1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 8 InterviewSolution

351.	The solubility of calcium sulphate is `0.67 g L^(-1)` . The value of `K_(sp)` for calcium sulphate will beA. `1.7xx10^(-6)`B. `3.5xx10^(-4)`C. `2.4xx10^(-5)`D. `9.3xx10^(-8)`
Answer» Correct Answer - C Calculate the number of moles of `CaSO_(4)` dissolved in 1 L of solution (formula mass of `CaSO_(4)` is `40+32+4xx16= 136 u)` `(0.67g CaSO_(4))/(1L "soln")xx(1mol CaSO_(4))/(136g CaSO_(4))= 4.9xx10^(-3)mol^(-1)` Form the solubility equilibrium `CaSO_(4)(s)hArrCa^(2+)(aq.)+SO_(4)^(2-)(aq.)` we observe that each mole of `CaSO_(4)` yields 1 mol of `Ca^(2+)` and 1 mol of `SO_(4)^(2)`. Thus, at equilibrium (saturated solution). `C_(Ca^(2+))=4.9xx10^(-3)mol L^(-1)` `C_(SO_(4)^(2-))= 4.9xx10^(-3)mol L^(-1)` `K_(sp)=C_(Ca^(2+))C_(SO_(4)^(2-))` `= (4.9xx10^(-3))(4.9xx10^(-3))` `= 2.4xx10^(-5)`

Discussion

352.	Why is BF3 termed as an acid though it does not contain H+ ions?
Answer» BF₃ is termed as an acid though it does not contain H⁺ ions. Because BF₃ is an electron deficient species. It has a tendency to accept electron pair. So, it behaves as Lewis acid.

Discussion

353.	In a reaction between `H_(2) and I_(2)` at a certain temperature, the amounts of `H_(2), I_(2) and HI` at equilibrium constant f werefound to be 0.45 mole, 0.39 mole and 3.0 moles respectively. Calculate the equilibrium constant for the reaction at the given temperature.
Answer» Thr reaction between `H_(2) and I_(2)` may be represented as : `H_(2) + I_(2) hArr 2 HI` Amounts of `H_(2), I_(2) and HI` at equilibrium are given to be `H_(2) = 0.45 " mole " , I_(2) = 0.39 " mole" and HI= 3.0 " mole"` Suppose the volume of the vessel ( i.e., reaction mixture) = V litres. Then the molar concentrations at equilibrium will be `[H_(2)] = (0.45)/V molL^(-1), [I_(2)]= (0.39)/V mol L^(-1) and [ HI] = (3.0)/V mol L^(-1)` Applying the law of chemical equilibrium to the above reaction, we get `K_(c) = [HI]^(2) /([H_(2)][I_(2)])=(3.0//V)^(2)/(0.45 xx 0.39) = 51.28`

Discussion

354.	For the equlibrium `CO+H_(2)OhArrCO_(2)+H_(2)` The relation between `K_(rho)and K_(c) at 25^(@)C` and at `100^(@)C` areA. `K_(rho)=K_(c),K_(rho)=K_(c)`B. `K_(rho)=K_(c)(RT)^(-1),K_(rho)=K_(c)`C. `K_(rho)=K_(c)(RT),K_(rho)=K_(c)(RT)`D. `K_(rho)=K_(c)(RT),K_(rho)=K_(c)`
Answer» Correct Answer - D

Discussion

355.	Give reasons for the following : (i) Magnesium is not precipitated from a solution of its salt by `NH_(4)OH` in the presence of `NH_(4)Cl`. (ii) Ammonium chloride is acidic in liquid ammonia solvent
Answer» (i) The solubility product of `Mg(OH)_(2)` is high. Presence of `NH_(4)Cl` suppresses the dissociation of `NH_(4)OH` due to common ion effect thus giving low giving low concentration of `[OH^(-)]` . The ionic product, therefore, cannot exceed the solubility product. (ii) In solution of `NH_(4)Cl` in liquid `NH_(3)`, the following reaction takes place : `NH_(4)^(+) + NH_(3) hArr NH_(3)+NH_(4)^(+)` Thus, `NH_(4)Cl` gives proton. Hence, it is acidic.

