1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 18 InterviewSolution

851.	Define chemical equilibrium. Give an example.
Answer» A chemical equilibrium is a system whose chemical composition of the system does not change when dynamic equilibrium is reached. E.g. N₂(g) + 3H₂(g) ⇌ 2NH₃ (g)

851.

Define chemical equilibrium. Give an example.

Answer»

A chemical equilibrium is a system whose chemical composition of the system does not change when dynamic equilibrium is reached.

E.g. N₂(g) + 3H₂(g) ⇌ 2NH₃ (g)

Discussion

852.	Which of the following is not true for solid-liquid equilibrium?A. It can be established at any given temperature.B. The mass of solid does not change with time.C. The mass of liquid does not change with time.D. There is no exchange of heat between the system and its surroundings.
Answer» Correct Answer - A It is established only at the melting point or freezing point of the substance.

Discussion

853.	Why is chemical equilibrium called dynamic equilibrium?
Answer» Chemical equilibrium is achieved when the rates of forward and backward reactions become equal. Although overall it is in equilibrium but inside the forward reaction and backward reaction both occur and so there is motion in the atomic and molecular level. Dynamic refers to anything having motion since, the equilibrium is not static so, the chemical equilibrium is called dynamic equilibrium.

Discussion

854.	Chemical equiluibrium fir the reaction `N_(2)O_(4)(g)hArr2NO_(2)(g)` can be achived in _________ different ways.A. twoB. threeC. fourD. just one
Answer» Correct Answer - B We can reach the equilibrium state by sarting with pure `N_(2)O_(4)` or with pure `NO_(2)` or with a mixture of `NO_(2)` and `N_(2)O_(4)`. In each case. We observe an initial change in color, either due to the formation of dark dark brown `NO_(2)` (if the color intensifies) or due to the depletion of `NO_(2)` (if the color fades), and then the final equilibrium state in which the color of `NO_(2)` no longer changes. Depending upon the temperature of the reacting system and on the initial amount of `NO_(2)` and `N_(2)O_(4)` the concentrations of `NO_(2)` and `N_(2)O_(4)` at equilibrium differ from system to system.

Discussion

855.	Which of the following equilibrium is dynamic?A. Solid `hArr` LiquidB. Liquid `hArr` VaporC. Solid `hArr` VaporD. All of these
Answer» Correct Answer - D At equilibrium, both the opposing processes occur simultaneously with equal rates. Hence, they are not static equilibria.

Discussion

856.	Define dynamic equilibrium.
Answer» When the reactants in a closed vessel at a particular temperature react to give products, the concentrations of the reactants keep on decreasing, while those of products keep on increasing for sometime after which there is no change in the concentrations of either the reactants or products. This stage of the system is the dynamic equilibrium.

Discussion

857.	The pH of black coffee is 5.0. Calculate corresponding hydrogen ion concentration.
Answer» [H⁺] of black coffee pH = 5.0 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 5.0 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 5.0 or [H⁺] = Antilog (-5.0) [H⁺] = Antilog\((\bar5)\) = 10^-5 M

857.

The pH of black coffee is 5.0. Calculate corresponding hydrogen ion concentration.

Answer»

[H⁺] of black coffee

pH = 5.0 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 5.0

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 5.0 or [H⁺] = Antilog (-5.0)

[H⁺] = Antilog\((\bar5)\) = 10^-5 M

Discussion

858.	The pH of egg white is 7.8. Calculate corresponding hydrogen ion concentration.
Answer» [H⁺] of egg white pH = 7.8 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 7.8 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog of H⁺ = Antilog (-7.8) [H⁺] = Antilog (8.20) = 1.58 x 10^-8 M

858.

The pH of egg white is 7.8. Calculate corresponding hydrogen ion concentration.

Answer»

[H⁺] of egg white

pH = 7.8 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 7.8

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog of H⁺ = Antilog (-7.8)

[H⁺] = Antilog (8.20) = 1.58 x 10^-8 M

Discussion

859.	The pH of lemon juice is 2.2. Calculate corresponding hydrogen ion concentration.
Answer» [H⁺] of lemon juice pH = 2.2 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 2.2 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 2.2 or [H⁺] = Antilog (-2.2) [H⁺] = Antilog \((\bar3-0.8)\) = 6.310 x 10^-3 M

859.

