1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 13 InterviewSolution

601.	What is the conjugate acid of `NH_(3)`?
Answer» Conjugate acid of ` overset(*****)(N)H_(3) " is " overset(+)(N)H_(4)` ion.

Discussion

602.	Identify acid-base according to Lewis conceptS2-, H+, OH–, BF3, Ni2+ , F–
Answer» According to Lewis concept, acids are electron pair acceptor and bases are electron pair donor. Lewis acid : H⁺, BF₃, Ni²⁺ Lewis basis : S^2-, OH^–, F^–

Discussion

603.	(a) Derive a relationship between `K_(p) " and " K_(c) ` (b) What is a buffer solution? (C) Define acid according to Lewis concept. (d) Calculate pH value of 0.01 m HCI.
Answer» (a) `K_(p) =K_(c) RT^(Deltang)`. (b) For the definition of buffer solution . ( C) According to Lewis concept, an acid is a substance which can accept a pair of electrons. (d) `pH =- log [H^(+)] =- log [10^(-2)] =(-) (-2) log^(10) =2`

Discussion

604.	Precipitation takes place when the ionic productA. equals their solubility productB. exceeds their solubility productC. is less than their solubility productD. is almost zero.
Answer» Correct Answer - B Precipiation of salt takes place when ionic products exceeds solubility product.

Discussion

605.	How much sodium acetate should be added to 0.1m solution of `CH_(3)C O O H` to give a solution of pH 5.5 `(pK_(a) "of" CH_(3)C O O H=4.5)`.A. 0.1 mB. 0.2 mC. 1.0 mD. 10.0 m
Answer» Correct Answer - C `pH=pK_(a)+log .([CH_(3)CO ONa])/([CH_(3)CO OH])` `5.5=4.5+log .([CH_(3)CO ONa])/(0.1)` `=4.5 + log [CH_(3)CO ONa]+1` or` log [CH_(3)CO ONa]=0` `:. [CH_(3)CO ONa]=1m`

Discussion

606.	An organic base X has Kb equal to 1 x 10-8 . In what amount should 0.01 M HCI and 0.01 M solution of X be mixed to prepare 1 litre of a buffer solution having pH 7?
Answer» K_b = 1 × 10⁻⁸ 0 ∙ 01 M HCl, 0 ∙ 01 M of solution × (Base) Total volume = 1 litre pH = 7 ∙ 0, X+ H₂O ⇌ XH+ + OH− K_b = \(\frac{[XH^+][OH^-]}{[X]}\) = 1 × 10⁻⁸ pOH = pK_b + log\(\frac{[XH^+]}{[X]}\) 7 = − log 10⁻⁸ + log\(\frac{[XH^+]}{[X]}\) 7 = 8 + log\(\frac{[XH^+]}{[X]}\) log\(\frac{[XH^+]}{[X]}\) = -1 \(\frac{[XH^+]}{[X]}\) =10^-1= 0.1 Suppose volume of HCl taken is x l. Volume of base = (1 - x) l After the reaction base left = (1 - x - x) of 0.01 M XH⁺ formed = x L of 0 ∙ 01 M \(\frac{[XH^+]}{[X]}\)= \(\frac{x}{1-2x}\) = 0.1 x = 0 ∙ 083 Volume of HCl = 0 ∙ 083 L Volume of Bases = 0 ∙ 917 L

606.

An organic base X has Kb equal to 1 x 10-8 . In what amount should 0.01 M HCI and 0.01 M solution of X be mixed to prepare 1 litre of a buffer solution having pH 7?

Answer»

K_b = 1 × 10⁻⁸

0 ∙ 01 M HCl, 0 ∙ 01 M of solution × (Base)

Total volume = 1 litre

pH = 7 ∙ 0,

X+ H₂O ⇌ XH+ + OH−

K_b = \(\frac{[XH^+][OH^-]}{[X]}\)

= 1 × 10⁻⁸

pOH = pK_b + log\(\frac{[XH^+]}{[X]}\)

7 = − log 10⁻⁸ + log\(\frac{[XH^+]}{[X]}\)

7 = 8 + log\(\frac{[XH^+]}{[X]}\)

log\(\frac{[XH^+]}{[X]}\) = -1

\(\frac{[XH^+]}{[X]}\) =10^-1= 0.1

Suppose volume of HCl taken is x l.

