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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.
| 1601. |
Assertion `(A ) :` When an acid or a base is added o water at constant temperature , the pH changes. Reason `(R ) : ` This is due to the change in ionic product of water.A. Both A and R true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - C | |
| 1602. |
Assertion : `0.20 M` solution of `NaCN` is more than basic than `0.20 M` solution of `NaF`. Reason : `K_(a)` of `HCN` is very much less than that of `HF`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-6B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 and Statement-2 is True |
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Answer» Correct Answer - b |
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| 1603. |
Assertion (A) `:` pH of aqueous solutin decreases when NaCN is added to it . Reason (R ) `:` NaCN provides a common ion to HCN.A. Both A and R true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - D | |
| 1604. |
Assertion : `0.20 M` solution of `NaCN` is more than basic than `0.20 M` solution of `NaF`. Reason : `K_(a)` of `HCN` is very much less than that of `HF`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-5B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 and Statement-2 is True |
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Answer» Correct Answer - a |
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| 1605. |
What is `[H^(+)]` of a solution that is 0.01 M in HCN and 0.02 M in NaCN (`K_(a)` for `HCN = 6.2 xx 10^(-10)`)A. `3.1 xx 10^(10)`B. `6.2 xx 10^(5)`C. `6.2 xx 10^(-10)`D. `3.1 xx 10^(-10)` |
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Answer» Correct Answer - D `K_(a) = ([H^(+)][CN^(-)])/([HCN^(-)]) rArr 6.2 xx 10^(-10) = ([H^(+)] [0.02])/([0.01])` `[H^(+)] = (6.2 xx 10^(-10) xx 0.01)/(0.02) = 3.1 xx 10^(-10)`. |
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| 1606. |
Which one of the following pairs of solution is not an acidic buffer?A. `HClO_(4)` and `NaClO_(4)`B. `CH_(3)COOH` and `CH_(3)COONa`C. `H_(2)CO_(3)` and `Na_(2)CO_(3)`D. `H_(3)PO_(4)` and `Na_(3)PO_(4)` |
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Answer» Correct Answer - A Strong acid with its salt cannot form buffer solution. Hence, `HClO_(4)` and `NaClO_(4)` is not an acidic buffer. |
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| 1607. |
`K_(a)` for ascorbic acid `(HASc)` is `5 xx 10^(-5)`. Calculate the `[H^(o+)]` in an aqueous solution in which the concentration of `Asc^(Theta)` ions is `0.02M`.A. `2 xx 10^(-6)`B. `2 xx 10^(-7)`C. `5 xx 10^(-9)`D. `5 xx 10^(-10)` |
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Answer» Correct Answer - C `Asc^(Theta) + H_(2)O hArr HASc^(Theta) + overset(Theta)OH` `[overset(Theta)OH] = Ch = C sqrt(((Kh)/(C))) = sqrt(K_(h)C)` ` = sqrt((K_(w).C)/(K_(a))) = ((10^(-14)xx0.02)/(5xx10^(-5)))^((1)/(2)` `[H_(3)O^(o+)] = (K_(w))/([overset(Theta)OH]) = (10^(-14))/(2xx10^(-16)) = 5 xx 10^(-9)` |
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| 1608. |
Assertion : `BaCO_(3)` is more soluble in `HNO_(3)` than in plain water. Reason : Carbonate is a weak base and reacts with the `H^(+)` from the strong acid, causing the barium salt to dissociate.A. If both assertion and reason are true and reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation is true but reason is false.C. If the assertion and reason both are false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A Barium carbonate is more soluble in `HNO_(3)` than in water because carbonate is a weak base and reacts with the `H^(+)` ion of `HNO_(3)` causing the barium salt to dissociate. `BaCO_(3) + HNO_(3) rarr Ba(NO_(3))_(2) + CO_(2) + H_(2)O` |
