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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.
| 851. |
Which of the following is a conjugate acid-base pairA. HCl, NaOHB. `NH_(4)Cl, NH_(4)OH`C. `H_(2)SO_(4), HSO_(4)^(-)`D. KCN, HCN |
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Answer» Correct Answer - C `H_(2)SO_(4)hArr H^(+) + HSO_(4)^(-)`. |
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| 852. |
When 100 ml of 1M NaOH solution is mixed with 10 ml of 10 M `H_(2)SO_(4)`, the resulting mixture will beA. AcidicB. AlkalineC. NeutralD. Strongly alkaline |
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Answer» Correct Answer - A `N_(NaOH) = 1 xx 1 = 1N` `N_(H_(2)SO_(4)) = 2 xx 10 = 20 N` M. eq. of `NaOH = 1 xx 100 = 100` M.eq. of `H_(2)SO_(4) = 20 xx 10 = 200` Thus M.eq. of acid are left and therefore `pH lt 7`, so the resulting mixture will be acidic. |
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| 853. |
Which of the following species is an acid and also a conjugate base of another acidA. `HSO_(4)^(-)`B. `H_(2)SO_(4)`C. `OH^(-)`D. `H_(3)O^(+)` |
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Answer» Correct Answer - A `H_(2)SO_(4) rarr H^(+) + HSO_(4)^(-) rarr H^(+) + SO_(4)^(--)` acid `H_(2)SO_(4)`. |
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| 854. |
Conjugate base of HBr isA. `H_(2)Br^(+)`B. `H^(+)`C. `Br^(-)`D. `Br^(-)` |
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Answer» Correct Answer - C `HBr + H_(2)O hArr H_(3)O^(+) + Br^(-)` |
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| 855. |
An aqueous solution of aluminium sulphate showsA. A basic natureB. An acidic natureC. A neutral natureD. Both acidic and basic nature |
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Answer» Correct Answer - B `Al_(2)(SO_(4))_(3) hArr 2Al^(3+) + 3 SO_(4)^(2-)` `underset("Weak base")(Al(OH)_(3)) + underset("Strong acid")(H_(2)SO_(4))`. |
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| 856. |
`Cl^(-)` is the conjugate base ofA. `HClO_(4)`B. HClC. HOClD. `HClO_(3)` |
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Answer» Correct Answer - B Acid `overset(-H^(+))rarr` conjugate base. Base `overset(+H^(+))rarr` conjugate acid. |
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| 857. |
The strongest Lewis base in the followingA. `CH_(3)^(-)`B. `F^(-)`C. `NH_(2)^(-)`D. `OH^(-)` |
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Answer» Correct Answer - A `CH_(4)` has almost no acidic nature and thus `CH_(3)^(-)` is strongest base. |
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| 858. |
Which of the following can give base `OH^(-)`A. `H_(2)O`B. `H_(3)O^(+)`C. `H_(2)`D. HCl |
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Answer» Correct Answer - A `H_(2)O + H_(2)O hArr H_(3)O^(+) + OH^(-)`. |
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| 859. |
10 mL of a strong acid solution of `pH=2.000` are mixed with 990mL of another strong acid solution of `pH=4.000`. The pH of the resulting solution will be:A. `4.002`B. `4.000`C. `4.200`D. `3.72` |
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Answer» Correct Answer - D `[H^(+)]` after mixing `= (10^(-2)xx10+10^(-4)xx990)/(1000)` `= (0.1+0.0990)/(1000)` `= (0.1990)/(1000)=1.9xx10^(-4)`, `:. pH=-4-0.28=3.72` |
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| 860. |
The pH of a buffer containing equal molar concentration of a weak base and its chloride `(K_(b)` weak base `=2 xx 10^(-5) , log 2 =0.3 )` isA. 5B. 9C. 4.7D. 9.3 |
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Answer» Correct Answer - D `pOH =pK_(b)=-log (2 xx 10^(-5))=5-0.36 =4.70` Hence `pH=14-4.7 =9.3` |
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| 861. |