Discussion

356.	Assertion. NaCl is precipitated when HCl gas is passed in a saturated solution of NaCl. Reason. HCl is a strong acid.A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If aasertion is true, but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - B Correct explanation. `NaCl hArr Na^(+) + Cl^(-)`. On passing HCl, `Cl^(-)` ion concentration increases. Equilibrium shifts backward.

Discussion

357.	What happens to ionic product of water if some acid is added to it?
Answer» If some acid is added to water, the ionic product of water remains unchanged.

Discussion

358.	`A+B hArr C+D` is 100. If intially the concentrations of A and B are equal but at equilibrium concentration of D will be twice that of A, then what will be the equilibrium constant of the reaction ?A. `4//9`B. `018`C. `017`D. `0*11`
Answer» Correct Answer - D `{:(,A,+,B,hArr,C,D,),(" Intial",a,,a,,,,),(" At eqm",(a-x),,(a-x),,x,,x):}` But we are given that `x= 2 (a-x)` or` 3 x = 2 a or a=3/2x = 1.5 x ` ` :. K= (x xx x)/(a-x^(2))= x^(2)/(1.5 x - x)^(2) = x^(2)/(0.25 x^(2))=4`

Discussion

359.	Under what conditions is a substance precipitated from its solution?
Answer» When ionic product exceeds the solubility product, then precipitation of a substance takes place.

Discussion

360.	For the reaction , ` N_(2) (g) + 3 H _(2) (g) hArr 2 NH_(3) (g),` the partial pressures of `N_(2) and H_(2)` are 0.80 and 0.40 atmosphere respectively at equilibrium . The total pressure of the system is 2.80 atmospheres. What is `K_(p)` for the above reaction ?
Answer» The reaction is : `N_(2) (g) + 3H_(2) (g) hArr 2 NH_(3) (g)` , We are given that at equilibrium , `p_(N_2) = 0.80 " atmosphere ", p_(H_2) = 0.40 " atmosphere "` ` p_(N_2) + p_(H_2) + p _(HN _(3)) = 2. 80 " atmosphere " :. p_(NH_(3)) = 2.80- (0.80 + 0.40) = 1.60 " atmosphere ".` Applying the law of chemical equilibrium , we get (taking pressures with respect with respect to standard state pressure of 1 atm) `K_(p) =(P_(NH_(3))^(2))/(P_(N_2) xx P_(H_(2))^(3)) = (1.60)^(2)/(0.80 xx(0.40)^(3))=50.0`

Discussion

361.	(a) A certain buffer solution contains equal concentration of X− and HX. The Kb for X− is 10-10. What is the pH of the buffer?(b) Assuming that the buffer in blood is CO2 −HCO3−, calculate the ratio of conjugate base to acid necessary to maintain blood at its proper pH, 7.4.Ka of H2CO3 = 4.5 × 10-7 .
Answer» (a) Given, K_b = 10⁻¹⁰ pK_b = − log[K_b ] = −log[10⁻¹⁰] = 10 log 10 = 10 \(\because\) pK_a+ pK_b = 14 or pK_a = 14 - pK_b = 14 - 10 = 4 For acidic buffer, pH = pK_a + log\(\frac{[Salt]}{[acid]}\) \(\because\) [X⁻] = [HX] \(\therefore\) pH = pK_a = 4 (b) Given, pH = 7.4 K_a = 4.5 × 10⁻⁷ CO₂ + H₂O ⇌ H⁺ + HCO₃⁻ K_a = \(\frac{[H^+][HCO_3^-]}{[CO_2]}\) Now pH = - log [H⁺] Or [H⁺] = 10^−pH = 10^−7.4 = 4.0 × 10⁻⁸ So, from equation (i), \(\frac{[HCO_3^-]}{[CO_2]}\) = \(\frac{K_a}{[H^+]}\) \(\frac{4.5\times10^{-7}}{4.0\times10^{-8}}\) = 11

361.

(a) A certain buffer solution contains equal concentration of X− and HX. The Kb for X− is 10-10. What is the pH of the buffer?(b) Assuming that the buffer in blood is CO2 −HCO3−, calculate the ratio of conjugate base to acid necessary to maintain blood at its proper pH, 7.4.Ka of H2CO3 = 4.5 × 10-7 .

