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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.
| 101. |
`20mL` of `0.2M NaOH` are added to `50mL` of `0.2M` acetic acid `(K_(a)=1.85xx10^(-5))` Take `log2=0.3,log3=0.48` (1) What is `pH` of solution? (2) Calculate volume of `0.2M NaOH` required to make the `pH` of origin acetic acid solution.`4.74`. |
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Answer» `{:(,NaOH+,CH_(3)COOHrarr,CH_(3)COONa+,H_(2)O),("Millimole added",20xx0.2,50xx0.2,,),(,=4,=10,0,0),("Millimole after reaction",0,6,4,4):}` `{"Molarity"}=("millimole")/("Total volume")` `[CH_(2)COOH]=(6)/(70)` &`[CH_(2)COONa]=(4)/(70)` rArr Buffer solution consisting of a weak acid & its salt with a strong base. (2) Let `V mL` of `0.2M NaOH` is required to make `pH=4.74`,Then `NaOH` should be completely used up (:. final solution is required to be acidic) `{:(,NaOH+,CH_(3)COOHrarr,CH_(3)COONa+,H_(2)O),("Millimole added",0.2xxV,50xx0.2,,),(,=0.2V,=10,0,0),("Millmole after reaction",0,(10-2V),0.2V,0.2V):}` `:.["Acid"]=(10-1.2V)/(50+V),["Salt"]=(0.2V)/(50+V)` `:.4.74=-log1.85xx10^(-5)+log""((0.2V)(50+V))/((10-0.2V)(50+V))` `:.V=25mL` |
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| 102. |
Which causes a largest change in pH on addition to 20ml 0.1 M acetic acid solution ?A. Additon of 20 mol of 0.01 `MCH_(3)COOH`B. Additon of 20 mL of 0.1 HClC. Additoin of 20 Ml of 0.1 M HClD. Additon of 20 mL of 0.1 M NaOH |
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Answer» Correct Answer - D Moles of AcOH `=20 xx 10^(-3) xx 0.1 =2 xx 10^(-3)` Moles of NaOH `=20 xx 10^(-3) xx 0.1 =2 xx 10^(-3)` pH will change from acidic side to basic. |
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| 103. |
The solubility of `Mg(OH)_(2)` in a buffer of pH=10 is found to be 0.0232 gm/lit. Molar solubility of `Mg(OH)_(2)`` in pure water would be (in mol/lit)A. `10^(-5)`B. `3.2xx10^(-8)`C. `3.2xx10^(-5)`D. `10^(-4)` |
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Answer» Correct Answer - d |
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| 104. |
What is the pH of a 0.15 M solution of hydrazine , `N_2H_4`? A. 3.41B. 6.82C. 10.59D. `11.00` |
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Answer» Correct Answer - c |
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| 105. |
Which is a Lewis base ?A. `B_(2)H_(6)`B. `LiAlH_(4)`C. `AiH_(3)`D. `NH_(3)` |
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Answer» Correct Answer - D Out of the four given choices only `NH_(3)` can act as a Lewis base. |
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| 106. |
In the equation `I_(2)+I^(-) rarr, I_(3)^(-)` which is Lewis baseA. `I_(2)`B. `I^(-)`C. `I_(3)^(-)`D. None of these |
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Answer» Correct Answer - B In the reaction `I_(2)+I^(-) rarr I_(3)^(-) , I^(-)` is electron pair donor , therefore , it is a base. |
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| 107. |
Which of the following is not a Lewis base ?A. `CN^(-)`B. `ROH`C. `NH_(3)`D. `AlCl_(3)` |
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Answer» Correct Answer - D `AlCl_(3)` is electron deficient and is a lewis acid. |
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| 108. |
The pH of 10M HCl aqueous solution isA. less than 0B. zeroC. 2D. 1 |
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Answer» Correct Answer - A A 10 M HCl solution will not be fully dissociated. Therefore, its pH will be less than zero but not`= -1`. [If complete dissociation in possible, For 1 M HCl, `H^(+)=10M=10^(@)MimpliespH=0` For `10 M HCl, H^(+)=10M=10^(1)M impliespH = -1]` |