Volume of base = (1 - x) l

After the reaction base left

= (1 - x - x) of 0.01 M

XH⁺ formed = x L of 0 ∙ 01 M

\(\frac{[XH^+]}{[X]}\)= \(\frac{x}{1-2x}\)

= 0.1

x = 0 ∙ 083

Volume of HCl = 0 ∙ 083 L

Volume of Bases = 0 ∙ 917 L

Discussion

607.	Calculate the pH of a solution obtained by mixing 50 ml of 0.1 M NaOH with 100 ml of 0.1 M CH3COOH, Ka for CH3COOH = 1.8 x 10-5 ?
Answer» 50 ml of 0 ∙ 1 M NaOH contains = 5 m mol NaOH 100 ml of 0∙1 M CH₃COOH contains = l0 m mol CH₃COOH CH₃COOH + NaOH → CH₃COONa + H₂O CH₃COOH left unreacted = 10 – 5 = 5 m mol CH₃COONa formed = 5 m mol Volume of solution = 150 [CH₃COOH] = \(\frac{5}{150}\) = \(\frac{1}{30}\)M [CH₃COONa] = \(\frac{5}{150}\) = \(\frac{1}{30}\) M This solution will behave as an acidic buffer. pH = pK_a + log\(\frac{[Salt]}{[Acid]}\) = − log (1∙8 × 10⁻⁵) + log\(\frac{\frac{1}{30}}{\frac{1}{30}}\) = 4∙74.

607.

Calculate the pH of a solution obtained by mixing 50 ml of 0.1 M NaOH with 100 ml of 0.1 M CH3COOH, Ka for CH3COOH = 1.8 x 10-5 ?

Answer»

50 ml of 0 ∙ 1 M NaOH contains

= 5 m mol NaOH

100 ml of 0∙1 M CH₃COOH contains

= l0 m mol CH₃COOH

CH₃COOH + NaOH → CH₃COONa + H₂O

CH₃COOH left unreacted = 10 – 5

= 5 m mol

CH₃COONa formed = 5 m mol

Volume of solution = 150

[CH₃COOH] = \(\frac{5}{150}\) = \(\frac{1}{30}\)M

[CH₃COONa] = \(\frac{5}{150}\) = \(\frac{1}{30}\) M

This solution will behave as an acidic buffer.

pH = pK_a + log\(\frac{[Salt]}{[Acid]}\)

= − log (1∙8 × 10⁻⁵) + log\(\frac{\frac{1}{30}}{\frac{1}{30}}\)

= 4∙74.

Discussion

608.	Will the PH of water is same at 277K and 298K?
Answer» No ,As the temperature increases ,dissociation of water take place to greater extent H⁺ ion concentration PH decreases.

Discussion

609.	What should be the ratio of concentrations of formate ion and formic acid in a buffer solution so that its pH should be 4? Around what pH will this buffer have maximum buffer capacity?Ka = 1.8 x 10-4?
Answer» pH = pK_a + log\(\frac{[Salt]}{[Acid]}\) 4 = − log(1.8 x 10^-4) + log\(\frac{[Formate]}{[Formic\,Acid]}\) 4 = 3∙74 + log\(\frac{[Formate]}{[Formic\,Acid]}\) log\(\frac{[Formate]}{[Formic\,Acid]}\) = 4- 3.47 = 0.26 \(\frac{[Formate]}{[Formic\,Acid]}\) = 1.8 The buffer capacity of this solution would be maximum near the pK_a of the acid. \(\therefore\) For maximum buffer capacity pH = pK_a = − log K_a = − log (1∙8 × 10⁻⁴) = 3∙74.