Answer»

(a) Given, K_b = 10⁻¹⁰

pK_b = − log[K_b ] = −log[10⁻¹⁰]

= 10 log 10 = 10

\(\because\) pK_a+ pK_b = 14

or pK_a = 14 - pK_b = 14 - 10 = 4

For acidic buffer,

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

\(\because\) [X⁻] = [HX]

\(\therefore\) pH = pK_a = 4

(b) Given, pH = 7.4

K_a = 4.5 × 10⁻⁷

CO₂ + H₂O ⇌ H⁺ + HCO₃⁻

K_a = \(\frac{[H^+][HCO_3^-]}{[CO_2]}\)

Now pH = - log [H⁺]

Or [H⁺] = 10^−pH

= 10^−7.4

= 4.0 × 10⁻⁸

So, from equation (i),

\(\frac{[HCO_3^-]}{[CO_2]}\) = \(\frac{K_a}{[H^+]}\)

\(\frac{4.5\times10^{-7}}{4.0\times10^{-8}}\)

= 11

Discussion

362.	What is the conjugate base of `OH^(-)`?A. `O_(2)`B. `H_(2)O`C. `O^(-)`D. `O^(2-)`
Answer» Correct Answer - D

Discussion

363.	The conjugate base of `OH^(-)` isA. `O^(2-)`B. `O^(-)`C. `H_(2)O`D. `O_(2)`
Answer» Correct Answer - A Conjugate base `hArr` Conjugate acid `-H^(+)` `:.` Conjugate base of `OH^(-)=OH^(-)-H^(+)=O^(2-)`

Discussion

364.	Two acids 'A' and 'B' with Pka value 6.3 and 5.1 , which one is the stronger acid?
Answer» Pka = -logKa, So hiher is the value Ka lower is its Pka greater is the acidic strength.

Discussion

365.	Which one of the following species cannot act as both Bronsted acid and base ?A. `H_(2) O`B. `HCO_(3)^(-)`C. `HSO_(4)^(-)`D. `NH_(2)^(-)`
Answer» Correct Answer - D `NH_(2)^(-)` can act only as a Bronsted base and not as Bronsted acid.

Discussion

366.	What is the minimum pH of a solution of 0.1 M in `Mg^(2+)` from which `Mg(OH)_(2)` will not precipitate `K_(sp) =1.2 xx 10^(-11) M^(3).`
Answer» `K_(sp) =[Mg^(2+)][OH^(-)]^(2)` `1.2 xx 10^(11) =[0.1][OH^(-)]^(2)` `[OH^(-)]^(2) =(1.2 xx10^(-11))/(0.1) =1.2 xx 10^(-10)` `[OH^(-)]=(1.2 xx 10^(-10))^(1//2) = 1.0954 xx 10^(-5) M` `pOH =- log (OH^(-)) =- log (1.0954 xx 10^(-5)) = (5 -0.04) =4.96` `pH =14 -4.96 = 9.04`

Discussion

367.	The solubility of Fe(OH)3 is x mol L-1. Its Ksp would be(a) 9x3 (b) 3x4(c) 27x4 (d) 9x4
Answer» The correct answer is (c) 27x⁴

Discussion

368.	Is it possible to get precipitate of Fe(OH)3 at pH = 2.2? Give reason.
Answer» No, it is not possible to get precipitate of Fe(OH)₃ at pH = 2.2. Because, Fe(OH)₃ will dissolve in strongly acidic medium.

Discussion

369.	Write the Ksp expression for AI(OH)3.
Answer» \(\begin{matrix}Al(OH)_3 \\S\end{matrix}\) ⇌ \(\begin{matrix}Al^{3+} \\S\end{matrix}\) + \(\begin{matrix}3OH^- \\3S\end{matrix}\) K_sp = [Al³⁺][OH^-]₃ = (S) (3S)³ = S x 27S³ = 27S⁴

369.

Write the Ksp expression for AI(OH)3.