`HOCl(aq)toH^(+)(aq)+OCl^(-)(aq)` The ionization of hypochlorous acid represented above has `K=3.0xx10^(-8)` at `25^@C` .What is K for this reaction ? `Ocl^(-)(aq)+H_2O(l)hArr HOCl(aq)+OH^(-)(aq)`A. `3.3xx10^(-7)`B. `3.0xx10^(-8)`C. `3.0xx10^(6)`D. `3.3xx10^(7)` |
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Answer» Correct Answer - a |
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| 862. |
Which of the following are not Lewis acids ?A. `Ag^(+)`B. `ZnCl_(2)`C. `H_(2)O`D. ` Ch_(3)OH` |
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Answer» Correct Answer - C,D `H_(2)O` and `CH_(3)OH` are Lewis bases as they have Ione paires of electrons on O atom |
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| 863. |
The compound that is not a Lewis acids isA. `BF_(3)`B. `AICI_(3)`C. `BeCI_(2)`D. `SnCI_(4)` |
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Answer» Correct Answer - D Only is `SnCI_(4)`, the octet of the central atom is complete. |
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| 864. |
The correct order of acid strength isA. `HClOltHCIO_(2)ltHClO_(3)ltHClO_(4)`B. `HClO_(4)ltHClO_(3)ltHClO_(2)ltHClO`C. `HClOltHClO_(4)ltHClO_(3)ltHClO_(2)`D. `HClO_(4)ltHClO_(2)ltHClO_(3)ltHClO` |
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Answer» Correct Answer - A Higher is the oxidation number of central non-metal, more is acidic nature. |
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| 865. |
The conjugate acid of amide ion `(NH_(2)^(-))` isA. `NH_(3)`B. `NH_(2)OH`C. `NH_(4)^(o+)`D. `N_(2)H_(4)` |
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Answer» Correct Answer - A Base `+H^(o+) rarr` Conjugate acid `NH_(2)^(Theta) + H^(o+) rarr NH_(3)` |
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| 866. |
`CHCl_(3)` does not give white ppt. with `AgNO_(3)` because it:A. is a covalent compoundB. does not give `CI^(-)` ions in solutionC. is not dissociated in waterD. None of these |
| Answer» Correct Answer - A::B::C | |
| 867. |
Which is the weakest acid ?A. ascorbic acid `(K_a=8.0xx10^(-5))`B. boric acid `(K_a=5.8xx10^(-10))`C. butyric acid `(K_a=1.5xx10^(-5))`D. hydrocyanic acid `(K_a=4.9xx10^(-10))` |
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Answer» Correct Answer - d |
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| 868. |
A solution of `0.1 M NaZ` has `PH = 8.90`. The `K_a` of `HZ` is.A. `1.6xx10^(-4)`B. `1.6xx10^(-5)`C. `6.3xx10^(-10)`D. `6.3xx10^(-11)` |
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Answer» Correct Answer - b |
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| 869. |
A 0.1 M solution of which salt will have a pH less than 7 ?A. NaClB. `NH_4Br`C. KFD. `NaO_2C CH_3` |
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Answer» Correct Answer - b |
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| 870. |
Which of the following salt is basic?A. `HOCI`B. `NaOCI`C. `NaHSO_(4)`D. `NH_(4)NO_(3)` |
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Answer» Correct Answer - B a. `HOCI`, it is `S_(A)`. So it is acidic b. `naOCI`, it is salt of `NaOH + HOCI`(salt of `S_(B)//W_(A))`. `:.` It is basic. c. `NaHSO_(4)` (Salt of `S_(A)//S_(B) NaOH +H_(2)SO_(4))`, it is neutral d. `NH_(4)NO_(3)` (Salt of `W_(B)//S_(A) NH_(3) +HNO_(3))`, is is acidic. |
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| 871. |
Percentange ionisation of weak acid can be calculated using the formula:A. `100sqrt((K_(a))/(C ))`B. `(100)/(1+10^((pK_(a)-pH))`C. both (a) and (b)D. none of these |
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Answer» Correct Answer - C For weak acid dissociation equilibria, degree of dissociation `alpha` is given as: `alpha = sqrt((K_(a))/(C ))` `:. %alpha = 100sqrt((K_(a))/(C ))` Also `K_(a)= ([H^(+)][A^(-)])/([HA])= ([H^(+)].Calpha)/(C(1-alpha))` `= ([H^(+)].alpha)/((1-alpha))` `log K_(a)= log H^(+)+log (alpha)/((1-alpha))` or `pK_(a)=pH+log(1-alpha)/(alpha)` or `pK_(a)-pH=log (1-alpha)/(alpha)` `:. (1-alpha)/(alpha)=10^(pK_(a)-pH)` or `(1)/(alpha)=10^(pK_(a)-pH)+1` or `alpha=(1)/([1+10^(pK_(a)-pH))]` |