609.

What should be the ratio of concentrations of formate ion and formic acid in a buffer solution so that its pH should be 4? Around what pH will this buffer have maximum buffer capacity?Ka = 1.8 x 10-4?

Answer»

pH = pK_a + log\(\frac{[Salt]}{[Acid]}\)

4 = − log(1.8 x 10^-4) + log\(\frac{[Formate]}{[Formic\,Acid]}\)

4 = 3∙74 + log\(\frac{[Formate]}{[Formic\,Acid]}\)

log\(\frac{[Formate]}{[Formic\,Acid]}\) = 4- 3.47 = 0.26

\(\frac{[Formate]}{[Formic\,Acid]}\) = 1.8

The buffer capacity of this solution would be maximum near the pK_a of the acid.

\(\therefore\) For maximum buffer capacity

pH = pK_a = − log K_a

= − log (1∙8 × 10⁻⁴)

= 3∙74.

Discussion

610.	Account for the following: (a) A solution of Na2CO3 is alkaline. (b) Ba(OH)2 is soluble in water while Baso4 is almost insoluble.
Answer» (a) Na₂CO₃ + H₂O ⇌ H₂CO₃ + 2 NaOH or CO₃²⁻ + 2 Na⁺ + 2H₂O ⇌ 2Na⁺ +2OH⁻+ H₂CO₂ Or CO₃²⁻ + 2H₂O ⇌ H₂CO₃ + 2OH⁻ Since N₂CO₃ produces OH⁻ ions in solution, hence N₂CO₃solution is alkaline is nature. (b) The lattice enthalpy of BaSO₄ is much more than its hydration enthalpy and hence it is insoluble in water. But hydration enthalpy of Ba(OH)₂ is much more than lattice enthalpy. Therefore, Ba(OH)₂ is soluble in water.

610.

Account for the following: (a) A solution of Na2CO3 is alkaline. (b) Ba(OH)2 is soluble in water while Baso4 is almost insoluble.

Answer»

(a) Na₂CO₃ + H₂O ⇌ H₂CO₃ + 2 NaOH

or CO₃²⁻ + 2 Na⁺ + 2H₂O ⇌ 2Na⁺ +2OH⁻+ H₂CO₂

Or CO₃²⁻ + 2H₂O ⇌ H₂CO₃ + 2OH⁻

Since N₂CO₃ produces OH⁻ ions in solution, hence N₂CO₃solution is alkaline is nature.

(b) The lattice enthalpy of BaSO₄ is much more than its hydration enthalpy and hence it is insoluble in water. But hydration enthalpy of Ba(OH)₂ is much more than lattice enthalpy. Therefore, Ba(OH)₂ is soluble in water.

Discussion

611.	Calculate the pH of a solution which is 0.1 M in HA and 0.5 M in NaA. Ka for HA is 1.8 × 10-6 ?
Answer» HA + H₂O ⇌ H₃O⁺ + A⁻ NaA → Na⁺ + A^- pH = pK_a + log\(\frac{[Salt]}{[acid]}\) = − log (1∙8 × 10⁻⁶) + log \(\frac{[0.5]}{[0.1]}\) = 5∙744 + 0∙6990 = 6∙44.

611.

Calculate the pH of a solution which is 0.1 M in HA and 0.5 M in NaA. Ka for HA is 1.8 × 10-6 ?

Answer»

HA + H₂O ⇌ H₃O⁺ + A⁻

NaA → Na⁺ + A^-

pH = pK_a + log\(\frac{[Salt]}{[acid]}\)

= − log (1∙8 × 10⁻⁶) + log \(\frac{[0.5]}{[0.1]}\)

= 5∙744 + 0∙6990

= 6∙44.