The pH of lemon juice is 2.2. Calculate corresponding hydrogen ion concentration.

Answer»

[H⁺] of lemon juice

pH = 2.2 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 2.2

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 2.2 or [H⁺] = Antilog (-2.2)

[H⁺] = Antilog \((\bar3-0.8)\) = 6.310 x 10^-3 M

Discussion

860.	The pH of tomato juice is 4.2. Calculate corresponding hydrogen ion concentration.
Answer» [H⁺] of tomato juice pH = 4.2 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 4.2 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 4.2 or [H⁺] = Antilog (-4.2) [H⁺] = Antilog\((\bar5.80)\) = 6.31 x 10^-5 M

860.

The pH of tomato juice is 4.2. Calculate corresponding hydrogen ion concentration.

Answer»

[H⁺] of tomato juice

pH = 4.2 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 4.2

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 4.2 or [H⁺] = Antilog (-4.2)

[H⁺] = Antilog\((\bar5.80)\) = 6.31 x 10^-5 M

Discussion

861.	The pH of milk is 6.8. Calculate corresponding hydrogen ion concentration.
Answer» [H⁺] of milk pH = 6.8 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 6.8 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 6.8 or [H⁺] = Antilog(-6.80) [H⁺] = Antilog\((\bar7.20)\) = 1.58 x 10^-7 M

861.

The pH of milk is 6.8. Calculate corresponding hydrogen ion concentration.

Answer»

[H⁺] of milk

pH = 6.8 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 6.8

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 6.8 or [H⁺] = Antilog(-6.80)

[H⁺] = Antilog\((\bar7.20)\) = 1.58 x 10^-7 M

Discussion

862.	Calculate the hydrogen ion concentration in the biological fluids whose pH is Human saliva, 6.4.
Answer» [H⁺] of human saliva pH = 6.4 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 6.4 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 6.4 or [H⁺] = Antilog(-6.4) [H⁺] = Antilog\((\bar7.20)\) = 1.58 x 10^-7 M

862.

Calculate the hydrogen ion concentration in the biological fluids whose pH is Human saliva, 6.4.

Answer»

[H⁺] of human saliva

pH = 6.4 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 6.4

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 6.4 or [H⁺] = Antilog(-6.4)

[H⁺] = Antilog\((\bar7.20)\) = 1.58 x 10^-7 M

Discussion

863.	Calculate the hydrogen ion concentration in the biological fluids whose pH is Human blood, 7.38.
Answer» [H⁺] of human blood pH = 7.38 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 7.38 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 7.38 or [H⁺] = Antilog(-7.38) [H⁺] = Antilog\((\bar8.62)\) = 4.168 x 10^-8 M

863.

Calculate the hydrogen ion concentration in the biological fluids whose pH is Human blood, 7.38.

Answer»

[H⁺] of human blood

pH = 7.38 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 7.38

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 7.38 or [H⁺] = Antilog(-7.38)

[H⁺] = Antilog\((\bar8.62)\) = 4.168 x 10^-8 M

Discussion

864.	On the basis of the equation pH = – log [H+], the pH of 10–8 mol dm–3 solution of HCl should be 8. However, it is observed to be less than 7.0. Explain the reason.
Answer» Concentration of 10^–8mol dm^–3 indicates that the solution is very dilute. Hence, the contribution of H₃O⁺ concentration from water is significant and should also be included for the calculation of pH.

864.

On the basis of the equation pH = – log [H+], the pH of 10–8 mol dm–3 solution of HCl should be 8. However, it is observed to be less than 7.0. Explain the reason.

Answer»

Concentration of 10^–8mol dm^–3 indicates that the solution is very dilute.

Hence, the contribution of H₃O⁺ concentration from water is significant and should also be included for the calculation of pH.

Discussion

865.	Calculate the hydrogen ion concentration in the biological fluids whose pH is Human stomach fluid, 1.2.
Answer» [H⁺] of human stomach fluid pH = 1.2 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 1.2 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 1.2 or [H⁺] = Antilog (-1.2) [H⁺] = Antilog \((\bar8.62)\) = 4.168 x 10^-8 M

865.