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| 872. |
Percentange ionisation of weak acid can be calculated using the formula:A. `100sqrtK_a/C`B. `100/(1+10^((pK_a-pH)))`C. Both (a) and (b)D. None of these |
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Answer» Correct Answer - c |
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| 873. |
Which of the following is true ?A. `pK_b` for `OH^-` is -1.74 at `25^@C`B. The equilibrium constant for the reaction between HA (`pK_a=4`) and NaOH at `25^@C` will be equal to `10^10`C. The pH of a solution containing 0.1 M HCOOH `(K_a=1.8xx10^(-4))` and 0.1 M HOCND. All the above are correct |
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Answer» Correct Answer - d |
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| 874. |
Physical and chemical equilibrium can respond to a change in their pressure, temperature, and concentration of reactants and products. To describe the change in the equilibrium we have a principle named Le Chatelier principle. According to this principle, even if we make some changes in equilibrium, then also the system even re-establishes the equilibrium by undoing the effect. If we add `SO_(4)^(2-)` ion to a saturated solution of `Ag_(2)SO_(4)`, it will result in a//anA. Result in an increase in `Ag^(o+)` concentrationB. Result in a decrease in `Ag^(o+)` concentrationC. Shift `Ag^(o+)` ions from solid `Ag_(2)CrO_(4)` into solution.D. Result in a decrease the `CrO_(4)^(2-)` ion concentration in the solution. |
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Answer» Correct Answer - B Due to common ion `(CrO_(4)^(2-))`, the ionisation of `Ag_(2)CrO_(4)` is supressed and solubility of `Ag_(2)CrO_(4)` decreases i.e., `[Ag^(o+)]` decreases. |
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| 875. |
In a buffer solution containing equal concentration of `B^(-)` and `HB`, the `K_(b)` for `B^(-)` is `10^(-10)`. The `pH` of buffer solution isA. 10B. 7C. 6D. 4 |
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Answer» Correct Answer - D Key Idea - (i) For basic buffer, `pOH=pK_(b)+log.(["salt"])/(["base"])` (ii) `pH+pOH = 14` Given, `K_(b)=1xx10^(-10)`, [salt]=[base] `pOH=-log K_(b)+log.(["salt"])/(["base"])` `therefore pOH=-log(1xx10^(-10))+log1=10` `pH+pOH=14 [because "concentration of "[B^(-)]=[HB]` `pH=14-10=4` |
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| 876. |
0.5 M ammonium benzoate is hydrolyzed to 0.25 percent, hence its hydrolysis constant isA. `2.5 xx 10^(-5)`B. `1.5 xx 10^(-4)`C. `3.125 xx 10^(-6)`D. `6.25 xx 10^(-14)` |
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Answer» Correct Answer - C `{:(C_(6)H_(5)COONH_(4),rarr,C_(6)H_(5)COO^(-),+,NH_(2)^(+)),(0.5 - (.25)/(100),,(.25)/(100),,(.25)/(100)):}` According to Ostwald dilution law. `K = (alpha^(2)C)/(1-alpha) " " ( :. alpha = (.25)/(100))` `K = alpha^(2)C " " ( because 1- alpha = "Very small")` `K = (0.25)/(100) xx (0.25)/(100) xx 0.5 , K = 3.125 xx 10^(-6)` |
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| 877. |
Calculate the minimum mass of `AB_(2) (s)` which must be added to 100 mL water (in mg) to form a saturated solution. `K_(sp) (AB_(2)) = 3.2 xx 10^(-11)` `M_(w.t) [AB_(2)(s)] = 100 g//"mole"` |
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Answer» Correct Answer - `0002` |
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| 878. |
What concentration of dichloroacetic acid gives `[H^(+)]=8.5xx10^(-3)M` ? `("Given": K_(a)"of acid" = 5.0xx10^(-2))`. |