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| 109. |
The solubility of AgCl will be miniumum 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 The concentration of common ion `Cl^(-)` is maximum in 0.01 M `CaCl_(2)`. Hence, it will supress the ionisation of AgCl and the solubility will be least. |
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| 110. |
The solubility of AgCl will be miniumum inA. `0.01 M Na_(2)SO_(4)`B. `0.1 M CaCl_(2)`C. Pure waterD. `0.001 M AgNO_(3)` |
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Answer» Correct Answer - B `CaCl_(2)` is strong electrolyte and common ion effect |
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| 111. |
Which of the following is not a conugate acidbase pairA. `HSO_(4)^(-), SO_(4)^(2-)`B. `H_(2)PO_(4)^(-), HPO_(4)^(2-)`C. `H_(2)PO_(4)^(-), H_(3)PO_(4)`D. `H_(2)PO_(4^(-)), PO_(4^(-3))` |
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Answer» Correct Answer - D Conjugate acid-base pair is differ by only one proton |
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| 112. |
If first dissociation of `X(OH)_(32)` is `100%` where as second dissociation is `50%` and third dissciation is negligible then the pH of `4 xx 10^(-3) M, X(OH)` is :A. `11.78`B. `10.78`C. `2.5`D. `2.22` |
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Answer» Correct Answer - A First dissociation `X(OH)_(3) rarr X(OH)_(2)^(+) + OH^(-)` Second dissociation : `X(OH)_(2)^(+) rarr X(OH)^(2+) + OH^(-)` Total `[OH^(-)] = 4 xx 10^(-3) + 2 xx 10^(-3) = 6 xx 10^(-3)` `pOH = 3 log 6 = 2.22` `:. pH = 11.78` |
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| 113. |
Calculate the pH of `6.66xx10^(-3)` M solution of `Al(OH)_3`, its first dissociation is 100% where as second dissociation is 50% and third dissociation is negligible.A. 2B. 12C. 11D. 3 |
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Answer» Correct Answer - b |
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| 114. |
Acetic acid tends to form dimer due to formation of intermolcular hydrogen bonding. `2CH_(2)COOHhArr(CH_(3)COOH)_(2)` The equilibrium constant for this reaction is `1.5xx10^(2)M^(-1)`in benzene solution and `3.6xx10^(-2)` in water. In benzene, monomer does not dissociate but it water, monomer dissociation simultaneously with acid dissociation constant `2.0xx10^(-5)`M. Dimer does not dissociate in benzene as well as water. The molar ratio of dimer to monmer for 0.1M acetic acid in water (neglecting the dissciation of acetic acid in water ) is equal to :A. `250:1`B. `1:250`C. `9:2500`D. `2500:9` |
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Answer» Correct Answer - c |
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| 115. |
Acetic acid tends to form dimer due to formation of intermolcular hydrogen bonding. `2CH_(2)COOHhArr(CH_(3)COOH)_(2)` The equilibrium constant for this reaction is `1.5xx10^(2)M^(-1)`in benzene solution and `3.6xx10^(-2)` in water. In benzene, monomer does not dissociate but it water, monomer dissociation simultaneously with acid dissociation constant `2.0xx10^(-5)`M. Dimer does not dissociate in benzene as well as water. The molar ration of dimer to monomer for 0.1 M acetic acid in benzene is equal to :A. `150:1`B. `1:150`C. `5:2`D. `2:5` |
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Answer» Correct Answer - c |