Calculate the hydrogen ion concentration in the biological fluids whose pH is Human stomach fluid, 1.2.

Answer»

[H⁺] of human stomach fluid

pH = 1.2 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 1.2

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 1.2 or [H⁺] = Antilog (-1.2)

[H⁺] = Antilog \((\bar8.62)\) = 4.168 x 10^-8 M

Discussion

866.	On the basis of the equation `ph=-log[H^(+)]`,the ph of `10^(-8)` mol `dm^(-3)` solution of HCI should be 8. However, it is observed to be less than 7.0. Explain the reason.
Answer» The pH of an acidic solution cannot be more than 7.0. whrneas the expected Ph is 8. To explain this it is suggested that `H^(+)` or `H_(3)O^(+)` ions are also released by the self ionisation of water. Since water is neutral, `[H_(3)O^(+)]=10^(-7)M`. `" therefore Total " [H_3O^(+)]" in solution "=10^(-8)M+10^(-7)M=10^(-8)[1+10]=11xx10^(-8)M` `ph=-log[H_3O^(+)]=-[11xx10^(-8)]=[8-log11]` `ph=[8-1.0411]=6.9586=6.96`.

Discussion

867.	Calculate the hydrogen ion concentration in the biological fluids whose pH is Human muscle-fluid, 6.83.
Answer» [H⁺] of human muscles fluid pH = 6.83 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 6.83 \(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 6.83 or [H⁺] = Antilog (-6.83) [H⁺] = Antilog \((\bar7.17)\) = 1.48 x 10^-7 M

867.

Calculate the hydrogen ion concentration in the biological fluids whose pH is Human muscle-fluid, 6.83.

Answer»

[H⁺] of human muscles fluid

pH = 6.83 or log\(\bigg[\frac{1}{H^+}\bigg]\) = 6.83

\(\bigg[\frac{1}{H^+}\bigg]\) = Antilog 6.83 or [H⁺]

= Antilog (-6.83)

[H⁺] = Antilog \((\bar7.17)\) = 1.48 x 10^-7 M

Discussion

868.	At `25^(@)C`, the solubility product of `Mg(OH)_(2)` is `1.0xx10^(-11)`. At which pH will `Mg^(2+)` ions start precipitating in the form of `Mg(OH)_(2)` from a solution of 0.0001 M `Mg^(2+)` ions ?A. 11B. 8C. 9D. 10
Answer» Correct Answer - D `K_(sp) ` for `Mg(OH)_(2)=[Mg^(2+)][OH^(-)]^(2)` `:. 1.0xx10^(-11)=(0.001)[OH^(-)]^(2)` or `[OH^(-)]^(2)=10^(8) or [OH^(-)]=10^(-4)` i.e, `pOH=4 or pH = 10`

Discussion

869.	At some temperature and under a pressure of 4 atm, `PCl_(5)` is 10% dissociated . Calculate the pressure at which `PCl_(5)` will be 20% dissociated , temperature remaining same.
Answer» `{:(,PCl_(5) ,hArr,PCl_(3),+,Cl_(2)), (" Atm eqm. " ,1-01=09"mole",,01 "mole",,01"mole"):}` Total no. of moles ` = 09 + 01 + 01 = 1 1 "mole " ` ` p_(PCl_(5))= (09)/(1 1) xx 4"atm " , p_(PCl_(3)) = (01)/(11) xx 4 " atm " , p_(Cl_(2)) = (01)/(1.1) xx 4 " atm " ` ` :. K_(p) = (p_(PCl_(3)) xx p_(Cl_(2)))/ (p_(PCl_(5))) = ((04)/(11) xx (04)/(11))/((09 xx 4)/(11) )=00404 ` 2nd case When `PCl_(5)` is 20% dissociated . Suppose total pressure = P atm . Then `{: (,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Intial " ,1 "mole",,,,),("At equilibrium " ,1-02=08,,02,,02 "mole"),(,,,,," Total no. of moles"= 08 + 02 + 02 = 12 "moles"):}` ` p_(PCl_(5)) = (08)/(12) xxP atm , p_(PCl_(3)) = (02)/(12) xx P atm , p_(Cl_(2))= (02)/(12) xx P atm ` ` K_(p) = ((02 P)/(12) xx(02)/(12))/(( 08P)/(12))= (02)/(12) xx (02)/(08) P = 00404 ("calculated above")` which gives `P= 0 97 ` atm