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Answer» `CHCI_(2)COOH hArr CHCI_(2)COO^(-)+H^(+)` `[H^(+)]=[CHCI_(2)COO^(-)]= 8.5xx10^(-3)M= C alpha` and `CHCI_(2)COOH=C(1-alpha)=C-C alpha` `K_(a)= ([H^(+)][CHCI_(2)COO^(-)])/([CHCI_(2)COOH])` `=((8.5xx10^(-3)xx8.5xx10^(-3)))/((C- 8.5xx10^(-3)))= 5xx10^(-2)` `:. C = 9.95xx10^(-3)M` |
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| 879. |
Which has maximum solubility `AB,AB_(2),AB_(3)` and `AB_(4)` if `K_(SP)` for all the salts are `10^(-10) :`A. `AB`B. `AB_(2)`C. `AB_(3)`D. `AB_(4)` |
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Answer» Correct Answer - 4 `{:(For,AB,,,K_(SP)=S^(2),rArr,S=10^(-5)),(For,AB_(2),hArr,A^(2+),+2B^(-),,),(,,,S," 2S",,):}` `K_(SP)=4S^(3)=10^(-10)` `S=((K_(SP))/(4))^(1//3)` For `AB_(3)" "K_(SP)=27S^(4)` `S=((K_(SP))/(27))^(1//4)` For `AB_(4)" "K_(CP)=(4)^(4)S^(5)` `s=((k_(sp))/((4)^(4)))^(1//5)` Hence, solubility will be maximum for `AB_(4)`. |
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| 880. |
`0.1 M H_(2)S` has `K_(1)=10^(-5) & K_(2)=1.5xx10^(-12)`. What will be the concentration of `S^(-2)` in the solution.A. `~~10^(-8)`B. `~~10^(-9)`C. `~~1.5xx10^(-12)`D. `1.2xx10^(-13)` |
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Answer» Correct Answer - 3 `[S^(2-)]=K_(2)=1.5xx10^(-12)` |
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| 881. |
`K_(SP)` of `BaSO_(4)` is `1.5xx10^(-9)` . Calculate is solubility in: (a) Pure water, , (b) `0.10MBaCI_(2)`. |
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Answer» `BaSO_(4)hArr underset(S)(Ba^(2+))+underset(S)(SO_(4)^(2-))` (where S in m/litre solubility of `BaSO_(4)`) (i) `K_(SP)=SxxS` `:. S=sqrt(K_(SP))=sqrt(1.5xx10^(-9))` `= 3.87xx10^(-5)` mol `litre^(-1)` (ii) In presence of `0.10M BaCI_(2)`, let S mol/litre `BaSO_(4)` is dissolved `K_(SP)=[Ba^(2+)][SO_(4)^(2-)]` `1.5xx10^(-9)=(0.1+S)(S)` `[(0.1+S)=0.1 as S ltltlt0.1]` `:. S=1.5xx10^(-8)M` |
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| 882. |
The addition of a polar solvent to a solenoid electrolyte results inA. PolarizationB. AssociationC. IonizationD. Electron transfer |
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Answer» Correct Answer - C Polar solvent facilitate ionisation of strong electrolytes due to dipole-ion attraction. |
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| 883. |
An ionizing solvent hasA. Low value of dielectric constantB. High value of dielectric constantC. A dielectric constant equal to 1D. Has a high melting point |
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Answer» Correct Answer - B Higher the dielectric constant of a solvent more of its ionising power. |
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| 884. |
Vant hoff factor of `BaCl_(2)` of conc. 0.01 M is 1.98. Percentage dissociation of `BaCl_(2)` on this conc. Will beA. 49B. 69C. 89D. 98 |
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Answer» Correct Answer - A `{:(,BaCl_(2),hArr,Ba^(2+),+,2Cl^(-)),("Initially",1,,0,,0),("After dissociation",1-alpha,,alpha,,2alpha):}` Total `= 1 - alpha + alpha + 2alpha = 1 + 2 alpha` `alpha = (i-1)/(m-1) = (1.98 - 1)/(3 - 1) = (0.98)/(2) = 0.49` m = number of particles in solution For a mole `alpha = 0.49` For 0.01 mole `alpha = (0.49)/(0.01) = 49`. |
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| 885. |
Accumulation of lactic acid `(HC_(3)H_(5)O_(3))`, a monobasic acid in tissues leads to pain and a feeling of fatigue. In a 0.10 M aqueous solution, lactic acid is `3.7%` dissociates. The value of dissociation constant Ka, for this acid will beA. `1.4 xx 10^(-5)`B. `1.4 xx 10^(-4)`C. `3.7 xx 10^(-4)`D. `2.8 xx 10^(-4)` |
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Answer» Correct Answer - B `alpha = sqrt((K_(a))/(C))` `alpha^(2) = (K_(a))/(C)` or `K_(a) = C alpha6(2)` `K_(a) = 0.10 xx ((3.7)/(100))^(2)` `K_(a) = 1.4 xx 10^(-4)` |