Discussion

612.	Why is the PH of blood remains almost constant?
Answer» Blood is a buffer solution and resist the change in PH.

Discussion

613.	SO2-3 is Bronsted base or acid and why?
Answer» SO^2-₃ is Bronsted base because it can accept proton (H⁺).

Discussion

614.	In a vessel containing SO2, SO3 and O2 at equilibrium some helium gas is introduce so that The pressure increases while temperature and volume remain constant. According to Le-Chatelier’s principal, the dissociation of SO3(a) Increases (b) Decreases(c) Remains unaltered (d) Changes unpredictably.
Answer» (c) Remains unaltered

Discussion

615.	Why pH of our blood remains almost constant at 7.4 though we quite often eat spicy food?
Answer» pH of our blood remains almost constant at 7.4 because of the buffer system present in the blood. The buffer for maintaining acid-base balance in the blood is the carbonic acid bicarbonate buffer. So, pH of blood remains constant after eating spicy food.

Discussion

616.	The dissociation constants for acetic acid and HCN at `25^(@)C` are `1.5xx10^(-5)` and `4.5xx10^(-10)` , respectively. The equilibrium constant for the equilibirum `CN^(-) + CH_(3)COOHhArr HCN + CH_(3)COO^(-)` would beA. `3.3xx10^(-5)`B. `3.3xx10^(-4)`C. `3.3xx10^(4)`D. `3.3xx10^(5)`
Answer» Correct Answer - C

Discussion

617.	The active mass for any pure liquid or pure solidA. is always taken as 1B. is always infinityC. is always zeroD. would depend on the nature of the solid
Answer» Correct Answer - A Since pure solids and pure liquids do not appear in the equilibrium constant expression (because their active masses are incorporated into equilibrium constant), their active mass is taken as 1.

Discussion

618.	A reversible chemical reaction is said to be at equilibrium whenA. equal amounts of reactants and products are presentB. reactants are completely converted into productsC. the rates of the forward and backward reactions become equalD. products cannot be converted into reactants and vice versa.
Answer» Correct Answer - C Chemical equilibrium exists when two opposing reactions of a reversible process occur simultaneoulsy at the same rate.

Discussion

619.	Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases ? `OH^(_), RO^(-), CH_(3)CO O^(-), Cl^(-)`
Answer» Conjugate acids of given bases are : `H_(2)O, ROH, CH_(3)CO OH, HCl`. Their acidic strength is in the order `HCl gt CH_(3)CO OH gt H_(2)O gt ROH`. Hence, their conjugate bases have strength in the order : `Cl^(-) lt CH_(3)CO O^(-) lt OH^(-) lt RO^(-)` or `RO^(-) gt OH^(-) gt CH_(3)CO O^(-) Cl^(-)`.

Discussion

620.	Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?OH–, RO– , CH3COO– , Cl–
Answer» Conjugate acids of given bases are H₂O, ROH, CH₃ COOH, HCl. Their acidic strength is in the order HCl > CH₃ COOH > H₂O >ROH Hence, basic strength is in the order RO^-> OH^- > CH₃COO^- >Cl^-

620.

Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?OH–, RO– , CH3COO– , Cl–

Answer»

Conjugate acids of given bases are H₂O, ROH, CH₃ COOH, HCl.

Their acidic strength is in the order

HCl > CH₃ COOH > H₂O >ROH Hence, basic strength is in the order RO^-> OH^- > CH₃COO^- >Cl^-

Discussion

621.	Give the formulae of the conjugate acids of the following : , `(i) NH_(3) (ii) HS^(-) (iii) HSO_(4)^(-) (iv) H_(2)O`
Answer» Correct Answer - `(i) NH_(4)^(+) (ii) H_(2)S (iii) H_(2)SO_(4) (iv) H_(3)O^(+)`

Discussion

622.	An equilibrium mixture for the reaction `2H_(2)S(g) hArr 2H_(2)(g) + S_(2)(g)` had 1 mole of `H_(2)S, 0.2` mole of `H_(2)` and 0.8 mole of `S_(2)` in a 2 litre flask. The value of `K_(c)` in mol `L^(-1)` isA. `0.004`B. `0.08`C. `0.016`D. `0.016`
Answer» Correct Answer - C