Answer»

\(\begin{matrix}Al(OH)_3 \\S\end{matrix}\) ⇌ \(\begin{matrix}Al^{3+} \\S\end{matrix}\) + \(\begin{matrix}3OH^- \\3S\end{matrix}\)

K_sp = [Al³⁺][OH^-]₃

= (S) (3S)³

= S x 27S³

= 27S⁴

Discussion

370.	The solubility of A2X3 is y mol dm-3.Calculate its solubility product.
Answer» A₂X₃(y) ⇋ 2A³⁺ (2y) + 3x^2- (3y) K_sp = [A³⁺][x^2-]³ = (2y)²(3y)³ = 4y² x 27y³ = 108y⁵

370.

The solubility of A2X3 is y mol dm-3.Calculate its solubility product.

Answer»

A₂X₃(y) ⇋ 2A³⁺ (2y) + 3x^2- (3y)

K_sp = [A³⁺][x^2-]³

= (2y)²(3y)³

= 4y² x 27y³

= 108y⁵

Discussion

371.	Write the significance of equilibrium constant?
Answer» The numerical value of equilibrium constant is a measure of the extent to which the reactants have been converted into products. A large value of K indicates large amount of reactants converted into products while its low value indicates less amount of reactants converted into products before achieving the equilibrium state.

Discussion

372.	Applying Le- Chatelier's principle, predict the effect of temperature and pressure on evaporation of water.
Answer» Evaporation of water can be represented as water + heat ----> water vapours: i) With increase in pressure- the equilibrium tend to shift in direction where there is decrease in the vol. i.e. towards left. ii) with increase in temperature- the equilibrium shifts to right as it is an endothermic reaction.

372.

Applying Le- Chatelier's principle, predict the effect of temperature and pressure on evaporation of water.

Answer»

Evaporation of water can be represented as water + heat ----> water vapours:

i) With increase in pressure- the equilibrium tend to shift in direction where there is decrease in the vol. i.e. towards left.

ii) with increase in temperature- the equilibrium shifts to right as it is an endothermic reaction.

Discussion

373.	Classify H+ species into Lewis acids and Lewis bases and show how these act as Lewis acids/base.
Answer» H⁺ :- Lewis acid because H⁺ can accept a pair of electrons.

Discussion

374.	Classify BCl3 species into Lewis acids and Lewis bases and show how these act as Lewis acids/base.
Answer» BCl₃ :- Lewis acid because BCl₃ can accept a pair of electrons.

Discussion

375.	Which one of the following is a hard base ?A. `Ag^(+)`B. `Cr^(3+)`C. `I^(-)`D. `F^(-)`
Answer» Correct Answer - D Hard base is an anion small in size and hence not easily polarizable.

Discussion

376.	The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10-3 M. What is its pH?
Answer» [H⁺] = 3.8 x 10^-3 M pH = -log [H⁺] - log (3.8 x 10^-3) = -log 3.8 - log 10^-3 = 0.5798 + 3 = 2.42

376.

The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10-3 M. What is its pH?

Answer»

[H⁺] = 3.8 x 10^-3 M

pH = -log [H⁺] - log (3.8 x 10^-3)

= -log 3.8 - log 10^-3

= 0.5798 + 3 = 2.42

Discussion

377.

The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 x 10-4, 1.8 x 10-4 and 4.8 x 10-4 respectively. Calculate the ionization constants of the corresponding conjugate base.

Answer»

For a conjugate acid-base pair

K_aK_b = K_w

Acid	K_a	Conjugate	K_b = K_w/K_a
HF	6.8 x 10^-4	F^-	\(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 1.5 x 10^-11
HCOOH	1.8 x 10^-4	HCOO^-	\(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 5.6 x 10^-11
HCN	4.8 x 10^-4	CN^-	\(\frac{1\times 10^{-14}}{6.8\times 10^{-14}}\) = 2.08 x 10^-11

Discussion

378.	A weak acid HX has the dissociation constant `1xx10^(-5)M`. It forms a salt NaX on reaction with alkali. The degree of hydrolysis of 0.1 M solution of NaX isA. `0.0001%`B. `0.01%`C. `0.1%`D. `0.15%`
Answer» Correct Answer - B Hydrolysis reaction is `X^(-) + H_(2)O hArr HX+OH^(-)` For a salt of weak acid with strong base, `K_(h)=(K_(w))/(K_(a))=ch^(2)` `:. (10^(-14))/(10^(-5))=0.1xxh^(2) or h^(2) = 10^(-8) or h = 10^(-4)` % hydrolysis `= 10^(-4) xx 100 = 10^(-2) = 0.01`