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| 116. |
The decreasing order of strength of the bases, `OH^(-), NH_(2)^(-), H-C-=C^(-)` and `CH_(3)-CH_(2)^(-)`:A. `CH_(3)CH_(2)^(-) gtNH_(2)^(-) gt H-C-=C^(-) gt OH^(-)`B. `H-C-=C^(-)gtCH_(3)-CH_(2)^(-)gtNH_(2)^(-) gtOH^(-)`C. `OH^(-) gt NH_(2)^(-) gt H-C-=C^(-)gtCH_(3)CH_(2)^(-)`D. `NH_(2)^(-)gtH-C-=C^(-)gtOH^(-)CH_(3)CH_(2)^(-)` |
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Answer» Correct Answer - A A weak acid has strong conjugate base. Since the acidic nature decrease in the order `H_(2)O gt H-C-=C-HgtNH_(3)gtCH_(3)-CH_(3)` Therefore , the conjugate base strength follows the order given in (A ). |
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| 117. |
What is the decreasing order of strength of the bases `OH^(-), NH_(2)^(-), H - C -= C^(-)` and `CH_(3) - CH_(2)^(-)`A. `CH_(3) - CH_(2)^(-) gt NH_(2)^(-) gt H - C^(-) gt OH^(-)`B. `H - C -= C^(-) gt CH_(3) - CH_(2)^(-) gt NH_(2)^(-) gt OH^(-)`C. `OH^(-) gt NH_(2)^(-) gt H - C -= C^(-) gt CH_(3) - CH_(3)^(-)`D. `NH_(2)^(-) gt H - C -= C^(-) gt OH^(-) gt CH_(3) - CH_(2)^(-)` |
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Answer» Correct Answer - A order of basic strength `CH_(3) - CH_(2)^(-) gt bar(N)H_(2) gt HC -= C^(-) gt OH^(-)` because their acids have following order of acidic strength `H - OH gt HC -= CH gt NH_(3) gt CH_(3) - CH_(3)` (conjugate base of strong acid is weaker while conjugate base of weak acid is stronger). |
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| 118. |
Which of the following will show common ion effect and from a buffer solutionA. `CH_(3)COONH_(4)` and `CH_(3)COOH`B. `NH_(4)Cl + NH_(4)OH`C. `H_(2)SO_(4) + Na_(2)SO_(4)`D. `NaCl + NaOH` |
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Answer» Correct Answer - A::B (a) is buffer because it contains weak acid and its salt and they will also show common ion effect. (b) Is also a buffer because it contains a weak acid and its salt. They will show common ion effect. (c) Is not buffer solution because they contains strong acids and its salt. They will not show common ion effect (d) Is not a buffer solution because it contains strong base and its salt. They will not show common ion effect. |
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| 119. |
The `pK_(a)` of an acid HA is 4.77 and `pK_(b)` of a base of BOH is 4.75 . The pH of 0.1 M aqueous solution of the salt AB isA. `7.02`B. `7.01`C. `6.99`D. `7.00` |
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Answer» Correct Answer - B AB is a salt of a weak acid and weak base. Hence `pH=7+(1)/(2)(pK_(a)-pK_(b))` `=7+(1)/(2)(4.77-4.75)=7.01` |
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| 120. |
`pK_(a)` of weak acid (HA) and `pK_(b)` of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution isA. 6.9B. `7.0`C. `1.0`D. 7.2 |
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Answer» Correct Answer - A Salt of weak acid and weak base `pH = (1)/(2)(pk_(w)+ pK_(a) - pK_(b))` `= (1)/(2)(14 + 3.2 - 3.4)` = 6.9 |
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| 121. |
How many litres of water must be added to 1 litre an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2A. 0.1 LB. 0.9 LC. `2.0 L`D. `9.0 L` |
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Answer» Correct Answer - D `pH = 1 [H^(+)] = 10^(-1) = 0.1 M` `pH = 2 [H^(+)] = 10^(-2) = 0.01` for dilution of HCl `M_(1)V_(1) = M_(2)V_(2)` `0.1 xx 1 = 0.01 xx V_(2)` `V_(2) = 10 l t` Volume of water added = 10 - 1 = 1litre. |