Discussion

623.	For the reaction `CH_(4)(g)+2O_(2)(g)hArr CO_(2)(g)+2H_(2)O(I)` `Delta_(r)H= - 170.8kJ mol^(-1)` Which of the following statements is not true ?A. Addition of `CH_(4)(g)` or `O_(2)(g)` at equilibrium will cause a shift to the right.B. The reaction is exothermic.C. At equilibrium, the concentrations of `CO_(2)(g)` and `H_(2)O` are not equal.D. The equilibrium constant for the reaction is given by `K_(p)= ([CO_(2)])/([CH_(4)][O_(2)])`
Answer» Correct Answer - D `K_(P)` is referred to as pressure equilibrium constant. Thus, according to the law of chemical equilibrium, `K_(p)= (P_(CO_(2)))/(P_(CH_(4))P_(O_(2))^(2))`

Discussion

624.	A large value of equilibrium constant shows thatA. the reaction is taking place at high temperatureB. the reaction is very little in the forward as well as backward directionC. the reaction is less in the forward direction and more in the backward directionD. the forward reaction occurs to a greater extent than the reverse reaction
Answer» Correct Answer - D The numberical value of the equilibrium constant for a reaction indicates the extent of the reaction. The magnitude of `K_(eq)` is directly proportional to the concentration of products but inversely proportional to the concentration of reactants.

Discussion

625.	A solution containing `Na_(2)CO_(3)` and NaOH requires 300 mL of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 M HCl is required. The amount of NaOH present in the original solution isA. 0.5 gB. 1gC. 2gD. 4g
Answer» Correct Answer - B 300 ml of 0.1 N HCl will neutralize all the NaOH and half of `Na_(2)CO_(3)` (with phenolphthalein as indicator). The remaining half will be neutralized by 25 ml of 0.2 N HCl (with methyl orange as indicator). Thus, 0.2 N HCl required for half neutralization of `Na_(2)CO_(3) = 25 ml` or 0.1 N HCl required will be 50 ml. Hence, 0.1 N HCl used up for NaOH = 300 - 50 = 250 ml . Gram equivalents present in 250 ml of 0.1 N HCl `=(0.1)/(1000)xx250=0.025`. It will neutralize 0.025 g eq. of NaOH, i.e., `0.025xx40g` = 1 g.

Discussion

626.	Adding a catalyst to a reaction at equilibriumA. changes QB. changes `K_(eq)`C. changes both `Q` and `K_(eq)`D. changes neither `Q` nor `K_(eq)`
Answer» Correct Answer - D Adding a catalyst to a reaction at equilibrium has no effect on the value of Q and `K_(aq)` because a catalyst affects the activation energy of both forward and reverse reactions equally. It changes both rate constants `(K_(f)` and `K_(b))` by the same factor. So their ratio, `K_(aq)`, does not change.

Discussion

627.	What is the effect of increasing pressure on the equilibrium of the equilibrium `N_(2) + 3H_(2) hArr 2NH_(3)?`
Answer» On increasing pressure, equilibrium shifts in the direction in which pressure decreases, i.e., having lesser number of moles, Hence, equilibrium will shift in the forward direction forming more of ammonia.

Discussion

628.	How many litres of water must be added to `1 L` of an aqueous solution of `HCl` with a `pH` of `1` to create an aqueous solution with `pH` of `2`?A. `9.0 L`B. `0.1 l`C. `0.9 l`D. `2.0 L`
Answer» Correct Answer - A Initial conc. Of HCI solution `(pH=1) =10^(-1) M` Final conc. Of HCI solution `(pH =2) =10^(-2) M` `M_(1)V_(1) =M_(2)V_(2)` `0.1 xx 1L 0.01 xx V_(2)` `V_(2) =((0.1) xx (1L))/(0.01) =10L` Volume of water to be added `=(10 -1) =9 L`

Discussion

629.	What is the effect of catalyst on the equilibrium state in a chemical reaction ?
Answer» Correct Answer - No. effect

Discussion

630.	Classify F- species into Lewis acids and Lewis bases and show how these act as Lewis acids/base.
Answer» F^- :- Lewis base because it is capable of donating an unshared pair of electrons.