Discussion

870.	In a reaction, A + 2B ⇋ 2C, 2.0mole of ‘A’ 3.0mole of ‘B’ and 2.0mole of ‘C’ are placed in a 2.0L flask and the equilibrium concentration of ‘C’ is 0.5mole/L. The equilibrium constant (K) for the reaction is –(a) 0.073 (b) 0.147 (c) 0.05 (d) 0.026
Answer» (c) The equilibrium constant (K) for the reaction is 0.05.

Discussion

871.	`0.16 g N_(2)H_(4)` is dissoolved in `H_(2)O` and total volume is made upto `500 mL`. Calculate the percentage of `N_(2)H_(4)` that has reacted with `H_(2)O` in this solution. `K_(b)` for `N_(2)H_(4) = 4.0 xx 10^(-6)M`.
Answer» Molarity of `N_(2)H_(4)` solution `=("Mass of "N_(2)H_(4))/("Molar mass of "N_(2)H_(4)) xx (100)/(500)` `=((0.16 g))/(32 g " mol"^(-1)) xx (100)/(500) =0.01M` Let x mole on `N_(2)H_(4)` be dissolved in water . `{:(,N_(2)H_(4)+H_(2)O,hArr,N_(2)H_(5)^(+) ,+, OH^(-)),("Initial conc", 0.01 "mol" ,,0,,0),("Eqm. conc",(0.01 -x)"mol",, x"mol",,x"mol"):}` `K_(b) =[[N_(2)H_(5)^(+)][OH^(-)]]/[[N_(2)H_(4)]] =(x xx x)/((0.01 -x))` `4.0 xx 10^(-6) =(x^(2))/((0.01 -x)) =(x^(2))/(0.01)` `x^(2) =4.9 xx 10^(-6) xx 10^(-2) " or " x= (4.0 xx 10^(-8))^(1//2)" "[:. 0.01 -x~~ 0.01]` `= 2.0 xx 10^(-4) mol` `:.`Precentage of `N_(2)H_(4)` reacted with water `=((2.0 xx 10^(-4) "mol"))/((0.01 "mol")) xx 100 =2.0%`

Discussion

872.	For a reaction, the value of K increases with increases with increase in temperature. The H forThe reaction would be(a) +ve (b) –ve(c) Zero (d) None of these.
Answer» (a)The reaction would be +ve.

Discussion

873.	Match Column (I) with Column (II).Column IColumn II(i) Equilibrium(a) ΔG > 0, K < 1(ii) Spontaneous reaction(b) ΔG = 0(iii) Non spontaneous reaction(c) ΔGΘ = 0(d) ΔG < 0, K > 1
Answer» (i) → (b) and (c) (ii) → (d) (iii) → (a)

Discussion

874.	The value of ΔG for a reaction, having K= 1, would be(a) –RT (b) -1(c) 0 (d) + RT
Answer» (c)The value of ΔG is 0.

Discussion

875.	solubility product of radium sulphate is `4xx10^(-9)`. What will be the solubility of ` Ra^(2+)` in ` 0.10 M` ` NaSO_(4)`?A. `xx10^(-10)M`B. `2xx10^(-5)M`C. `4xx10^(-5)M`D. `2xx10^(-10)M`
Answer» Correct Answer - A `RaSO_(4)hArrRa^(2+)+SO_(4)^(2-)` `K_(sp)=[Ra^(2+)][SO_(4)^(2-)]` Concentration of `SO_(4)^(2-)` from `Na_(2)SO_(4)=0.10M` `Ra^(2+)=(4xx10^(-11))/(0.10)=4xx10^(-10)`

Discussion

876.	A buffer solution is one which hasA. reserved acidB. reserved baseC. reserved acid and reserved baseD. pH equal to 7
Answer» Correct Answer - C A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. If `0.01` mol of hydrochloric acid is added to 1 L of pure water, the pH decreases form `7.0` to `2.0-a` pH change of `5.0` units. By contrats, the addition of this amount of hydrolichloric acid to 1 L of buffered solution might change the pH only `0.1` unit. A buffer solution resists changes in pH through its ability to combine with both `H^(+)` and `OH^(-)` ions. The reseverd base of buffer neutralizes the added `H^(+)` ions while the reserved acid of buffer neutralizes the added `OH^(-)` ions.