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| 1609. |
Determine the volume of 0.125 M NaOH required to titrate to the equivalence point 25.0 mL of a 0.175 M solution of a monoprotic weak acid that is 20 % ionized :A. 7.00 mLB. 17.9 mLC. 28.0 mLD. 35.0 mL |
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Answer» Correct Answer - d |
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| 1610. |
Choose the correct statement(s)A. pH of 0.1 M `CH_(3)` COOH solution decrease with increase in temperatureB. `pH of 0.1 M NH_(4)` OH solution decrease with increse in temperature.C. `10^(-4)` M HCI solution is more acidic than 0.1 M HCN`(Ka_(HCN)=10^(-5)`solutionD. On dilution degree of dissociation of waek electrolytes increase. |
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Answer» Correct Answer - a,d |
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| 1611. |
100 mL of 1 M HCl is mixed with 50 ml of 2M HCl Hence , `[H_3O^+]` is :A. 1.00 MB. 1.50 MC. 1.33 MD. 3.00 M |
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Answer» Correct Answer - c |
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| 1612. |
12.2 gm of sodium salt of weak monbasic acid `(K_(a)=10^(-6))` is dissolved in 100 ml of water and pH of resulting solution is found to be 10 at `25^(@)` C. Then whichg of the following is corrrect?A. Molar mass of salt is `122 gm//mole`B. pH of solution when 1 gm of same weak acid is `10^(-8)`C. Molar mass oof conjugate base of same weak acid is 99D. |
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Answer» Correct Answer - a,b,c,d |
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| 1613. |
The total number of different kind of buffers obtained during the titration of `H_(3)PO_(4)` with `NaOH` are:A. 3B. 1C. 2D. zero |
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Answer» Correct Answer - A `NaH_(2)PO_(4)+H_(3)PO_(4), NaH_(2)PO_(4)+Na_(2)HPO_(4)`, `Na_(2)HPO_(4)+Na_(3)PO_(4)` |
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| 1614. |
A salt `X` is dissolved in water having `pH =7`. The resulting solution has a `pH` more than `7`. The salt is made by neutralisation ofA. A strong acid and strong baseB. A strong acid and weak baseC. A weak acid and weak baseD. A weak acid and strong base. |
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Answer» Correct Answer - D `pH gt 7` implies the solution is basic. Since salt on hydrolysis in water give basic solution. Therefore, it is a salt of weak acid strong base and undergoes anionic hydrolysis in aqueous solution `A^(-) +HOH hArr HA +OH^(-)` |
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| 1615. |
A salt `X` is dissolved in water having `pH =7`. The resulting solution has a `pH` more than `7`. The salt is made by neutralisation ofA. A strong acid and strong baseB. A strong acid and strong weak baseC. A weak acid and weak baseD. A weak acid and strong base |
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Answer» Correct Answer - D Salt of `W_(A)//S_(B)` hydrolyse and the solution is basic, i.e., `pH gt 7`. `pH = (1)/(2) (pK_(w) + pK_(a) + logC)` |
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| 1616. |
Calcium hydroxide is soluble in water with a `K_(sp)` of `1.3xx10^(-6)`. What is the pH of a saturated solution of calcium hydroxide at `25^(@)` C?A. 12.34B. 12.14C. 12.04D. 11.84 |
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Answer» Correct Answer - b |
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| 1617. |
What is the [`OH^(-)`] in a suspension of the antacid `Mg(OH)_(2)`? (`K_(sp)=2.06xx10^(-13)`)A. `7.4xx10^(-5)`MB. `5.9xx10^(-5)`MC. `4.7xx10^(-5)`MD. `3.7xx10^(-5)`M |