Discussion

379.	`CO_(2)` gas is much more soluble in aqueous NaOH solution than in water. Justify.
Answer» `CO_(2)` dissolves in water to form carbonic acid `(H_(2)CO_(3))` `CO_(2)(g) + H_(2)O(g) ltimplies underset("Weak acid")H_(2)CO_(3)(aq)` As the reaction is reversible, `H_(2)CO_(3)` dissociats to give `CO_(2)` and `H_(2)O` in the backward reaction. However, with aqueous NaOH solution, `H_(2)O_(2)` reacts as follows: `H_(2)CO_(3)(aq) + 2NaOH(aq) to Na_(2)CO_(3)(aq) + 2H_(2)O(l)` As a result, more of `CO_(2)` dissolve in water to form carbonic acid. This means that the gas is more soluble in the aqueous alkali solution than in water alone.

Discussion

380.	Give reason for the following: Co2 is more soluble in aqueous NaOH solution than in water.
Answer» In water, CO₂ dissolves to form carbonic acid( H₂CO₃): CO₂(g) + H₂O(l)----> H₂CO₃(aq) As the reaction is reversible, the solubility is low. In aqueous NaOH, H₂CO₃ reacts with sodium hydroxide as follows: 2 NaOH(aq)+ H₂CO₃---> Na₂ CO₃(aq)+ H₂O(l) As a result, the equilibrium involving dissolution of CO₂ in H₂O shifts forward, i.e. solubility increases.

380.

Give reason for the following: Co2 is more soluble in aqueous NaOH solution than in water.

Answer»

In water, CO₂ dissolves to form carbonic acid( H₂CO₃):

CO₂(g) + H₂O(l)----> H₂CO₃(aq)

As the reaction is reversible, the solubility is low.

In aqueous NaOH, H₂CO₃ reacts with sodium hydroxide as follows:

2 NaOH(aq)+ H₂CO₃---> Na₂ CO₃(aq)+ H₂O(l)

As a result, the equilibrium involving dissolution of CO₂ in H₂O shifts forward, i.e. solubility increases.

Discussion

381.	What will be the ionisation constant of formic acid if its 0.01 M solution is `14.5%` ionised ?A. `2.1xx10^(-4)`B. `14.5`C. `0.145`D. `1.45xx10^(-4)`
Answer» Correct Answer - A `K_(a)=calpha^(2)` `=(14.5)/(100)=0.145` `K_(a)=0.01xx(0.145)^(2)=2.1xx10^(-4)`

Discussion

382.	What is the percentage dissociation of 0.1 M solution of acetic acid ? `(K_(a)=10^(-5))`A. `10%`B. `100%`C. `1%`D. `0.01%`
Answer» Correct Answer - C `alpha=sqrt((K_(a))/(c))oralphasqrt((10^(-5))/(0.1))=10^(-2)` `%` dissociation `=10^(-2)xx100=1%`

Discussion

383.	For poly basic acid , the dissociation constant have a different valves for each step. `H_(3)AhArrH^(+)+A,Keq=Ka_(1)` `H_(2)AhArrH^(+)+A^(-2),Keq=Ka_(2)` `HA^(2)hArrH^(+)+A^(3-),Keq=Ka_(3)` What is the observed trend of dissociation constant im successive stages ?A. `K_(a_(1))gtK_(a_(2))gtK_(a_(3))`B. `K_(a_(1))=K_(a_(2))=K_(a_(3))`C. `K_(a_(1))ltK_(a_(2))ltK_(a_(3))`D. `K_(a_(1))=K_(a_(2))+K_(a_(3))`
Answer» Correct Answer - A The value of dissociation constants for successive stages decrease.