Discussion

877.	0.365 g of HCl gas was passed through `100 cm^(3)` of 0.2 M NaOH solution. Th e pH of the resulting solution would beA. 1B. 5C. 8D. 13
Answer» Correct Answer - D `0.365 g HCl =(0.365)/(36.5) "mole" = 0.01 "mole"` `100 cm^(3) "of" 0.2 M NaOH=(0.2)/(1000)xx100` `=0.02 ` mole NaOH left unneutralized = 0.01 mole Volume of solution = 100 ml `:.` Molarity of NaOH in the solution `=(0.01)/(100)xx1000=0.1M = 10^(-1)M` `:. [H^(+)]=(10^(-14))/(10^(-1)M)=10^(-13)M :. pH=13`

Discussion

878.	In which of the following reactions, the equilibrium reamins unaffected on addition of small amount of argon at constant volume?A. `H_(2(g))+I_(2(g))hArr2HI_((g))`B. `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`C. `N_(2(g))+3H_(2(g))hArr2NH_(3(g))`D. The equilibrium will remain unaffected in all the three cases.
Answer» Correct Answer - D If the volume is kept constant and an inert gas such as argon is added which does not take part in the reaction, the equilibrium remains undisturbed.

Discussion

879.	Which of the following equilibria remains unaffected by a change in pressure (or volume)?A. `2NOCl(g)hArr2NO(g)+Cl_2(g)`B. `H_2(g)+CO_2(g)hArrH_2O(g)+CO(g)`C. `2PbS(s)+3O_2(g)hArr2PbO(s)+2SO_2(g)`D. `PCl_5(g)hArrPCl_3(g)+Cl_2(g)`
Answer» Correct Answer - B If there is no change in the total number of moles of gases in a reaction, a pressure (volume) change does not affect the position of equilibrium, i.e., Q is unchanged bythe volume (or pressure) change. In reaction (2), the number of moles of gas products is equal to the number of moles of gas reactants. Thus, a change in pressure has no effect on the equilibrium.

Discussion

880.	Amongest the following hydroxides, the one which has the lowest value of `K` sp at ordinary temperature is:A. `Mg(OH)_(2)`B. `Ca(OH)_(2)`C. `Ba(OH)_(2)`D. `Be(OH)_(2)`
Answer» Correct Answer - D Solubility of hydroxider of group 2 elements increases down the group. Thererfore `Be(OH)_(2)` has lowest solobility and hence lowest solubility product.

Discussion

881.	Amongst the following hydroxides, the one which has the lowest value of `K_(sp)` is:A. `Ba(OH)_(2)`B. `Mg(OH)_(2)`C. `Be(OH)_(2)`D. `Ca(OH)_(2)`
Answer» Correct Answer - C All these compounds have identical composition. Thus, `K_(sp)` is directely related to the solubility of the compound. Since the solubility of the group 2 hydroxides increases on moving down the group, the order of solubility of these hydroxides is `Ba(OH)_(2)gtCa(OH)_(2)gtMg(OH)_(2)gtBe(OH)_(2)` `i.e., Be(OH)_(2)` has the lowest value of `K_(sp)`.

Discussion

882.	Which of the following gives the maximum number of ions per mole when dissolved in water ?A. `K_(2)MgI_(4)`B. `CuSO_(4)`C. `FeCI_(3)`D. `KI_(3)`
Answer» Correct Answer - C `FeCI_(3)(aq.)rarr Fe^(3+)(aq.)+3CI^(-1)(aq.)` `K_(2)HgI_(4)(aq.)rarr 2K^(+)(aq.)+HgI_(4)^(2-)(aq.)` `CuSO_(4)rarr Cu^(2+)(aq.)+SO_(4)^(2-)(aq.)` `KI_(3)rarr (aq.)+I_(3)^(-)(aq.)` 1 mmol of `FeCI_(3)` gives4 mol of ions. The strength of an electrolyte depends on the number of ions in solution and also on the charges on these ions.