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Answer» Correct Answer - a |
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| 1618. |
When a salt of weak acid and weak base is dissolved in water, the pH of the resulting solution will be :A. 7B. depends on the value of `K_a and K_b`C. less than 7D. greater than 7 |
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Answer» Correct Answer - b |
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| 1619. |
Which salt gives the most acidic 0.1 M solution in water ?A. NaClB. `NaNO_3`C. `NH_4Cl`D. `NH_4NO_2` |
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Answer» Correct Answer - c |
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| 1620. |
One litre of a saturated solution of `AgBrO_3` is prepared by adding just sufficient `AgBrO_3` at `27^@C`. To this solution copper wire is dipped such that following redox reaction occures to completion. `Cu(s)2Ag^+(aq)hArr2Ag(s)+Cu^+2(aq)` On drying the wire it was observed that it containde 7.04mg less copper than original. [Atomic weight of Cu =64] Equivalents of copper reacted in the above conditon:A. `1.1xx10^-4`B. `2.2xx10^-4`C. `5.5xx10^-5`D. `0.11` |
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Answer» Correct Answer - b |
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| 1621. |
One litre of a saturated solution of `AgBrO_3` is prepared by adding just sufficient `AgBrO_3` at `27^@C`. To this solution copper wire is dipped such that following redox reaction occures to completion. `Cu(s)2Ag^+(aq)hArr2Ag(s)+Cu^+2(aq)` On drying the wire it was observed that it containde 7.04mg less copper than original. [Atomic weight of Cu =64] What will be the solubility of `AgBrO_3` in another aqueous solution Ph=10 if `K_b` of AgOH is `10^-8`?A. `2.2xx10^-2 M`B. `4.84xx10^-8 M`C. `4.84xx10^-4 M`D. `1.1xx10^-2 M` |
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Answer» Correct Answer - a |
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| 1622. |
Stroniyum fluoride `(SrF_(2.))` is a sparingly soluble slat. Let`s_(1)` be its solubility (in mol/lt.) in pure water at `25^@C`, assuming no hydrolysis of `F^(-)` ions. Also let `s_(2)` be its solubility (in hydrolysis of `F^(-)`ion and no complex formation. However,it is known that `s_(1):s_(2)=10^(6):256`. The `K_(sp)` value of `SrF_(2)` at `25^(@)` C is :A. `2.048xx10^(-9)`B. `1.372xx10^(-9)`C. `1.864xx10^(-9)`D. `2.916xx10^(-9)` |
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Answer» Correct Answer - a |
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| 1623. |
Stroniyum fluoride `(SrF_(2.))` is a sparingly soluble slat. Let`s_(1)` be its solubility (in mol/lt.) in pure water at `25^@C`, assuming no hydrolysis of `F^(-)` ions. Also let `s_(2)` be its solubility (in hydrolysis of `F^(-)`ion and no complex formation. However,it is known that `s_(1):s_(2)=10^(6):256`. The mass of NaF to be added to 100 ml solution of `0.0011MSr^(+2)` ions to reduce its concentration to `2xx10^(-4)`M is : [Assume no hydrolysis of `F^(-)` ions.A. 0.42 gB. 0.063gC. 0.021 gD. 0.084 g |
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Answer» Correct Answer - c |
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| 1624. |
Assertion : The proton transfer reaction between `NH_(3)` and `H_(2)O` proceeds only to a slight extent. Reason : Proton transfer reaction is virtually complete in the case of `HCl` in dilute solution.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-2B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 and Statement-2 is True |