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| 122. |
Assertion (A): A is very dilute acidic solution of `Cd^(2+)` and `Ni^(2+)` gives yellow precipitate of `CdS` on passing hydrogen sulphide. Reason (R) : Solubility product of `CdS` is more than that of `NiS`.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
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Answer» Correct Answer - C (A) is true. (R) is false: Correct reason: `K_(sp)` of `CdS lt K_(sp) NiS`. Since compound with lower `K_(sp)` is precipitated first. |
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| 123. |
`18mL` of mixture of `CH_(3)COOH` and `CH_(3)COONa` required `6mL` of `0.1M NaOH` for neutralisation of the acid `12mL` of `0.1M HC1` for reaction with salt, separately. If `pK_(a)` of the acid is `4.75`, what is the `pH` of the mixtureA. `4.5`B. `4.6`C. `4.75`D. `5.05` |
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Answer» Correct Answer - D For neutralisation of `CH_(3)COOH` with `NaOH` mmol of `CH_(3)COOH` = mmol of `NaOH` `= 6 xx 0.1 = 0.6` For neutralisation `CH_(3)COONa` with `HC1` mmol of `CH_(3)COONa` mmol of `HC1` `12 xx 0.1 = 1.2` Concentration of acid and slat in `18mL` of mixture `[CH_(3)COOH] = (0.6 "mmol")/(18mL)` `[CH_(3)COONa] = (1.2 "mmol")/(18mL)` `pH = 4.75 + log ((1.2)/(0.6))` `= 4.75 + log 2 = 4.75 + 0.3 = 5.05` |
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| 124. |
`2M` solution of `Na_(2)CO_(3)` is boiled in a closed container with excess of `CaF_(2)` .Very little amount of `CaCO_(3)` and `NaF` are formed.If the solubility product `(K_(sp))` of `CaCO_(3)` is `x` and molar solubility of `CaF_(1)` is `y`.Find the molar concentration of `F^(-)` in resulting solution after equilibrium is attained. |
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Answer» `{:(,Na_(2)CO_(3)+,CaF_(2)(s)hArr,2NaF(aq)+,CaCO_(3)(s)),(t=0,2,-,0,-),(t=eq,2-a,-,2a,-):}` where is very small For `CaCO_(3),K_(sp)=x=[Ca^(2+)][CO_(3)^(2-)]=[Ca^(2+)]xx2( :.CO_(3)^(2-)` mainly coming from `Na_(2)CO_(3)` `[Ca^(2+)]=(x)/(2)` For `CaF_(2),K_(sp)=4y^(3)=((x)/(2))[F^(-)]^(2)rArr [F^(-)]=sqrt((8y^(3))/(x))` |
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| 125. |
`2M` solution of `Na_(2)CO_(3)` is boiled in a closed container with excess of `CaF_(2)`. Very small amount of `CaCO_(3)` and `NaF` are formed. If `K_(sp)` of `CaCO_(3)` is `x` and molar solubility of `CaF_(2)` is `y`, find the molar after cocentration of `F^(Theta)` in the resulting solution after equilibrium is attained. |
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Answer» `{:(,Na_(2)CO_(3)+,CaF_(2)(s)rarr,CaCO_(3)+,2NaF),("Mole taken",2,,0,0),("Mole left",(2-a),,a,2a):}` where `a` is very-very small and thus assume that `CaCO_(3)` is in soluble form Now, `K_(sp)` of `CaCO_(3) = x = [Ca^(2+)] [CO_(3)^(2-)]` Also `[CO_(3)^(2-)] =2 -a+a =2 :. [Ca^(2+)] = (x)/(2)` For `{:(CaF_(2) hArr,Ca^(2+)+,2F^(Theta),,),(,y,2y,,):}` `K_(sp(CaF_(2))) = [Ca^(2+)] [F^(Theta)]^(2) = (y) (2y)^(2) = 4y^(3)` Further for `[F^(Theta)]`, we can have `[F^(Theta)] = [F^(Theta)]` frm `CaF_(2) + [F^(Theta)]` from `NaF` `[F^(Theta)] = sqrt((K_(sp(CaF_(2))))/([Ca^(2+)])) +` Negligible value `[F^(Theta) = sqrt((4y^(3))/(x//2))sqrt((8y^(3))/(x))` |
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| 126. |
When solid `SrCO_(3)` is equilibrated with a `pH 8.60` buffer, the solution was found to have `[Sr^(2+)] = 2.2 xx 10^(-4)`. What is the `K_(sp)` of `SrCO_(3)`. `(K_(2)` of `H_(2)CO_(3) = 4.7 xx 10^(-11))` |