Discussion

631.	Explain why the gas fizzes out when soda water bottle is opened?
Answer» In soda water bottle carbon dioxide gas is dissolved under pressure. As soon as the bottle is opened, the pressure inside the bottle tends to decrease, so the solubility decreases and gas escapes out.

Discussion

632.	Classify OH- species into Lewis acids and Lewis bases, and show how these act as Lewis acids/base.
Answer» OH^- :- Lewis base because it is an electron pair donor.

Discussion

633.	Consider the following equilibrium system: `SO_2(g)+1/2O_2(g)hArr SO_3(g)` set up in a cylinder fitted with a piston. Some inert gas is added and the piston is moved outwards to keep the total gaseous pressure constant. Predict which of the following is true?A. Addition of inert gas does not affect the equilibrium.B. Less `SO_3(g)` is product.C. More `SO_3(g)` is produced.D. The system moves to new equilibrium position which cannot be predicted theoretically.
Answer» Correct Answer - B Whenever we add an inert gas at constant total pressure, the equilibrium shifts in that direction which tends to increase the total number of moles of gas. For this system, it is the reverse direction.

Discussion

634.	The solubility of `A_(2)X_(3)` is y mol `dm^(-3)` . Its solubility product isA. `6 y^(4)`B. `64 y^(4)`C. `36 y^(5)`D. `108 y^(5)`
Answer» Correct Answer - D `A_(2)X_(3)rarr2A^(3+)+3X^(-2)`, `K_(ap)=(2y)^(2)xx(3y)^(3)=108y^(5)`.

Discussion

635.	The solubility of AgCl (s) with solubility product `1.6xx10^(-10)` in 0.1 M NaCl solution would beA. `1.26xx10^(-5) M`B. `1.6xx10^(-9) M`C. `1.6xx10^(-11)M`D. zero
Answer» Correct Answer - B `{:(AgCl(s),overset(+aq)hArr ,Ag^(+)(aq),+,Cl^(-)(aq)),(S,,S,,S+0.M"(from NaCl)"),(,,,,~=0.1M (" as " S lt lt 0.1)):}` `K_(sp)=[Ag^(+)][Cl^(-)]` `1.6xx10^(-10)=S(0.1) "or" S=1.6xx10^(-9)M`

Discussion

636.	Which of the following is the best conductor of electricity ?A. `1M H_(3)PO_(4)`B. `1M H_(2)SO_(4)`C. `1M H_(2)CO_(3)`D. `1M HCI`
Answer» Correct Answer - B Carbonic acid `(H_(2)CO_(3))` and phosphoric acid `(H_(3)PO_(4))` are weak acids and, hence, weak electrolytes while hydrochloric acid (HCI) and sulphuric acid `(H_(2)SO_(4))` are strong acids and, hence, strong electrolytes. Strong electrolytes conduct electrity well in dilute aqueous solution while weak electrolytes conduct electricity poorly in dilute aqueous solultion. `1 M H_(2)SO_(4)` produces 3 mol of ions `(2H^(+)` and `SO_(4)^(2-))` while `1M HCI` gives 2 mol of ions `(1H^(+)` and `1CI^(-))`.

Discussion

637.	Which of the following equilibrium is favored by a temperature increase?A. C (graphite) `hArr` C (diamond)B. `H_2(g)+1/2O_2(g) hArr H_2O(g)`C. `2O_3 hArr 3O_2`D. `2SO_2+O_2 hArr 2SO_3`
Answer» Correct Answer - A The value of any equilibrium constant changes as the temperature changes. The `K_(aq)` values of exothermic reactions decreas with increasing T, and the `K_(aq)` values of enthermic reactions increase with increasing T. The conversion ofgraphite (more stable) to diamond (less stable) is an endothermic equilibrium while all others are exothermic equilibria. Note that the decomposition of ozone is an exothermic equilibrium.