Discussion

384.	Among the following the one which does not represent a conjugate acid-base pair isA. `HCl and Cl^(-)`B. `HOH and OH^(-)`C. `SO_(2)and H_(2)SO_(4)`D. `NH_(4)^(+)and NH_(3)`
Answer» Correct Answer - C

Discussion

385.	The pH of `0.05` M solution of a strong dibasic acid isA. `0.0`B. 1C. 2D. 5
Answer» Correct Answer - B

Discussion

386.	The `pH` of `10^(-8)M` solution of `HCl` in water isA. 8B. 6C. Between 6 and 7D. Between 7 and 8
Answer» Correct Answer - C

Discussion

387.	Which of the following factors will favour the reverse reaction in a chemical equilibrium?A. Increasing the concentration of one of the reactantsB. Removal of at least one of the products at regular intervalsC. Increasing the concentration of one or more of the productsD. none of these
Answer» Correct Answer - C

Discussion

388.	If `0.561 g` of `(KOH)` is dissolved in water to give. `200 mL` of solution at `298 K`. Calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its `pH`?
Answer» Correct Answer - `[K^(+)] = [OH^(-)] = 0.05M, [H^(+)] = 2.0 xx 10^(-13) M` `[KOH_(aq)] = (0.561)/((1)/(5)) g//L` `= 2.805 g//L` `= 2.805 xx (1)/(56.11) M` `= .05 M` `KOH_((aq)) rarr K_((aq))^(+) + OH_((aq))^(-)` `[OH^(-)] = .05 M = [K^(+)]` `[H^(+)] [H^(-)] = K_(w)` `[H^(+)] = (K_(w))/([OH^(-)])` `= (10^(-14))/(0.05) = 2 xx 10^(-13) M` `:. pH = 12.70`

Discussion

389.	Calculate the ph of a solution formed by mixing equal volumes of two solutions A and B of a strong acids having `ph=6" and "ph=4` respectively.
Answer» ph of solution `A=6,[H^(+)]` "of solution" `[A]=10^(-6)mol L^(-1)`. ph of solution `B=4,[H^(+)]` "of solution" `[B]=10^(-4)mol L^(-1)`. Total `[H^(+)]` of the solution `=(10^(6)+10^(-4)) mol L^(-1)` Volme of the solution upon mixing `=(1+1)=2L` Total `[H^(+)]` on mixing solution `=(10^(-6)+10^(-4))/(2)mol L^(-1)= (10^(-4)(0.01+1))/(2) mol L^(-1)` `=(0.01xx10^(-4))/(2)mol L^(-1)=0.5xx10^(-4)mol L ^(-1)` `=5xx10^(-5)mol L^(-1)` ph of solution `=-log[H^(+)]=-[5xx10^(-5)]` `=[5-log5]=[5-0.6990]=4.3010=4.3`

Discussion

390.	What will the PH of a solution formed by mixing equal volumes of two solutions A and B having PH=6 and PH=4
Answer» PH=6 [H⁺ ] = 10^-6 M PH=2 [H⁺ ] = 10^-2 M Resultant [H⁺ ] M₁xV₁+M₂xV₂=M₃ xV₃ 10^-6 x1 +10^-6 x¹=M₃x2 M₃=101x10^-6 /2 = 50. 5 x 10^-6 PH=- log(50.5x10^-7 ) =4.296

390.

What will the PH of a solution formed by mixing equal volumes of two solutions A and B having PH=6 and PH=4

Answer»

PH=6

[H⁺ ] = 10^-6 M

PH=2

[H⁺ ] = 10^-2 M

Resultant [H⁺ ]

M₁xV₁+M₂xV₂=M₃ xV₃ 10^-6 x1 +10^-6 x¹=M₃x2

M₃=101x10^-6 /2 = 50. 5 x 10^-6

PH=- log(50.5x10^-7 )

=4.296

Discussion

391.	Calculate the pH of a solution obtained by mixing equal volumes of the solutions with pH = 3 and pH = 5.
Answer» pH = 3 means `[H^(+)]=10^(-3)M and pH = 5` means `[H^(+)]=10^(-5)`M. On mixing equal volumes, in the final solution, `[H^(+)]=((10^(-3)+10^(-5))/(2)) = (10^(-3)(1+10^(-2)))/(2) = (1.01xx10^(-3))/(2) = 0.505xx10^(-3) M = 5.05 xx 10^(-4) M ` `:. pH = - log (5.05xx10^(-4))=4-0.7033~~3.3`