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Answer» First Method by using direct formula Solubility of salt of weak acid `(H_(2)CO_(3))` in buffer (refersection) is given by the formula `S_("buffer") = sqrt(K_(sp)) [1+([H^(oplus)])/(K_(a))]^(1//2) ….(i)` `S_("buffer") = 2.2 xx 10^(-4)M, pH = 8.60`, `:.[H^(oplus)] =- log (8.6) = 2.51 xx 10^(-9)` `K_(2) = 4.7 xx 10^(-11)` Substituting the value in equation (i), we get `2.2 xx 10^(-4) = (K_(sp))^(1//2) [1+(2.51xx10^(-9))/(4.7xx10^(-11))]^(1//2)` `=(K_(sp))^(1//2) (1+53.4)^(1//2)` `:.K_("sp") = ((2.2 xx 10^(-4))^(2))/(54.4) = 8.9 xx 10^(-10)` Second method: `K_(sp) = [Sr^(2+)] [CO_(3)^(2-)]` `CO_(3)^(2-) + H_(2)O hArr HCO_(3)^(Θ) + overset(Θ)OH` `K_(h) = (K_(w))/(K_(2)) = ([overset(Θ)OH][HCO_(3)^(Θ)])/([CO_(3)^(2-)] ) = (K_(w))/(4.7 xx 10^(-11))` `[H_(3)O^(oplus)] =- (log 8.6) = 2.5 xx 10^(-9)`. `[overset(Θ)OH] = (K_(w))/(2.51 xx 10^(-9))` `((K_(w))/(2.51xx10^(-9)))[(HCO_(3)^(Θ))/(CO_(3)^(2-))] = (K_(w))/(4.7xx10^(-1))` `([HCO_(3)^(Θ)])/([CO_(3)^(2-)]) = (2.51 xx 10^(-9))/(4.7 xx 10^(-11)) = 53.4` The `CO_(3)^(2-)` ion which dissolves forms `HCO_(3)^(Θ)` in a `1:1` mol ratio of remains unreacted, thus, by electroneutrality, `[Sr^(2+)] = [HCO_(3)^(Θ)] + [CO_(3)^(2-)]` `2.2 xx 10^(-4) = 53.4 [CO_(3)^(2-)] + [CO_(3)^(2-)]` `2.2 xx 10^(-4) = 54.4 [CO_(3)^(2-)]` `:.[CO_(3)^(2-)] = (2.2xx10^(-4))/(54.4)` `K_(sp) = (2.2 xx 10^(-4)) ((2.2xx10^(-4))/(54.4)) = 8.9 xx 10^(-10)`. |
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| 127. |
A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant isA. `1.0 xx 10^(-3)`B. `1.0 xx 10^(-6)`C. `1.0 xx 10^(-8)`D. `1.0 xx 10^(-11)` |
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Answer» Correct Answer - D `:.` Monoprotic acid HA `HA hArr H^(+) + A^(-)` Ionisation constant = ? `alpha = 0.001 % = (0.001)/(100) = 10^(-5) rArr K = (alpha^(2))/(V) = ([10^(-5)]^(2))/(10) = 10^(-11)`. |
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| 128. |
The solubility of silver chromate in 0.01 M `K_(2)CrO_(4)` is `2 xx 10^(-8) mol m^(-3)`. The solublity product of silver chromate will beA. `8 xx 10^(-24)`B. `16 xx 10^(-24)`C. `1.6 xx 10^(-18)`D. `16 xx 10^(-18)` |
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Answer» Correct Answer - D `{:("Given " K_(2)CrO_(4),hArr,2K^(+),+,CrO_(4)^(2-)),("Conc. "0.01,,0.02,,0.01),("and for "Ag_(2)CrO_(4),,,,):}` `{:(Ag_(2)CrO_(4),hArr,2Ag^(+)+,CrO_(4)^(2-)),(S,,2S,S+0.1):}` `K_(sp) = [Ag^(+)][CrO_(4)^(2-)]` `K_(sp) = (2S)^(2)(S + 0.01) rArr 4S^(2) [S + 0.01]` ? `K_(sp) = 4[2 xx 10^(-8)]^(2) xx 0.01 [[2 xx 10^(-8)]^(2) lt lt 0.01]` `K_(sp) = 4 xx 4 xx 0.01 xx 10^(-16), K_(sp) = 16 xx 10^(-18)` |
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| 129. |
Ionic product of water increases, ifA. Pressure is reducedB. Temperature increasesC. `OH^(-)` is addedD. Temperature increases |
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Answer» Correct Answer - D `K_(w)` increases with increase in temperature. |
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| 130. |
If the concentration of `CrO_(4)^(2-)` ions in a saturated solution of silver chromate is `2 xx 10^(-4)`. Solubility product of silver chromate will beA. `4 xx 10^(-8)`B. `8 xx 10^(-12)`C. `12 xx 10^(-12)`D. `32 xx 10^(-12)` |
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Answer» Correct Answer - D `K_(sp)` of `Ag_(2)CrO_(4) = [Ag^(+)]^(2)[CrO_(4)^(--)]` `CrO_(4)^(--) = 2 xx 10^(-4)` then `Ag^(+_ = 2 xx 2 xx 10^(-4)` `K_(sp) = (4 xx 10^(-4)) (2 xx 10^(-4)) = 32 xx 10^(-12)`. |