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Answer» Correct Answer - b |
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| 1625. |
At a temperature and under high pressure, `K_w(H_2O)=10^(-10)`.A solution of pH 5.4 under those conditions is said to be :A. acidicB. basicC. neutralD. amphitonic |
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Answer» Correct Answer - b |
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| 1626. |
Assertion (A): Solubility of `AgCN` in acidic solutions is greater than in pure water. Reason (R) : Solubility equilibrium of `AgCN` is shifted in formwed direction due to the formation of `HCN`.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 - A `AgCN(s) hArr Ag^(o+) + Cn^(Theta)` `CN^(Theta) + H^(o+) rarr HCN`. As `CN^(Theta)` ios are consumed by `H^(o+)` ions to form a `W_(A), HCN`, more `AgCN` is dissolved. |
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| 1627. |
The pH of pure water at `50^@C` is 6.63. What is the value of `K_w` at `50^@C`?A. `1.8xx10^(-15)`B. `1.0xx10^(-14)`C. `5.5xx10^(-14)`D. `2.2xx10^(-13)` |
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Answer» Correct Answer - c |
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| 1628. |
At `20^@C` water has `K_w=6.807xx10^(-15)`.What is the pH of pure water at this temperature ?A. 6.667B. `6.920`C. `7.000`D. `7.084` |
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Answer» Correct Answer - d |
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| 1629. |
Liquid `NH_(3)` ionises to a slight extent. At `-50^(@)C`, its ionic product `K_(NH_(3)) = [overset(Θ)Nh_(4)] [overset(Θ)NH_(2)]` is `10^(-30)`. How many amide ions, `overset(o+)NH_(2)` are present per `mm^(3)` of pure liquid `NH_(3)`?A. `6xx10^(6)ions`B. `6xx10^(5)ions`C. `6xx10^(-5)ions`D. `6xx10^(-6)ions` |
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Answer» Correct Answer - B `2NH_(3)hArrNH_(4)^(+)+NH_(2)^(-)` (self ionisation) and `K = [NH_(4)^(+)][NH_(2)^(-)]` `:. [NH_(4)^(+)]=[NH_(2)^(-)]` `:. [NH_(2)^(-)]=sqrt(K)= sqrt(10^(-30))=10^(-15)M` Number of amide ions in `10^(3) cm^(3)` `=10^(-15)xx6xx10^(23)` `:.` Number of amide ions in `10^(3)cm^(3)` `=(10^(-15)xx6xx10^(23))/(10^(3))` `= 6xx10^(5)ions` |
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| 1630. |
Liquid `NH_(3)` ionises to a slight extent. At `-50^(@)C`, its ionic product `K_(NH_(3)) = [overset(Θ)Nh_(4)] [overset(Θ)NH_(2)]` is `10^(-30)`. How many amide ions, `overset(o+)NH_(2)` are present per `mm^(3)` of pure liquid `NH_(3)`? |
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Answer» `2NH_(3) hArr overset(oplus)(N)H_(4)+ overset(oplus)(N)H_(2)` `K_(NH_(3)) = [overset(o+)NH_(4)] [overset(Θ)NH_(2)] = 10^(-30)` Since each ion ins produced in equal number of moles `:. [overset(Θ)NH_(2)] = [overset(o+)NH_(4)] = 10^(-15)M` Number of `"ions/ mm"^(3)` `= ((10^(-15) "mol")/(L)) ((1L)/(10^(6)mm^(3))) ((6xx10^(23) ions)/("mol"))` `= (10^(-15)xx1xx6 xx 10^(23))/(10^(6)) = 600 "ions mm"^(-3)` |
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| 1631. |
Hydrogen peroxide ionises as `H_2O_2 hArr H^+ + HO_2^-` if pH of `H_2O_2` is 5.7 at `25^@C`, the ionic product of `H_2O_2` is :A. `2xx10^(-12)`B. `4xx10^(-12)`C. `1xx10^(-14)`D. `1.4xx10^(-12)` |
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Answer» Correct Answer - b |
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| 1632. |
Solutions X and Y contain one mole and two mols of `CH_(3)XOONH_(4)` per litre respectively. The extent of hydrolysisA. More in XB. More in YC. Same in bothD. unpredictable |
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Answer» Correct Answer - C Degree of dissociation is independent of concentration. |