Discussion

638.	Which of the following is not true for the equilibrium reaction `N_2(g)+O_2(g)hArr 2NO(g)`, `DeltaH=180kJmol^(-1)`A. The formation of `NO` is increased at higher temperature.B. The volume change at constant pressure does not affect the equilibrium.C. The pressure change at constant volume does not affect the equilibrium.D. The formation of `NO` is decreased at higher temperature.
Answer» Correct Answer - D It is endothermic (in the forward direction). Thus, a temperature increase favors the formation of NO. The pressure change at constant volume is possible by adding an inert gas. Similarly, the volume change at constant pressure can be achived by adding an inert gas. For this system, there is no effect of the addition of inert gas at constant volume or at constant pressure because `Delta n_(g)` is zero for this system, i.e., there is no change in the total number of moles of hases in a reaction.

Discussion

639.	In which of the following solvents silver chloride easily soluble ?A. `0.1 mol dm^(-3)AgNO_(3)` solutionB. `0.1 mol dm^(-3)HCI` solutionC. `H_(2)O`D. Aqueous ammonia
Answer» Correct Answer - D AgCI is most soluble in aqueous ammonia solution due to the formation of a complex.

Discussion

640.	STATEMENT-1: Buffer capacity is maximum when concentrtion of salt is equal concentrtion of acid. STATEMEN T-2: pH of the buffer is given by pH=`pK_(a)+log""(["salt"])/(["acid"]).`A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-11B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-11C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True
Answer» Correct Answer - B

Discussion

641.	Match Column-I with Colummn-II
Answer» Correct Answer - A(q), B(p, s ), C(p, s), D(r )

Discussion

642.	A weak monobasic acid is 1% ionized in 0.1 M solution at `25^(@)`C. The percentage ionization in its 0.025 M solution isA. 1B. 2C. 3D. 4
Answer» Correct Answer - B `HAhArrH^(+)+A^(-)`, `K=Calpha^(2)=0.1xx((1)/(100))^(2)=10^(-5)` When `C=0.025,alpha=sqrt((K)/(C))=sqrt((10^(-5))/(0025))=002` `:.%" ionization "=2%`.

Discussion

643.	Which of the following statements is incorrect ?A. In equilibrium mixture of ice and water kept in perfectly insulated flask, mass of ice and water does not change with time.B. The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.C. On addition of catalyst, the equilibrium constant value is not affected.D. Equilibrium constant for a reaction with negative `Delta H` value decreases as the temperature increases.
Answer» Correct Answer - B `Fe^(3+)+SCN^(-) hArr FeSCN^(2+) ` (Red). When oxalic acid is added it combines with `Fe^(3+)` ions. Hence, equilibrium shifts backward and intensity of red colour decreases.

Discussion

644.	When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. On the basis of this information mark the correct answer. `{:([Co(H_(2)O)_(6)]^(3+)(aq)+4Cl^(-)(aq)hArr[CoCl_(4)]^(2-)(aq)+6H_(2)O(l)),(" (pink)" " " "(blue)"):}`A. `Delta H gt 0` for the reactionB. `Delta lt 0` for the reactionC. `Delta H = 0` for the reactionD. The sign of `DeltaH` cannot be predicted on the basis of this information.
Answer» Correct Answer - A On cooling , the equilibrium shifts backward or on heating , the equilibrium shifts forward. Hence, reaction is endothermic, i.e., `Delta H gt 0`.

Discussion

645.	Assertion : concentration of the reactant and product does not change with time at equilibrium for a a chemical reaction. Reason : The rate of a reaction is zero at equilibrium .A. if both assertion and reason are correct and reason is correct explanation for assertion.B. if both assertion and reason are correct but reason is not correct explanation for assertion.C. if assertion is correct but reason is incorrect.D. if assertion and reason boht are incorrect.
Answer» Correct Answer - A Reason is the correct explanation for assertion.