Discussion

883.	Among the following hydroxides, the one which has the lowest value of `K_(sp)` at ordinary temperature (about `25^(@)` C ) isA. `Mg(OH)_(2)`B. `Ca(OH)_(2)`C. `Ba(OH)_(2)`D. `Be(OH)_(2)`
Answer» Correct Answer - D The solubility increases down the group due to increase in size of the ion and decrease in lattice energy. Lower the solubility, lower is the `K_(sp)`.

Discussion

884.	Which of the following is a strong electrolyte ?A. `Ba(OH)_(2)`B. `Sr(OH)_(2)`C. `Ca(OH)_(2)`D. `Mg(OH)_(2)`
Answer» Correct Answer - D Hydroxides of the heavier members of group 2 `(Ca, Sr, and Ba)` are strong base while `Be(OH)_(2)` and `Mg(OH)_(2)` are weak bases.

Discussion

885.	Write the expression for the equilibrium constant, KC forCH3COOC2H5(aq) + H2O(l) ⇌ CH3COOH(aq) + C2H5OH(aq)
Answer» K_C = \(\frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5][H_2O]}\)

Discussion

886.	Write the expression for the equilibrium constant, KC forFe3+(aq) + 3OH-(aq) ⇌ Fe(OH)3(s)
Answer» K_C = \(\frac{[Fe(OH)_3(s)]}{[Fe^{3+}(aq)][OH(aq)]^3}\)

Discussion

887.	Write the expression for the equilibrium constant, KC forI2(s) + 5F2 ⇌ 2IF5
Answer» K_P = \(\frac{p^2IF_5}{p^5_{F_2}}\)

Discussion

888.	Find out the value of Kc from the value of Kp:2NOCl(g) ⇌ 2NO(g) + Cl2(g), Kp = 1.8 x 10-2 at 500 K
Answer» K_C = K_p(RT)^-Δn R = 0.082 L atm mol^-1 K^-1 For this reaction Δn = (2 + 1) - (2) = +1 K_c = K_p(RT)^-1 = 1.8 x 10^-2 x (0.082 x 500)^-1 = 4.4 x 10^-4

888.

Find out the value of Kc from the value of Kp:2NOCl(g) ⇌ 2NO(g) + Cl2(g), Kp = 1.8 x 10-2 at 500 K

Answer»

K_C = K_p(RT)^-Δn

R = 0.082 L atm mol^-1 K^-1

For this reaction Δn = (2 + 1) - (2) = +1

K_c = K_p(RT)^-1

= 1.8 x 10^-2 x (0.082 x 500)^-1 = 4.4 x 10^-4

Discussion

889.	Write the expression for the equilibrium constant, KC for2NOCl(g) ⇌ 2NO(g) + Cl2(g)
Answer» K_P = \(\frac{p^2_{NO}.p_{O_2}}{p^2_{NaCl}}\)

Discussion

890.	Calculate the approximate pH of a 0.100 M aqueous `H_(2)S` solution. `K_(1) and K_(2)` for `H_(2)S` are `1.00xx10^(-7) and 1.30xx10^(-13)` respectively at `25^(@)C`.
Answer» `K_(2) lt lt K_(1)`. Hence `H^(+)` ions are mainly from 1st dissociation, i.e., `H_(2)S hArr H^(+)+HS^(-)` `K_(1)=([H^(+)][HS^(-)])/([H_(2)S])=([H^(+)]^(2))/([H_(2)S]) or [H^(+)]=sqrt(K_(1)[H_(2)S]):. [H^(+)]=sqrt(10^(-7)xx10^(-1))=10^(-4)`. Hence, pH = 4