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| 886. |
In acid buffer solution `(pH = 4.4)`, the ratio of concentrations of acid to salt is `2 : 1`. The value of dissociation constant of weak acid may beA. `1.8 xx 10^(-4)`B. `2 xx 10^(-7)`C. `4 xx 10^(-5)`D. `2 xx 10^(-8)` |
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Answer» Correct Answer - D `p^(H) = p^(ka) + "log" (["salt"])/(["acid"])` |
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| 887. |
What is the [`Mg^(2+)`] in 0.10 M NaF that is saturated with `MgF_(2)` at `25^(@)` C ? A. 0.050 MB. `1.9xx10^(-3)` MC. `1.2xx10^(-3)` MD. `6.4xx10^(-7)` M |
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Answer» Correct Answer - d |
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| 888. |
The solubility of `A_(2)B_(3)` is "x mol dm"^(-3)`. Its `K_(sp)` isA. `6x^(4)`B. `64x^(4)`C. `36x^(5)`D. `108x^(5)` |
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Answer» Correct Answer - D `K_(sp) = (2)^(2).(3)^(2).x^(2+3) = 108 x^(5)` |
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| 889. |
Assertion (A): `pH` of neutral solution is always `7`. Reason (R) : `pH` of solution does not depend upon 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 both (A) and (R) are incorrect. |
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Answer» Both `(A)` and `(R)` are wrong. Since it depends on temperature. |
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| 890. |
Which one has pH 12A. 0.01 M KOHB. 1 N KOH mlC. 1 NaOH mlD. 1 N `Ca(OH)_(2)ml` |
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Answer» Correct Answer - A `[OH^(-)] = 10^(-2)M, pOH = 2` `pH + pOH = 14, pH = 14 - pOH` `pH = 14 - 2 = 12`. |
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| 891. |
In a solution of acetic acid, sodium acetate is added, then its pH valueA. DecreasesB. IncreasesC. Remains unchangedD. (a) and (b) both are correct |
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Answer» Correct Answer - B Solution of `CH_(3)COONa` on addition to acid shows a decrease in dissociation of acid due to common ion effect. To decrease in `[H^(+)]` or increase pH. |
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| 892. |
What is the `pH` of the solution when `0.20 mol` of `HCI` is added to `1L` of a solution containing a. `1M` each of acetic acid and acetate ion. b. `0.1M`each of aceta acid and acetate ion. Assume the total volume is `1L. K_(a)` for acetic acid is `1.8 xx 10^(-5)`. |
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Answer» Correct Answer - A::B::D a. A mixture of acetic acid and acetate ions forms an acidic buffer. `pH = pK_(a) +"log" (["Salt"])/(["Acid"])` Rule `A A A`: When `HCI` is added, the amount of acid increases because acetate ions react with hydorgen ions to form undissociated salt. Therefore, the amount of acetate decreases. New `["Acid"] = 1 + 0.2 = 1.2` New `["Salt"] = 1- 0.2 = 0.8` `pH = pK_(a) + "log" (["Salt"])/(["Acid"])` `=- log (1.8 xx 10^(-5)) +"log" (0.8)/(1.2)` `=4.47 + "log" (2)/(3) =4.568` b. When `0.2M` hydrochloric acid is added in the second case, `0.1M` aceate is consumed totally in the reaction with hydrogen ion. only `0.1M HCI` is left. Also there is acetic acid. So there is no buffer, `pH` is calculated according to strong acid `HCI` only. `pH =- log [H^(o+)] =- log 0.1 = 1` |
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| 893. |
`K_(sp)` value of `Al(OH)_(3)` and `Zn(OH)_(2)` are `8.5 xx 10^(-23)` and `1.8 xx 10^(-14)` respectively. If `NH_(4)OH` is added in a solution of `Al^(3+)` and `Zn^(2+)`, which will precipitate earlierA. `Al(OH)_(3)`B. `Zn(OH)_(2)`C. Both togetherD. None |
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Answer» Correct Answer - A Solubility of `Al(OH)_(3)` is lesser than `Zn(OH)_(2)`. |