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| 131. |
Some chemists at wished to perpare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compound would they use ? `K_(sp) (AgCI) = 1.8 xx 10^(-10), K_(sp) (AgBr) = 5.0 xx 10^(-13)`, `K_(sp) (Ag_(2)CrO_(4)) = 2.4 xx 10^(-12) ["Use" 3sqrt(0.6) = 0.84]`A. `AgCI`B. `AgBr`C. `Ag_(2)CrO_(4)`D. all of these |
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Answer» Correct Answer - C Find the `[Ag^(o+)]` in all the saturated solutions. `[Ag^(o+)]_("in" AgCI) = sqrt(K_(sp) AgCI) = sqrt(1.8 xx 10^(-10))M`, `[Ag^(o+)]_("in" AgBr) = sqrt(K_(sp)AgBr) = sqrt(5.0 xx 10^(-13))M` `[Ag^(o+)]_("in" Ag_(2)CrO_(4)) = 2 3root(3)((K_(sp)Ag_(2)CrO_(4))/(4))` `=2root(3)((2.4xx10^(-12))/(4))M` `rArr [Ag^(o+)]` is maximum is staurated solution of `Ag_(2)CrO_(4)`. |
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| 132. |
`0.1` millie moles of `CdSO_(4)` are present in `10ml` acid solution of `0.08 N HCI`. Now `H_(2)S` si passed to precipitate all the `Cd^(2+)` ions. What would be the `pH` of solution after filtering off percipitate, boilling of `H_(2)S` and making the solution `100ml` by adding `H_(2)S`? |
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Answer» `{:(,CdSO_(4)+,HC1+H_(2)Srarr,CdS+H_(2)SO_(4),),("mmol addedr",rArr0.1,10xx0.08,,),("mmol after reaction",rArr0,=0.8,0.1,0.1):}` mmoles of `H^(o+)` = [From HC1] +[From `H_(2)SO_(4)`] `= 0.8 +0.1xx2 = 1.0` Total volume `=100mL` `:. [H^(o+)] = 1//100 = 10^(-2) M :. pH = 2` |
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| 133. |
A mixture of aqueous solution of sodium acetat and sodium propanota forms a buffer solution. |
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Answer» Correct Answer - A Since on hydrolysis some weak acid is also formed. So mixture of salts of `W_(A)//S_(B)` in aqeuous solution may from buffer. |
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| 134. |
The solubility of a springly soluble salt `AB_(2)` in water is `1.0xx10^(-5) mol L^(-1)`. Its solubility product is:A. `4 xx 10^(-15)`B. `4 xx 10^(-10)`C. `1 xx 10^(-15)`D. ` 1 xx 10^(-10)` |
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Answer» Correct Answer - A `AB_(2)hArr A^(2+)+2B^(-)` At eqm. ` 1.0 xx 10^(-5)" "2.0 xx 10^(-5)` `K_(sp)=[A^(2+)][B^(-)]^(2)` `=(1 xx 10^(-5))(2 xx 10^(-5))^(2)=4 xx 10^(-15)` |
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| 135. |
For the reaction `C(s)+CO_(2)(g) rarr 2CO(g)` an equilibrium mixture has parital pressure, for `CO_(2)` and CO, of 4.0 and 8.0 atm respectively . `K_(p)` for the reaction isA. `0.5`B. `2.0`C. `16.0`D. `32.0` |
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Answer» Correct Answer - C `K_(p)=((p_(co))^(2))/((p_(co_(2))))=p_(co)` for moles `=2` `:. K_(p)=(8xx 8)/(4)=16 atm` |
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| 136. |
A solution containing 75 mL of `0.2M` HCI and 25 mL of `0.2M` NaOH. Calculate the pH of solution. |
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Answer» Correct Answer - 1 `{:(HCI+,NaOHrarr,NaCI+,H_(2)O),(75xx0.2,25xx0.2,0,),(=15,=5, ,),(10,0,5,5):}` `[H^(+)]= (10)/(100)=0.1` , `:. pH=1` |
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| 137. |
Passing `H_(2)S` gas into a mixture of `Mn^(2+), Ni^(2+), Cu^(2+)` and `Hg^(2+)` ions in an acidified aqueous solution precipitatesA. `CuS` and `HgS`B. `MnS` and `CuS`C. `MnS` and `NiS`D. `NiS` and `HgS` |