Discussion

891.	A weak monobasic acid is found to be 4% ionided at 0.1 M concentration. Calculate the value of the ionisation constant.
Answer» Let the weak monobasic acid be H.A . Its ionisation in a aqueous solution can be represented as : `{:(,HA(aq),hArr,H^(+)(aq),+,A^(-)(aq)),("Initial molar concentration",C,,0,,0),("Equilibrium molar concentration",C(1-alpha),,alpha,,alpha):}` `Here " "C=0.1 M,alpha =4% 4/(100) =4 xx 10^(-2)` `"At equilibrium, " " [HA]=C(1-alpha)=0.1 Mxx (1-4xx 10^(-2)) =9.6 xx 10^(-2) M` `[H^(+)]=[A^(-)]=Calpha=0.1Mxx4xx10^(-2)=4xx10^(-3) M` `K_(a) =[[H^(+)][A^(-)]]/[[HA]]=((4xx10^(-3))xx(4xx10^(-3)))/(9.6xx10^(-2))=1.67 xx 10^(-4)`

Discussion

892.	A sample of hard water contains 96 ppm of `SO_(4)^(2-)` and 183 ppm of `HCO_(3)^(-) "with" Ca^(2+)` as the only cation. How many moles of CaO will be required to remove `HCO_(3)^(-)` from 1000 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual `Ca^(2+)` ions ? (Assume `CaCO_(3)` to be completely insoluble in water). If the `Ca^(2+)` ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (One ppm means one part of the substance in one million parts of water , weight/weight).
Answer» `SO_(4)^(2-)` present in 1000 kg of water `=(96)/(10^(6))xx1000 kg = 96 g = (96)/(96) = 1 ` mole `HCO_(3)^(-)` present in 1000 kg of water `= (183)/(10^(6))xx1000 kg = 183 g = (183)/(61)=3 ` moles `Ca^(2+)` present along with `SO_(4)^(2-)` ions = 1 mole `Ca^(2+)` present along with `HCO_(3)^(-) ` as `Ca(HCO_(3))_(2) = (3)/(2) ` mole `:.` Total `Ca^(2+)` present in 1000 kg of water ` = 1 +(3)/(2) = 2.5` moles CaO added will react with `Ca(HCO_(3))_(2)` as follows : `Ca(HCO_(3))_(2) + CaO rarr 2 CaCO_(3) harr + H_(2)O` But `Ca(HCO_(3))_(2)` present `= (3)/(2)` mole (calculated above) `:.` CaO required `= (3)/(2) ` mole = 1.5 moles After treatment with CaO i.e., removal of `Ca(HCO_(3))_(2)`, amount of `Ca^(2+) ` left (due to `CaSO_(4)` only ) in 1000 kg of water = 1 mole = 40 g `:.` Concentration of residual `Ca^(2+)` (in ppm) = 40 ppm Now 1000 kg of water contain `Ca^(2+) = 10^(-3)` mole No. of mole of `H^(+)` exchanged `= 2 xx 10^(-3)` mole `:. pH = - log (2xx10^(-3))= 2.7`

Discussion

893.	When 0.1 mole of `NH_(3)` is dissolved in water to make 1.0 L of solution, the `[OH^(-)]`of solution is `1.30xx10^(-3)M.` Calculate `K_(b)` for ` NH_(3).`
Answer» `1.8xx10^(-5)=K_(b)`

Discussion

894.	What is the approximate `OH^(-)` ion concentration of a `0.150M NH_(3)` solution? `(K_(b)=1.75xx10^(-5))`A. `2.62xx10^(-6)`B. `4.6xx10^(-6)`C. `1.62xx10^(-3)`D. `3.6xx10^(-3)`
Answer» Correct Answer - C

Discussion

895.	What will be the pH of `0.05`M barium hydroxide solution ?A. `8`B. `9`C. `7`D. `13`
Answer» Barium hydroxide is a strong electrolyte, i.e., it dissociates completely: `Ba(OH)_(2)(aq.)rarr Ba^(2+)(aq.)+2OH^(-)(aq.)` According to the equation, `C_(OH^(-))=2C_(Ba(OH)_(2))=(2xx0.05)M` `=0.1M` `=10^(-1)M` `:. C_(OH^(-))= 10^(-pOH)M` We have `pOH = 1` For any aqueous solution at 298 K, we have `pH+pOH=14` `pH=14-pOH` `=14-1` `=13`