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| 1633. |
`pH` of a solution solution at `25^(@)C` is `2`. If the pH is to be doubled , the hydronium ion concentration of the solution should beA. doubledB. halvedC. decreased by 100 timesD. increased by 100 times. |
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Answer» Correct Answer - C `pH=2 :. [H_(3)O^(+)]=10^(-2)` For `pH =4, [H_(3)O^(+)]=10^(-4)` This means , `[H^(+)]` conc. Should decrease from `10^(-2)` to `10^(-4)`, i.e., by 100 times. |
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| 1634. |
Which of the following would not change the pH of `10 c.c.` of dil `HCl` when added ?A. `5c.c.` of pure waterB. `20c.c` of pure waterC. `10c.c.` of pure waterD. `20c.c.` of dil `HCl` of same conentration |
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Answer» Correct Answer - D Addition of water causes change in `p^(H)` due to change in concentration |
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| 1635. |
`pH` of a solution solution at `25^(@)C` is `2`. If the pH is to be doubled , the hydronium ion concentration of the solution should beA. HalvedB. Increased to `100` timesC. DoubledD. Decreased to `100` times |
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Answer» Correct Answer - D Every `10` times dilution `p^(H)` is increased by one unit |
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| 1636. |
The compound whose `0.1 M` solution is basic isA. ammonium acetateB. ammonium chlorideC. ammonium sulphateD. sodium acetate |
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Answer» Correct Answer - D Sodium acetate undergoes anionic hydrolysis giving basic solution. `CH_(3)COO^(-)+HOH hArr CH_(3)COOH+OH^(-)` |
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| 1637. |
Which of the following statements do not require the assumption of a perfect gas in order to be true ?A. `Delta E=Q-W`B. `C_(p)=C_(v)+R`C. `Delta H =Delta E +RT . Delta n`D. `PV^(y)=` constant |
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Answer» Correct Answer - A The question are base on facts. |
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| 1638. |
A weak acid, HA, has a `K_(a)` of `1.00xx10^(-5)`. If `0.100` mol of the acid is dissolved in 1 L of water, the percentage of the acid dissociated at equilibrium is the closed toA. `99.0%`B. `1.00%`C. `99.9%`D. `0.100%` |
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Answer» Correct Answer - B `HA hArr H^(+)+A^(-)` At equilibrium `[H^(+)=A^(-)]` `K_(a)=([H^(+)][A^(-)])/([HA])=([H^(+)]^(2))/([HA])` `[H^(+)]=sqrt(K_(a)[HA])=sqrt(1xx10^(-5)xx0.1)` `=sqrt(1xx10^(-6))=1xx10^(-3)` `alpha = ("Actual ionisation")/("Molar concentration")=(10^(-3))/(0.1)=10^(-2)` % of acid dissociated `=10^(-2)xx1.00=1%=100%` |
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| 1639. |
A weak acid HA has a `K_(a)` of `1.00 xx 10^(-5)`. If 0.100 mol of this acid is dissolved in one litre of water the percentage of acid dissociated at equilibrium is closed toA. `99.0%`B. `1.00%`C. `99.9%`D. `0.100%` |
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Answer» Correct Answer - B `K_(a) = 1 xx 10^(-5)` `{:(HA,hArr,H^(+),+,A^(-)),(.1m//l,,0,,0),(.1-x,,x,,x):}` `10^(-5) = (x^(2))/(1-x)` Assume `x lt lt.1` `x^(2) = 10^(-6)` or `x = 10^(-3)`, % dissociation `= (10^(-3))/(.1) xx 100 = 1%` |
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| 1640. |
What is the pH of 1 M solution of acetic acid ? To what volume one litre of this solution be diluted so that pH of the resulting solution will be twice of the original value ? `(K_(a)=1.8xx10^(-5))` |
