = 8.5 xx 10^(-5))` `NH_(2)C_(2)H_(4)overset(o+)NH_(3) +H_(2)O hArr [NH_(3)C_(2)H_(4)NH_(3)]^(2+) + overset(Theta)OH (K_(2) = 2.7 xx 10^(-8))` |
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Answer» Case I: `[OH^(Theta)] = [NH_(2)C_(2)H_(4)NH_(3)]^(o+)` `=C xx sqrt((K_(b_(1)))/(c)) = sqrt(0.15 xx 8.5 xx 10^(-5)) = 3.57 xx 10^(-3)M` Case II: `{:(NH_(2)C_(2)H_(4)NH_(3)^(o+) +H_(2)O hArr,NH_(3)C_(2)H_(4)NH_(3)^(2+)+,OH^(Theta),,),(3.57 xx 10^(-3),0,1.275xx10^(-5),,),([3.57xx10^(-3)-X],X,(X+1.275xx10^(-5)),,):}` `2.7 xx 10^(-8) = (x.(X+3.57xx10^(-3)))/((3.57xx10^(-3)-X))` Neglecting `X^(2)`, also `3.57 xx 10^(-3) - X = 3.57 xx 10^(-3)` , (X s very small) `2.7 xx 10^(-8) = (3.57 xx 10^(-3)X)/(3.57 xx 10^(-3))` `:.X = 2.7 xx 10^(-8)` `:. [NH_(2)C_(2)H_(4)NH_(3)]^(2+) = 2.7 xx 10^(-8)M = K_(b_(2))` |
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| 685. |
Solubility of salt `A_(2)B_(3)` is `1 xx 10^(-4)`, its solubility product isA. `1.08 xx 10^(20)`B. `1.08 xx 10^(18)`C. `2.6 xx10^(-18)`D. `8 xx 10^(-15)` |
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Answer» Correct Answer - D `x=1x10^(-4)mol L^(-1)` `:. K_(sp)=(2xx10^(-4))(3xx10^(-4))^(3)` `=1.08 xx 10^(-18)` |
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| 686. |
The reverse process of neutralisation is :A. hydrolysisB. decompositionC. dehydrationD. synthesis |
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Answer» Correct Answer - a |
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| 687. |
The value of `K_(sp)` is `HgCI_(2)` at room temperature is `4.0 xx 10^(-15)`. The concentration of `CI^(Theta)` ion in its aqueous solution at saturation point isA. `1 xx 10^(-5)`B. `2xx10^(-5)`C. `2xx10^(-15)`D. `8xx10^(-15)` |
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Answer» Correct Answer - B `{:(HgCI_(2)hArr,Hg^(2+)+,2CI^(Theta),),(,2S,2S,):}` `K_(sp) = 4S^(3)`. `S = root3((K_(sp))/(4)) =root3((4xx10^(-15))/(4))10^(-5)` `:. 2S = 2 xx 10^(-5) = [CI^(Theta)]` |
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| 688. |
The value of `K_(sp)` is `HgCI_(2)` at room temperature is `4.0 xx 10^(-15)`. The concentration of `CI^(Theta)` ion in its aqueous solution at saturation point isA. `1xx10^(-5)`B. `2xx10^(-5)`C. `2xx10^(-15)`D. `8xx10^(-15)` |
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Answer» Correct Answer - B `HgCl_(2) rarr Hg^(2+) + 2Cl^(-)` If solubility is X mol `L^(-1)` then `[Cl^(-)]=2X` Now `4X^(3)=K_(sp)` `X=[(K_(sp))/(4)]^(1//3)=[(4xx10^(-15))/(4)]^(1//2)=1xx10^(-5)mol L^(-1)` or `2X =2 xx 10^(-5) mol L^(-1)` |
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| 689. |
Which of the following are true for an acid- base titration?A. Indicators catalyse the acid base reactions by releasing or accepting `H^(+)` ions.B. Indicators do not significantly affect the pH of the solution ti which they are added.C. Acid--base reactions do not occur in absence of indicators.D. Indicators have different colours in dissciated and undissociated forms. |
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Answer» Correct Answer - b,d |
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| 690. |
The solubility of `Sb_(2)S_(3)` in water is `1.0 xx 10^(-5)` mole/litre at 298 K. What will be its solubility productA. `108 xx 10^(-25)`B. `1.0 xx 10^(-25)`C. `144 xx 10^(-25)`D. `126 xx 10^(-24)` |
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Answer» Correct Answer - A `{:(Sb_(2) rarr,2Sb^(+2)+,3S^(-),,K_(sp) = (2x)^(2).(3x)^(3),),(,2x,3x,,):}` `K_(sp) = 108x^(5), K_(sp) = 108 xx (1 xx 10^(-5))^(5) = 108 xx 10^(-25)`. |
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| 691. |
a. At what `pH` does indicator change colour if the indicator is a weak acid with `K_(Ind) = 4.0 xx 10^(-4)`. b. For which of the following neutralisation would the indicator be useful? i. `HCl +NaOH` ii. `CH_(3)COOH +NaOH` iii. `HCl +NH_(3)` c. Name the indicators which can be used for such titration. |
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Answer» Correct Answer - A::B::C::D i. The mid-point of the colour change range of an indicator is the point at which its acid and conjugate base forms are present in equal concnetration. `pH = pK_(Ind) =- (log K_(Ind))` `=- log (4 xx 10^(-4))` `= - log 2^(2) + 4` `=- 0.6 + 4 = 3.4` ii. `pH` range of the indicator is `3.4` and is suitable for the titration of `S_(A)//W_(B).(HC1 + NH_(3))`. iii. The `pH` at the end point lies between `3.0` and `6.0`. For such titrations, methy`1` ornage `(pH` range `3.1 - 4.5)`, methy`1` red `(pH` range `4.2 - 6.3)`, and bromophenl blue `(3.1 - 4.6)` are thus suitable indicators. |