Discussion

896.	Calculate the pH at equivalence point when a solution of 0.10 M acetic acid is titrated with a solution of 0.10 M NaOH solution . `K_(a) ` for acetic acid `= 1.9xx10^(-5)`
Answer» At the equivalence point, `CH_(3)CO ON a` is formed asn its concentration `=(0.1)/(2) M = 0.05 M`. It is a slat of weak acid and strong base. The formula for finding the pH of such a salt is `pH = - (1)/(2) [log K_(w) + log K_(a) - log c]` `:. pH = - (1)/(2) [log 10^(-14)+log (1.9xx10^(-5))-log(5xx10^(-2))]` `=-(1)/(2) [-14 + (-5+0.2788)-(-2+0.6990)]=(1)/(2) (14+5-0.2788-2+0.6990)=(17.42)/(2) = 8.71`

Discussion

897.	When 0.1 mole of `NH_(3)` is dissolved in water to make 1.0 L of solution , the `[OH^(-)]` of solution is `1.34 xx 10^(-3)`M. Calculate `K_(b)` for `NH_(3)`.
Answer» Degree of dissociation of `NH_(3) " or " NH_(4)OH i.e. alpha = (1.34 xx 10^(-3))/(0.1) = 1.34 xx 10^(-2)` According to Ostwald Dilution formula `alpha=sqrt((K_(b))/(C)) " or " K_(b) =alpha^(2)C` `K_(b) = (1.34 xx 10^(-2))^(2)xx (0.1) =1.8 xx 10^(-5)`

Discussion

898.	The ionization constant of `NH_(4)^(+)` in water is `5.6 xx 10^(-10) ` at `25^(@)C`. The rate constant for reaction of `NH_(4)^(+) and OH^(-) ` to form `NH_(3) and H_(2)O "at" 25^(@)C "is " 3.4xx10^(10) ` litre ` "mol"^(-1) "sec^(-1)"`. Calculate the rate constant for proton transfer from proton transfer from water to `NH_(3)`.
Answer» `NH_(4)^(+) + H_(2)O hArr NH_(4)OH + H^(+) , K_(a) = 5.6xx10^(-10)` `NH_(3) + H_(2)O underset(k_(b))overset(k_(f))hArr NH_(4)^(+) + OH^(-) , k_(b) = 3.4xx10^(10)` Ai,=m, To find `k_(f)` we know that for a conjugate acid-base pair `K_("acid")xxK_("base") = K_(w), i.e., K_(a)xxK_(b)=K_(w)` `:. K_(b) = (k_(w))/(K_(a))=(10^(-14))/(5.6xx10^(-10))` But `K_(b) = (k_(f))/(k_(b))` `k_(f) = K_(b) xx k_(b)=(10^(-14))/(5.6xx10^(-10))xx 3.4xx10^(10)=0.607xx10^(6)=6.07xx10^(5)`

Discussion

899.	At `25^(@)C`, the ionisation constant of anilinium hydroxide is `4.6xx10^(-10)`. Taking ionic product of water as `1xx10^(-14)`, calculate (a) hydrolysis constant of anilinium chloride (b) the degree of hydrolysis and pH value of 0.2 molar solution of the salt.
Answer» Correct Answer - (a)`2.17xx10^(-5) (b) 1.042xx10^(-2) (c) 2.68`

Discussion

900.	A certain weak acid has `K_(a)=1.0xx10^(-4)`. Calculate the equilibrium constant for its reaction with a strong base.
Answer» `{:(HA,+,BOH,hArr,BA,+,H_(2)O),("weak",,"strong",,,,,):}`:}` or HA+B^(+)+OH^(-) hArr B^(+)+A^(-)+H_(2)O or HA + OH^(-) hArr A^(-)+ H_(2)O` `K=([A^(-)])/([HA][OH^(-)]) ` ...(i) Further, for the weak acid, `HA hArr H^(+)+A^(-), K_(a) = ([H^(+)][A^(-)])/([HA])` ...(ii) Also `K_(w)=[H^(+)][OH^(-)]` ...(iii) From eqns. (i), (ii) and (iii), `K=(K_(a))/(K_(w))=(10^(-4))/(10^(-14))=10^(10)`

Discussion

Explore topic-wise InterviewSolutions in .

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A certain weak acid has `K_(a)=1.0xx10^(-4)`. Calculate the equilibrium constant for its reaction with a strong base.

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