Discussion

646.	The pH of neutral water at `25^(@)`C is 7.0 . As the temperature increases, ionisation of water increases, however, the concentration of `H^(+)` ions and `OH^(-)` ions are equal. What will be the pH of pure water at `60^(@)`C ?A. Equal to 7.0B. Greater than 7.0C. Less than 7.0D. Equal to zero
Answer» Correct Answer - C At `25^(@)C, [H^(+)]=[OH^(-)]=10^(-7) and K_(w) = [H^(+)][OH^(-)]=10^(-14)`. On heating, `K_(w)` increases , i.e, `[H^(+)][OH^(-)] gt 10^(-14)`. As `[H^(+)]=[OH^(-)],[H^(+)]^(2) gt 10^(-14) ` or `[H^(+)] gt 10^(-7) M or pH lt 7`.

Discussion

647.	The Ph OF NEUTRAL WATER AT `25^(@)C` is `7.0.` As the temperature increases, ionisation of water increases, however the concentration of `H^(+)` ions nad `OH^(-)` ions equal. What will be the ph of puire water at `60^(@)C`?A. Equal to 7.0B. Greater than 7.0C. Less than 7.0D. Equal to zero
Answer» Correct Answer - C Since ionisation of water increases upon increasing the temperature, `[H^(+)]` increases. The ph of the solution therefore, decreases and beocomes less than 7.

Discussion

648.	What is pH? What is its value for acidic solution?
Answer» The pH of a solution is defined as the negative logarithm to base 10 of the activity (aH⁺) of hydrogen ion or concentration of hydrogen ion [H⁺] . pH = −log₁₀[H⁺] For acidic solution, the value of pH is less than 7.

648.

What is pH? What is its value for acidic solution?

Answer»

The pH of a solution is defined as the negative logarithm to base 10 of the activity (aH⁺) of hydrogen ion or concentration of hydrogen ion [H⁺] .

pH = −log₁₀[H⁺]

For acidic solution, the value of pH is less than 7.

Discussion

649.

What will be the conjugate bases for the following Bronsted acids: HE, H2SO4, H3PO4 and HCO3- ?

Answer»

Bronsted Acid	Conjugate Base
HF	F⁻
H₂SO₄	HSO₄⁻
H₃PO₄	H₂PO₄⁻
HCO₃⁻	CO₃²⁻

Discussion

650.	The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H+ ions and OH– ions are equal. What will be the pH of pure water at 60°C?(i) Equal to 7.0(ii) Greater than 7.0(iii) Less than 7.0(iv) Equal to zero
Answer» (iii) Less than 7.0

Discussion

Explore topic-wise InterviewSolutions in .

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Match Column-I with Colummn-II

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The pH of neutral water at `25^(@)`C is 7.0 . As the temperature increases, ionisation of water increases, however, the concentration of `H^(+)` ions and `OH^(-)` ions are equal. What will be the pH of pure water at `60^(@)`C ?A. Equal to 7.0B. Greater than 7.0C. Less than 7.0D. Equal to zero

The Ph OF NEUTRAL WATER AT `25^(@)C` is `7.0.` As the temperature increases, ionisation of water increases, however the concentration of `H^(+)` ions nad `OH^(-)` ions equal. What will be the ph of puire water at `60^(@)C`?A. Equal to 7.0B. Greater than 7.0C. Less than 7.0D. Equal to zero

What is pH? What is its value for acidic solution?

What will be the conjugate bases for the following Bronsted acids: HE, H2SO4, H3PO4 and HCO3- ?

The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H+ ions and OH– ions are equal. What will be the pH of pure water at 60°C?(i) Equal to 7.0(ii) Greater than 7.0(iii) Less than 7.0(iv) Equal to zero