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| 692. |
The acid from of an acid base indicator is yellow in acid and red in basic from. What is the change in `pH` in order to change the indicator form `80%` yellow to `80%` red. |
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Answer» Correct Answer - A::B `{:(,HInhArr,H^(o+)+,Ind^(Theta),),(,"Acid form",,"Base form",),("First case:",80%"yellow",,20%red,),("Second case:",20%yellow,,80%red,):}` First case: a. `pH_(1) = pK_(a) + "log" ([In^(Theta)])/(["Hin"])` `= pK_(a) + log ((20)/(80)) …(i)` ` = pK_(a) + log 1//4` Second case: b. `pH_(2) = pK_(a) + log ((80)/(20))` `= pk_(a) + log 4 ...(ii)` From equations (i) and (ii), we get `pH_(2) - pH_(1) = 2 log 4 = 4 log 2 = 4 xx 0.3 = 1.2` |
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| 693. |
The no.of hydroxyl ions produced by one molecule of `Na_(2)CO_(3)` on hydrolysis isA. `4`B. `2`C. `3`D. `0` |
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Answer» Correct Answer - B `Na_(2)CO_(3) + 2H_(2)O rarr H_(2)CO_(3) + 2NaOH` |
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| 694. |
The correct order of increasing `[H_(3)O^(o+)]` in the following aqueous solution isA. 0.01 M `H_(2) S lt` 0.01 M `H_(2)SO_(4) lt` 0.01 M NaCl `lt` 0.01 M `NaNO_(2)`B. 0.01 M NaCl `lt` 0.01 M `Na NO_(2) lt` 0.01 M `H_(2)S lt` 0.01 `MH_(2)SO_(4)`C. 0.01 M `NaNO_(2) lt` 0.01 M NaCl `lt` 0.01 M `H_(2)S lt ` 0.01 M `H_(2)SO_(4)`D. None |
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Answer» Correct Answer - C This is because the acidic nature of aqueous solution increases in the order `NaNO_(2) ltNaCl lt H_(2)S ltH_(2)SO_(4)` |
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| 695. |
Statement: `CCI_(4),C_(6)H_(6)` and liquid `SO_(2)` are aprotic solvents. Explanation: Aprotic solvents does not influence the aicdic or basic nature of solute.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 Both are facts. |
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| 696. |
The solubility of sodium hydroxide increases with increase of temperature. |
| Answer» The solubility of sodium hydroxide decreases with increase in temperature. | |
| 697. |
If `2H_(2)O_((g))hArr2H_(2(g))+O_(2(g)), K_(1)=2.0xx10^(-13)` `2CO_(2(g))hArr2CO_((g))+O_(2(g)), K_(2)=7.2xx10^(-12)` Find the equilibrium constant for the reaction `CO_(2(g))+H_(2(g))hArrCO_((g))+H_(2)O_((g))` |
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Answer» Correct Answer - 6 `K_(c)=sqrt((K_(2))/(K_(1)))=sqrt((7.2xx10^(-12))/(2.0xx10^(-13)))=sqrt(36)=6` |
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| 698. |
If, `H_(2)(g)+Cl_(2)(g)rarr 2HCl(g) , Delta H^(@)=-44` Kcal `2Na(s)+2HCl(g)rarr 2NaCl(s)+H_(2)(g), Delta H=-152` Kcal Then, `Na(s)+0.5Cl_(2)(g)rarr NaCl(s) , Delta H^(@) = ?`A. `-108 kcal`B. `-196kcal`C. `-98 kcal`D. `-54 kcal`. |
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Answer» Correct Answer - C Multiply eq. (i) by `1//2` and also `(ii)` by `1//2` and add the two reactions. |
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| 699. |
In a titration `20 cm^(3)` of 0.1 N oxalic acid solution requires `20 cm^(3)` of sodium hydroxide for complete neutralization. The mass of sodium hydroxide in `250 cm^(3)` solution isA. 12.5 gB. 1.25 gC. 0.125 gD. 125 g |
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Answer» Correct Answer - B Now `underset("Acid")(N_(1)V_(1))=underset("Base")(N_(2)V_(2))` `N xx 25=.1 xx 25` or `N_(1)=(0.1 xx 25)/(25)=0.1 N` Now `N=("Mass of solute")/("Eq mass")xx(1000)/("Volume in mL")` `0.1 =("Mass of solute")/(63)xx(1000)/(200)` `:. ` Mass of solute`=(63 xx0.1)/(5)=1.25g` |
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| 700. |
If solutbility of `Ca(IO_(3))_(2)` in water at `20^(@)C` is `3.9gL^(-1)`. Calculate the `K_(sp)`. Given `Mw Ca(IO_(3))_(2) = 390`. |
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Answer» Correct Answer - A::C::D `M = (3.9)/(390) = 0.01M = S` `{:(Ca (IO_(3))_(2) hArr,Ca^(2+)+,2IO_(3)^(Theta),,),(,S,2S,,):}` `K_(sp) = 4S^(3) = 4 xx (0.01)^(3) = 4 xx 10^(-6) M^(3)` |
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