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| 894. |
The `P^(H)` of saturated aqueous solution of `Ba(OH)_(2)` is `10`. If the `K_(sp)` of `Ba(OH)_(2)` is `5 xx 10^(13)`, then the concentration of `Ba^(2+)` ions in the solution isA. `1 xx 10^(-5)`B. `1 xx 10^(-3)`C. `5 xx 10^(-5)`D. `1 xx 10^(-2)` |
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Answer» Correct Answer - C `Ba(OH)_(2) hArr Ba^(+2) + 2OH^(-)` given `P^(H) = 10` then `P^(OH) = 4` `2S = [OH^(-)] = 10^(-4)` `[Ba^(+2)] [2S^(2)] = K_(sp)` `[Ba^(+2)] = (5 xx 10^(-13))/(10^(-5)) = 5 xx 10^(-5)` |
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| 895. |
`NH_(4)OH` is weak base but it becomes still weaker in the aqueous solutions ofA. `0.1 MHCl`B. `0.1MNH_(4)Cl`C. `0.1MH_(2)SO_(4)`D. `0.1 MH_(2)SO_(4)` |
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Answer» Correct Answer - B Due to common ion effect |
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| 896. |
In the eqantitative analysis `Bi^(3+)` is detected precipitation of `[BiO (OH)(s)]` [bismuthy`1` hydroxide). Calculate the `pH` when the following equilibria exists: `BiO (OH)(s) hArr BiO^(o+)(aq) + overset(Theta)OH (aq) (K = 4 xx 10^(-10))` |
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Answer» `[overset(Theta)OH] [BiO^(o+)] = 4 xx 10^(-10)` `:. [overset(Theta)OH]^(2) = 4 xx 10^(-10) :. [overset(Theta)OH] = 2 xx 10^(-5)` `:. pOH = 4.6989 :. pH = 9.3010` |
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| 897. |
An equimolar solution of `CH_3COOH` and `C_2H_5COONa` will be :A. acidicB. neutralC. alkalineD. insufficient data to predict |
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Answer» Correct Answer - a |
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| 898. |
If the salts `M_(2)X,QY_(2)`, and `PZ_(3)` have the same solubilities `(lt(4)/(27))`, their `K_(sp)` values are relatedA. `k_(sp)(M_(2)x)=k_(sp)(Qy_(2)) lt k_(sp)(Pz_(3))`B. `k_(sp)(M_(2)x) gt k_(sp)(Qy_(2))=k_(sp)(Pz_(3))`C. `k_(sp)(M_(2)x) lt k_(sp)(Qy_(2)) lt k_(sp)(Pz_(3))`D. `k_(sp)(M_(2)x)gt k_(sp)(Qy_(2)) gt k_(sp)(Pz_(3))` |
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Answer» Correct Answer - A `M_(s)x` and `Qy_(2)` are salts of the type `A_(2)B` for which `K_(sp)=4 s^(3)`. If s is same, `K_(sp)` will be equal . For `P_(Fz_(3)),K_(sp)=s(3s)^(3)=27s^(4)`i.e., much higher . Hence , A is the corrct option. |
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| 899. |
If the salts `M_(2)X,QY_(2)`, and `PZ_(3)` have the same solubilities `(lt(4)/(27))`, their `K_(sp)` values are relatedA. `K_(sp) (M_(2)X) = K_(sp)(QY_(2)) gt K_(sp) (PZ_(3))`B. `K_(sp)(M_(2)X) gt K_(sp) (QY_(2)) =K_(sp) (PZ_(3))`C. `K_(sp) (M_(2)X) = K_(sp) (QY_(2)) = K_(sp) (PZ_(3))`D. `K_(sp)(M_(2)X) gt K_(sp) (QY_(2)) gt K_(sp) (PZ_(3))` |
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Answer» Correct Answer - A Let `x =` solubility of each slat and `x` is small. `{:(i.,M_(2)XhArr,2M^(o+)+,X^(2-),[K_(sp)=4x^(3)]),(,,2x,x,),(ii.,QY_(2)hArr,Q^(2+)+,2Y^(Theta),[K_(sp)=4x^(2)]),(,,x,2x,),(iii.,PZ_(3)hArr,P^(3+)+,3X^(Theta),[K_(sp)=27x^(4)]),(,,x,3x,):}` `:. K_(sp) (M_(2)X) = K_(sp) (QY_(2)) gt K_(sp) (PZ_(3))`. |
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| 900. |
How many millilitres of 6.0 M hydrochloric acid should be used to prepare 150 mL of a solution which is 0.30 M in hydrogen ionA. `3.0`B. 7.5C. 9.3D. 30 |
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Answer» Correct Answer - B `M_(1) = 6.0 M` of HCl , `V_(1) = ?` `M_(2) = 0.30 M` is `H^(+)` concentration in solution. `V_(2) = 150 ml` of solution. `M_(1)V_(1) = M_(2)V_(2), 6.0 xx V_(1) = 0.30 xx 150` `V_(1) = (0.30 xx 150)/(6) = 7.5 ml`. |
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