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Answer» Case I `{:(,CH_(3)COOH,hArr,CH_(3)COO^(Theta)+,H^(oplus)),("Concentration before dissociation",1,,0,0),("Concentration after dissociation",1-alpha,,alpha,alpha):}` `[H^(o+)] = Calpha = C sqrt((K_(a)//C))` `=sqrt((K_(a)C)) = sqrt((1.8 xx 10^(-5)xx1))` `= sqrt((18 xx 10^(-6)) = 4.24 xx 10^(-3) M` `:. pH =- log H^(o+) =- log (4.24 xx 10^(-3)) = 2.3724`or Use formula for the `pH` of weak acid. `pH = (1)/(2) (pK_(a) - log C) = (1)/(2) (4.7447 - 0) = 2.3724` Case II: New `pH` is given as `= 2.3724 xx 2 = 4.7488` Let new concentration be `C_(1)` and dergee of dissociation be `alpha_(1)` `:. -log [H^(o+)] = 4.7448` `:. [H^(o+)] = 1.8 xx 10^(-5)`or `C_(1) alpha_(1) = 1.8 xx 10^(-5)` Now again `K_(a) = ([H^(o+)][CH_(3)COO^(Θ)])/([CH_(3)COOH])` `K_(a) = (C_(1)alpha_(1) xx C_(1)alpha_(1))/(C_(1)(1-alpha_(1))) = (C_(1)alpha_(1)alpha_(1))/((1-alpha_(1)))` `:. 1.8 xx 10^(-5) = (1.8 xx 10^(-5) xx alpha_(1))/((1-alpha_(1))) :. alpha_(1) = 0.5` Now `C_(1)alpha_(1) = 1.8 xx 10^(-5)` `:. C_(1) = (1.8 xx 10^(-5))/(alpha_(1)) = (1.8 xx 10^(-5))/(0.5) = 3.6 xx 10^(-5)M` Let `1L` of concentrated solution be diluted to `VL` Equivalent of dilute solution `=` Equivalent of concentrated solution `3.6 xx 10^(-5) xxV = 1xx1 [M_(1)V_(1) = M_(2)V_(2)]` `:. V = (1)/(3.6 xx 10^(-5)) = 2.77 xx 10^(4)L` Note: iIn `II` case `alpha` comes `0.5` by `K_(a) = (C_(1)alpha_(1)^(2))/((1-alpha_(1)))`, and thus, it is not advisible to assume `(1-alpha_(1)) ~~1`. |
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| 1641. |
The dissociation constant of a substituted benzoic acid at `25^(@)C` is `1.0 xx 10^(-4)`. The pH of a 0.01 M solution of its sodium salt is |
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Answer» Correct Answer - 8 Given `K_(a) = 10^(-4), p_(a) = 4, C = 0.01 M` `pH = 7 + (1)/(2) pK_(a) + (1)/(2)log C = 7 + (1)/(2) (4) + (1)/(2)(-2) = 8`. |
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| 1642. |
Solution of a mono basic acid has a pH =5 . If 1 mL of it is diluted to 1 litre , what will be the pH of the resulting solution ?A. 3.45B. 6.96C. 8.58D. 10.25 |
| Answer» Correct Answer - 2 | |
| 1643. |
A solution of `HCl` has a `HCl` has a `pH=5` if one mL of it is dilluted to `1` litre what will be `pH` of resulting solution. |
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Answer» `[HCl]_(f)=10^(-5)M` since `pH=5` since volume of original solution has been made `1000` times so concentrated of solution will decrease by `1000` times. `[HCl]_(f)=10^(-8)M` so `H^(+)` form water should also be considered (as done in solved example-8) then `pH` of resulting solution =`6.96`. |
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| 1644. |
Explain why `CoS` is more soluble than predicted by the `K_(sp)`. |
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Answer» b. Not all the sulphide which dissolves remains as `S^(2-)`, most of it hydrolyses: `S^(2-) +H_(2)O hArr HS^(Θ) + overset(Θ)OH`. |
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| 1645. |
The solubility of `CuS` in pure water at `25^(@)C` is `3.3 xx 10^(-4)g L^(-1)`. Calculate `K_(sp)` of `CuS`. The accurate value of `K_(sp)` of `CuS` was found to be `8.5 xx 10^(-36)` at `25^(@)C`. |
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Answer» c. `CuS rarr Cu^(2+) +S^(2-)` `S = (3.3 xx 10^(-4)gL^(-1))/(95.6g mol^(-1)) = 3.5 xx 10^(-6)M`. Apparent `K_(sp) = (3.5 xx 10^(-6))^(2) = 1.2 xx 10^(-11)`. |
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| 1646. |
Calculate the `[H^(o+)]` and `[overset(Theta)OH]` of `0.0315g` of `HNO_(3)` in `500 mL` of water. Calculate `pH` and `pOH` also. |