Discussion

392.	Calculate the pH of a solution obtained by mixing equal volumes of N/10 NaOH and N/20 HCl.
Answer» Correct Answer - 12.398 Suppose 1 L of N/10 NaOH is mixed with 1 L of N/20 HCl. 1 L of N/10 NaOH contains NaOH `=(1)/(10)" g eq." = 0.1 g eq`. 1 L of N/20 HCl contains HCl `=(1)/(20) g eq. = 0.05 g eq`. 0.05 g eq. of HCl will neutralise 0.05 g eq. of NaOH. `:. ` NaOH left unneutralised = 0.05 g eq. Volume of solution = 2 L `:. `Concentration of NaOH in the final solution `= (0.05)/(2)" g eq" L^(-1) = 0.025 N` `[NaOH]=[OH^(-)] = 0.025 M = 2.5 xx 10^(-2) :. [H_(3)O^(+)]=(10^(-14))/(2.5xx10^(-2))=4xx10^(-13) M`

Discussion

393.	The ionization constant of chloroacetic acid is `1.35xx10^(-3)`. What will be the `pH` of `0.1 M` acid and its `0.1M` sodium salt solution?
Answer» Correct Answer - (a) `pH` of acid solution `= 1.9` , (b) `pH ` of its salt solution `= 7.9` It is given that `K_(a)` for `CICH_(2)COOH" is "1.35xx10^(-3)`. `rArr K_(a)=calpha^(2)` `thereforealpha=sqrt((K_(a))/(c))` `=sqrt((1.35xx10^(-3))/(0.1))" "(therefore"concentration of acid = 0.1 m")` `alpha=sqrt(1.35xx10^(-2))=0.116` `therefore[H^(+)]=calpha=0.1xx0.116` = .0116 `rArrpH=-log[H^(+)]=1.94` `ClCH_(2)COONa` is the salt of weak acid i.e., `ClCH_(2)COOH` and a strong base i.e., NaOH. `ClCH_(2)COO^(-)+H_(2)OleftrightarrowCICH_(2)COOH+OH^(-)` `K_(h)=([ClCH_(2)COOH][OH^(-)])/([ClCH_(2)COO^(-)])` `K_(h)=(K_(v))/(K_(a))` `K_(h)=(10^(-14))/(1.35xx10^(-3))` `=0.740xx10^(-11)` Also, `K_(h)=(x^(2))/(0.1)` (Where x is the concentration of `OH^(-)` and `ClCH_(2)COOH`) `0.740xx10^(-11)=(x^(2))/(0.1)` `0.074xx10^(-11)=x^(2)` `rArr x^(2)=0.74xx10^(-12)` `x=0.86xx10^(-6)` `[OH^(-)]=0.86xx10^(-6)` `therefore [H^(+)]=(K_(w))/(0.86xx10^(-6))` `=(10^(-14))/(0.86xx10^(-6))` `[H^(+)]=1.162xx10^(-8)` `pH=-log[H^(+)]` = 7.94.

Discussion

394.	The pH of the solution formed on mixing 20 mL of `0.05 M H_(2)SO_(4)` with 5.0 mL of 0.45 M NaOH of 298 K isA. 6B. 2C. 12D. 7
Answer» Correct Answer - C `2 NaOH+H_(2)SO_(4)rarrNa_(2)SO_(4)+2H_(2)O` 20 mL of 0.05 M `H_(2)SO_(4)=20xx0.05` millimoles = 1 millimole 5 mL of 0.45 M NaOH `= 5 xx 0.45` millimoles `=2.25` millimole 2 millimoles of NaOH react with 1 millimole of `H_(2)SO_(4)` `:.` NaOH left in the solution = 0.5 millimole Volume of solution = 25 mL `:. [OH^(-)]= (0.5)/(25)M = 0.01 M = 10^(-2)M` `:. [H^(+)]=10^(-12) ` or pH = 12

Discussion

395.	A crystal of common slat of given mass is by kept in aqueous solution. After 12 hours, its mass ramanins the same, Is the crystal in equilibrium with the solution?
Answer» Yes, as the mass of the crystal remains the same, this means that the solution is saturated. Therefore, there is a physical equilibrium between the crystal of the common salt and its saturated solution.