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Answer» Correct Answer - A In acidic solution `[S^(2-)]` is low. So `CuS` and `HgS` (IInd group cations) will be precipitated more easily due to low `K_(sp)` value. `NiS` and `MnS` (IV group cation) will precipitate if `H_(2)S` is passed in basic medium. |
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| 138. |
Passing `H_(2)S` gas into a mixture of `Mn^(2+), Ni^(2+), Cu^(2+)` and `Hg^(2+)` ions in an acidified aqueous solution precipitatesA. CuS and `Hg_(2)S`B. MnS and CuSC. MnS and NiSD. NiS and `Hg_(2)S` |
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Answer» Correct Answer - A `H_(2)S` in acidic medium precipitates radicals of group of 2 of quantitative analysis, i.e., `Cu^(2+), Hg^(2+), ` and `Pb^(2+), Br^(3+),Cd^(2+),As^(3+),Br^(3+),Sn^(2+)` |
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| 139. |
If an acid-base reaction `HA(aq)+B^(-)(aq)hArr HB(aq)+A^(-)(aq)` has `K_(aq)=10^(-4)`.how many of the following statements are true? (i) `HB` is stronger acid than `HA` (ii) `HA` is stronger acid than `HB` (iii) `HA` and `HB` have the same acidic strength (iv) `B^(-)` stronger base than `A^(-)` (v) `A^(-)` is stronger base than `B^(-)` (vi) `B^(-)` and `HB` are conjugate acid-base pair (vii) `A^(-)` is the conjugate base of acid `HA`. (viii)`HA` can be`HSO_(4)^(-)` and `HB` can be `HCOOH`. `A^(-)` can be `F^(-)` and `B^(-)` can be `CN^(-)` |
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Answer» `HA(aq)+B^(-)(aq)hArr HB(aq)+A^(-)(aq)" "K_(eq)=10^(-4)` `K_(eq)` is very small .So reaction shift in backward direction. `(1) HB` is more stronger acid than `HA` (2) `A^(-)` is more stronger base than `B^(-)` (3) `B^(-)` and `HB` are conjugate acid-base pair. |
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| 140. |
The equilibrium `A^(-) + H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-5))`. The degree of hydrolysis of 0.001 M solution of the salt isA. `10^(-3)`B. `10^(-4)`C. `10^(-5)`D. `10^(-6)` |
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Answer» Correct Answer - A `K_(a) = 1.0 xx 10^(-5)` `K_(h)` = hydrolysis constant `K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(10^(-5)) = 10^(-9)` Degree of hydrolysis (h) `= sqrt((K_(h))/(C)) = sqrt((10^(-9))/(0.001))` `= sqrt(10^(-6)) = 10^(-3) rArr h = 10^(-3)`. |
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| 141. |
For a dibasic acid, `H_(2)A hArr HA^(-) + H^(+) (K_(1))` `HA^(-) hArr A^(2-) + H^(+)(K_(2))` `H_(2)A hArr 2H^(+) + A^(-2)(K)` thenA. `K = K_(1) + K_(2)`B. `K = K_(1) - K_(2)`C. `K = K_(1) //K_(2)`D. `K = K_(1).K_(2)` |
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Answer» Correct Answer - D `K = K_(1)K_(2)` |
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| 142. |
The equilibrium constant for the ionization of `R-NH_(2)(g)` in water as: `R-NH_(2)(g)+H_(2)O(l)hArr R-NH_(3)+(aq)+OH^(-)(aq)` is `10^(-6) at 25^(@)C.` Select correct Statement(s).A. pH of solution is 11 and `P_(RNH_(2)(g))=1bar`B. Forward reaction is favoured by additon of HCI(aq)C. Forward reaction is favoured by addition of `H_(2)O(l)`D. Forwared reaction is favoured by addition of `RNH_(2)(g)` |
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Answer» Correct Answer - a,b,c,d |
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| 143. |
In the reaction: `H_(2)S hArr 2H^(+) + S^(-)`, when `NH_(4)OH` is added, thenA. `S^(--)` is precipitateB. No reaction takes placesC. Concentration of `S^(--)` decreasesD. Concetration of `S^(-)` increases |
