Ncert Solutions Chemistry Class 11th
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Assertion and Reason:
Directions:
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
7.74. Assertion: OH– is called the conjugate base of an acid H2O.
Reason: NH4+ is called conjugate acid of the base NH3.
Assertion and Reason:
Directions:
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
7.74. Assertion: OH– is called the conjugate base of an acid H2O.
Reason: NH4+ is called conjugate acid of the base NH3.
(b) Both are correct statements but R is not the reason for A. The acid-base pair that differs only by one proton is called a conjugate acid-base pair. Therefore, OH– is called the conjugate base of an acid H2O and NH4+ is called conjugate acid of the base NH3. If Brönsted acid is a strong acid the
Upvote 
7.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?
7.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?
For precipitation to take place, it is required that the calculated ionic product exceeds the Ksp value. So, following data on Ksp values should have been provided to answer the question.
KspforFeS=6.3×10−18,
MnS=2.5×10−13,
ZnS=1.6×10−24
CdS=8.0×10−27
Before mixing:
[S2−]=1.0×10−19M and [M2+]=0.04M&nbs
Upvote 5.8. What will be the pressure of the gas mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in all vessel at 27 °C?
5.8. What will be the pressure of the gas mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in all vessel at 27 °C?
5.8. Calculation of partial pressure of H2 in 1L vessel:
P1= 0.8 bar, P2=? V1= 0.5 L, V2 = 1.0 L
As temperature remains constant, P1V1 = P2V2
=> (0.8 bar) (0.5 L) = P2 (1.0 L)
=>P2 = 0.40 bar, i.e., PH2 = 0.40 bar
Calculation of partial pressure of O2 in 1 L vessel
P1V1
Upvote 7.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 x 10−6).
7.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 x 10−6).
CaSO4 (s)↔Ca2+ (aq)+SO2−4 (aq)
Ksp= [Ca2+] [ SO2−4]
Let the solubility of CaSO4 be s.
[Ca2+] = [ SO2−4] = s
Then, Ksp=s2
9.1×10−6=s2
s =3.02×10−3mol/L
Molecular mass of CaSO4=136g/mol
Solubility of CaSO4 in gram/L= 3.02×10−3×136=0.41g/L
This means that we need 1L of water to dissolve 0.41g of CaSO4.
Therefor
Upvote 7.71. What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide?
(For iron sulphide, Ksp=6.3 x 10−18)
7.71. What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide?
(For iron sulphide, Ksp=6.3 x 10−18)
Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., x/2.
∴ [FeSO4]= [Na2S]=x / 2
Then, [Fe2+]= [FeSO4]=x/ 2
Also, [S2−]= [Na2S]=x/2
FeS (x)↔Fe2+ (aq)+S2− (aq)
Ksp= [Fe2+] [S2−]
= >6.3×10−18= (x/2) (x/2)
x2/
Upvote 7.70. The ionization constant of benzoic acid is 6.46×10−5 and Ksp for silver benzoate is 2.5×10−13. How many times is silver benzoate more soluble in a buffer of pH= 3.19 compared to its solubility in pure water?
7.70. The ionization constant of benzoic acid is 6.46×10−5 and Ksp for silver benzoate is 2.5×10−13. How many times is silver benzoate more soluble in a buffer of pH= 3.19 compared to its solubility in pure water?
Since pH=3.19,
[H3O+]=6.46×10−4M
C6H5COOH+H2O↔C6H5COO−+H3O
Ka= [C6H5COO−] [H3O+] / [C6H5COOH]
[C6H5COOH] / [C6H5COO−]= [H3O+] / Ka=6.46×10−4 / 6.46×10−5=10
Let the solubility of C6H5COOAg be xmol/L.
Then,
[Ag+]=x
[C6H5COOH]+ [C6H5COO−]=x
10 [C6H5COO−]+ [C6H5COO−]=x
[C6H5CO
Upvote 7.69. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, Ksp = 7.4 × 10–8).
7.69. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, Ksp = 7.4 × 10–8).
After mixing, the concentration of NaIO3? is 0.002 / 2? =0.001M.
After mixing, the concentration of Cu (ClO3? )2? is 0.002? / 2 =0.001M.
NaIO3? Na++IO3−?
[IO3−? ]=0.001M
Cu (ClO3? )2? Cu2++2ClO3−?
[Cu2+]=0.001M
The ionic product of Cu (IO3? )2? is
[Cu2+] [I−]
Upvote 7.68. The solubility product constant of Ag2CrO4 and AgBr are 1.1×10−12and 5.0×10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
7.68. The solubility product constant of Ag2CrO4 and AgBr are 1.1×10−12and 5.0×10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Silver chromate: Ag2CrO4? ? 2Ag+ + CrO42−?
[Ag+] = 2s1? , CrO42−? = s1?
Ksp? = (2s1? )2 (s1? ) = 4s3 = 1.1×10−12
s1? = 6.5 × 10−5 . (1)
Silver bromide: AgBr? Ag+ + Br−
[Ag+] = [Br−] = s2?
Ksp? = (s2? ) × (s2? ) = (s2)2? = 5.0 × 10−13
s2? =7.07×10−7. (2)
Divide equation (1) by equation
Upvote 7.67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.
7.67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.
1. Silver chromate:
Ag2CrO4→2Ag++CrO42−
Then, [Ag+] = (2s), [CrO42−] = s
Ksp= [Ag+]2 [CrO42−]
s = 0.65×10−4M
So, [Ag+] = 2s = 1.30 x 10−4M, [CrO42−] = 6.5 x 10-5M
2. Barium Chromate:
BaCrO4→Ba2++CrO42−
[Ba2+] = [CrO42−] = s
Then, Ksp= [Ba2+] [CrO42−] = s x s
= > 1.2 x 10-10M = s2
= > s = 1.09 x 10-5M
3. Ferric
Upvote