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Answer» Correct Answer - A::C `(Mw_(2)` of `HNO_(3) = 53 g mol^(-1))` `M_(HNO_(3)) = (W_(2) xx 1000)/(Mw_(2) xx "vol of sol. In mL")` `= (0.0315 xx 1000)/(63xx500) = 0.001= 10^(-3)M`. `[H^(o+)] = 10^(-3)M,pH = 3` `[overset(Theta)OH] = (K_(w))/([H^(o+)]) = (10^(-14))/(10^(-3)) = 10^(-11)M` `pOH = 11`. |
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| 1647. |
Which of the following consitute a set of atomspheric species?A. `H_(2)O,H_(2)PO_(3)^(Theta), HPO_(4)^(2-)`B. `HC_(2)O_(4)^(Theta),H_(2)PO_(4)^(Theta),SO_(4)^(2-)`C. `H_(2)O,HPO_(4)^(2-),H_(2)PO_(2)^(Theta)`D. `H_(3)O^(o+),H_(2)PO_(4)^(Theta),HCO_(3)^(Theta)` |
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Answer» Correct Answer - A a. Atmospheric species are those which can loose a proton and can accept a proton. a. `H_(3)O^(o+)hArr H_(2)O underset(-H^(o+))hArr Ooverset(Theta)H + H^(o+)` `H_(3)PO_(3) underset(+H^(o+))hArr underset("acid")underset("Dibasic")(H_(2)PO_(3)^(Theta))underset(-H^(o+))hArrHPO_(3)^(2-)+H^(o+)` `H_(2)PO_(4)^(Theta)underset(+H^(o+))hArrunderset("acid")underset("Tribasic")(HPO_(4)^(2-))underset(-H^(o+))hArrPO_(4)^(3-)+H^(o+)` b. `H_(2)C_(2)O_(4)overset(+H^(o+))hArrunderset("acid")underset("Dibasic")(HC_(2)O_(4)^(Theta))overset(-H^(o+))hArrC_(2)O_(2)^(2-)+H^(o+)` `H_(3)PO_(4)overset(+H^(o+))hArrunderset("acid")underset("Tribasic")(H_(2)PO_(4)^(Theta))overset(-H^(o+))hArrHPO_(4)^(2-)+H^(o+)` `HSO_(4)^(Theta) overset(+H^(o+))hArr underset("acid")underset("Monobasic")(SO_(4)^(2-))rarr "Cannot loose proton"` c. `H_(3)O^(o+) overset(+H^(o+))hArr H_(2)O overset(-H^(o+))hArr overset(Theta)OH +H^(o+)` `H_(2)PO_(4)^(Theta) overset(+H^(o+))hArr underset("Tribasic acid")(HPO_(4)^(2-))overset(-H^(o+))hArrPO_(4)^(3-)+H^(o+)` `H_(3)PO_(2)^(Theta)overset(+H^(o+))hArrunderset("Monobasic acid")(H_(2)PO_(2)^(Theta))rarr "Cannot loss proton"` d. `"Cannot accept proton" larr H_(3)O^(o+)overset(-H^(o+))hArrH_(2)O+H^(o+)` `H_(3)PO_(4)overset(+H^(o+))hArrunderset("Tribasic acid")(H_(2)PO_(4)^(Theta))overset(-H^(o+))hArrHPO_(4)^(2-)+H^(o+)` `H_(2)CO_(3)overset(+H^(o+))hArrunderset("Dibasic acid")(HCO_(3)^(Theta))overset(-H^(o+))hArrCO_(3)^(2-)+H^(o+)` Hence answer is (a). |
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| 1648. |
Which one of the following compounds is not a protoric acid?A. `SO_(2)(OH)_(2)`B. `B(OH)_(3)`C. `PO(OH)_(3)`D. `SO(OH)_(2)` |
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Answer» Correct Answer - B `B(OH)_(5)` in water gives `H^(+)` ions as follows `B(OH)_(3)+H_(2)O hArr [B(OH)_(4)^(-)]+H^(+)` Thus, it is not a protonic acid i.e., it does not give its hydrogen `H^(+)` ions. |
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| 1649. |
The basic character of the transition metal monoxide follows the orderA. `TiO gt VO gtCrO gtFeO`B. `VO gt CrO gt TiO gt FeO`C. `TiO gtFeO gtVOgt CrO`D. |
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Answer» Correct Answer - A In a period, metallic character decreases and as such basic character of monoxide also decrease, thus correct order is `TiOgtVOgtCrO gtFeO`. |
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| 1650. |
The solubility of AgCl in 0.2 M NaCl is `[K_(sp) AgCl =1.8 xx 10^(-10)]`A. `1.8 xx 10^(-11)M`B. `9.0 xx 10^(-10)M`C. `6.5 xx 10^(12)M`D. `5.6 xx 10^(-11)M` |
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Answer» Correct Answer - B `K_(sp)(AgCl)=[Ag^(+)][Cl^(-)]` Let the solubility of AgCl in 0.2 M NaCl be s mol `L^(-1)` `:. K_(sp)=(s) (Cl^(-))` `1.8 xx 10^(-10)=s xx 0.2` `:. s=(1.8 xx 10^(-10))/(0.2)=9 xx 10^(-10)M` |
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