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Answer» Correct Answer - D In `IV^(th)` group the `S^(2-)` concentration increase when added the `NH_(4)OH` because `NH_(4)OH hArr NH_(4)^(+) + OH^(-)` `H_(2)S hArr 2H^(+) + S^(2-)` `OH^(-) + H^(+) hArr H_(2)O`. So that `S^(2-)` is increased. |
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| 144. |
Solubility product of AgCl is `1 xx 10^(-6)` at 298 K. Its solubility in mole `"litre"^(-1)` would beA. `1 xx 10^(-6)` mol/ litreB. `1 xx 10^(-3)` mol/litreC. `1 xx 10^(-12)` mol/litreD. None of these |
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Answer» Correct Answer - B `{:("For",AgClrarr,Ag^(+)+,Cl^(-)),(,,x,x):}` `K_(sp) = x^(2), x = sqrt(K_(sp)), sqrt(1 xx 10^(-6)) = 1 xx 10^(-3)` mole/litre. |
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| 145. |
Zn salt is mixed with `(NH_(4))_(2)S` of molarity `0.021M`. What amount of `Zn^(2+)` will remain unprecipitated in 12 mL of the solution ? `(K_(SP) "of ZxnS" = 4.51xx10^(-24))` |
| Answer» Correct Answer - `1.677xx10^(-22)g//12mL;` | |
| 146. |
The mixed salt among the following is:A. `underset(CHOHCOONa)underset(|)(CHOHCOOK)`B. `NaKSO_(4)`C. `CaOCI_(2)`D. all of these |
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Answer» Correct Answer - D A mixed salt is one which furnishes two types of cations or anions. (other than `H^(+)` and `OH^(-))`. |
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| 147. |
What will happen if `C Cl_(4)` is treated with `AgNO_(3)`A. A white ppt. of AgCl will formB. `NO_(2)` will be evolvedC. `C Cl_(4)` will dissolved in `AgNO_(3)`D. Nothing will happen |
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Answer» Correct Answer - D Because `C Cl_(4)` is a organic solvent and `AgNO_(3)` is insoluble in organic solvent. |
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| 148. |
A saturated solution of `Ag_(2)SO_(4)` is `2.5 xx 10^(-2) M`, The value of its solubility product isA. `62.5 xx 10^(-6)`B. `6.25 xx 10^(-4)`C. `15.625 xx 10^(-6)`D. `3.125 xx 10^(-6)` |
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Answer» Correct Answer - A `{:("For",AgSO_(4)rarr,2Ag^(+)+,SO_(4)^(-)),(,,2x,x):}` `K_(sp) = (2x)^(2).x , K_(sp) = 4x^(3) , K_(sp) = 4 xx (2.5 xx 10^(-2))^(3)` `K_(sp) = 62.5 xx 10^(-6)`. |
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| 149. |
Which one of the following substances will be a mixed saltA. `NaHCO_(3)`B. Ca(OCl)ClC. `K_(2)SO_(4)Al_(2)(SO_(4))_(3).24H_(2)O`D. Mg(OH)Br |
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Answer» Correct Answer - B `CaOCl_(2)` has two anions `Cl^(-)` and `OCl^(-)` along with `Ca^(2+)` ions. |
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| 150. |
For which of the following sparingly soluble salt, the solubility (S) and solubility product `(K_(sp))` are related by the expression `S = (K_(sp)//4)^(1//3)`A. `BaSO_(4)`B. `Ca_(3)(PO_(4))_(2)`C. `Hg_(2)Cl_(2)`D. `Ag_(3)PO_(4)` |
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Answer» Correct Answer - C `BaSO_(4) hArr Ba^(2+) + SO_(4)^(2-) , K_(sp) = S^(2), S = sqrt(K_(sp))` `Ca_(3)(PO_(4))_(2) hArr 3Ca^(2+) + 2PO_(4)^(3-), K_(sp) = (3S)^(3) xx (2S)^(2) = 108 S^(5)` `:. S = ((K_(sp))/(108))^(1//5)` `Hg_(2)Cl_(2) hArr Hg_(2)^(2+) + 2Cl^(-), K_(sp) = S xx (2S^(2)) = 4S^(3)` `:. S = ((K_(sp))/(4))^(1//3)` `Ag_(3)PO_(4)rarr 3Ag^(+) + PO_(4)^(3-), K_(sp) = (3S)^(3) xx S = 27 S^(4)` `:. S = ((K_(sp))/(27))^(1//4)` `CuS hArr Cu^(2+) + S^(2-), K_(sp) = S^(2), S = sqrt(K_(sp))`. |
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