Discussion

396.	The ionic product of water at 310 K is `2.7xx10^(-14)`. What is the pH of neutral water at this temperature ?
Answer» `[H^(+)]=sqrt(K_(w))=sqrt(2.7xx10^(-14))=1.643xx10^(-7)M` `pH= - log [H^(+)]=-log (1.643xx10^(-7))=7-0.2156=6.78`

Discussion

397.	Ionic product of water at `310 K` is `2.7xx10^(-14)`. What is the `pH` of netural water at this temperature?
Answer» Correct Answer - `pH = 6.78` Ionic product, `K_(w)=[H^(+)][OH^(-)]` Let `[H^(+)]=x` ltBrgt Since `[H^(+)]=[OH^(-)], K_(w)=x^(2)` `rArrK_(w)"at310K is "2.7xx10^(-14)` `therefore 2.7xx10^(-14)=x^(2)` `rArrx=1.64xx10^(-7)` `rArr[H^(+)]=1.64xx10^(-7)` `rArrpH=-log[H^(+)]` `=-log[1.64xx10^(-7)]` = 6.78 Hence, the pH of neutral water is 6.78.

Discussion

398.	Explain with the help of an example, how an electrolyte undergo dissociation in water. what is the role play by water in formation of an aqueous solution of an electrolyte?
Answer» When strong electrolyte like NaCl is dissolve ion water, the negative part of the water molecule start attracting the Na⁺ ion and positive part start attracting the Cl- ion , as these interaction are stronger than lattice enthalpy of sodium chloride ,so sodium chloride get broken into ions in solution and formed aqueous solution of the electrolyte. NaCl ------> Na⁺ + Cl^-

398.

Explain with the help of an example, how an electrolyte undergo dissociation in water. what is the role play by water in formation of an aqueous solution of an electrolyte?

Answer»

When strong electrolyte like NaCl is dissolve ion water, the negative part of the water molecule start attracting the Na⁺ ion and positive part start attracting the Cl- ion , as these interaction are stronger than lattice enthalpy of sodium chloride ,so sodium chloride get broken into ions in solution and formed aqueous solution of the electrolyte.

NaCl ------> Na⁺ + Cl^-

Discussion

399.	Calculate the pH of a solution formed on mixing 0.2 M `NH_(4)Cl` and 0.1 M `NH_(3)`. The `pK_(b)` of ammonia solution is 4.75.
Answer» `NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-), pK_(b)=4.75` On mixing `NH_(3) ` with `NH_(4)Cl` solution, we have `{:(,NH_(3),+,H_(2)O,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",0.1M,,,,0.2M,,0),("At eqm.",0.1 -x,,,,0.2+x,,x):}` `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])` As `pK_(b) = 4.75 , i.e., -log K_(b) = 4.75 or log K_(b) = -4.75 = bar(5) .25 :. K_(b) = 1.77xx10^(-5)` `:. 1.77xx10^(-5) = ((0.2+x)(x))/(0.1-x)~=((0.2)x)/(0.1)` (neglecting x in comparison to 0.1 and 0.2) or, `x=010^(-5),i..e., [OH^(-)]=0.885xx10^(-5)=8.85xx10^(-6)M` or `[H^(+)]=(10^(-14))/(8.85xx10^(-6))=1.13xx10^(9) :. pH = - log [H^(+)]=-log (1.13xx10^(-9))=9-0.053=8.95` Alternatively, the given solution is a buffer solution of a weak base and is salt with a strong acid (discussed in sec. 7.31). For such a buffer, `pOH=pK_(b)+log.(["Salt"])/(["Base"])=4.75+log. (0.2)/(0.1) = 4.75+log 2 = 4.75 + 0.30 = 5.05 `[ Eqn . (vi)] `pH = 14 -pOH = 14 - 5.05=8.95`

Discussion

400.

What will be the correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?(i) Water < ether < acetone(ii) Water < acetone < ether(iii) Ether < acetone < water(iv) Acetone < ether < water

Answer»

(ii) Water < acetone < ether

Greater the boiling point, less is the vapour pressure. Hence, the correct order of vapour pressures will be:

Discussion

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Explain with the help of an example, how an electrolyte undergo dissociation in water. what is the role play by water in formation of an aqueous solution of an electrolyte?

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