Ncert Solutions Chemistry Class 11th
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Ka for CH3COOH is 1.8 × 10–5 and K b for NH4OH is 1.8 × 10–5 . The pH of ammonium acetate will be
(i) 7.005
(ii) 4.75
(iii) 7.0
(iv) Between 6 and 7
Ka for CH3COOH is 1.8 × 10–5 and K b for NH4OH is 1.8 × 10–5 . The pH of ammonium acetate will be
(i) 7.005
(ii) 4.75
(iii) 7.0
(iv) Between 6 and 7
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option(iii)
The equation for weak acid and weak base is given as-
pH=7+ ?(pKa?−pKb?)
Ka=1.8×10-5
pKa=−log (1.8×10-5) =5−log1.8 =4.744
pKb?=−log (1.8×10-5) =4.744
pH=7+ ?(pKa?−pKb?) =7
What will be the value of pH of 0.01 mol dm–3 CH3COOH (Ka = 1.74 × 10–5 )?
(i) 3.4
(ii) 3.6
(iii) 3.9
(iv) 3.0
What will be the value of pH of 0.01 mol dm–3 CH3COOH (Ka = 1.74 × 10–5 )?
(i) 3.4
(ii) 3.6
(iii) 3.9
(iv) 3.0
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: option (i)
pH for weak acid is given by the following equation:
pH= [pKa−logC]
= [−log (1.74×10−5)−log0.01]
=3.4
Write one chemical reaction for the preparation of D2O2.
Write one chemical reaction for the preparation of D2O2.
This is a short answer type question as classified in NCERT Exemplar
Peroxodisulphate, obtained by electrolytic oxidation of acidified sulphate solutions at high current density, on hydrolysis yields hydrogen peroxide.
2HSO4- (aq) → HO3SOOSO3 2HSO4- (aq)+2H+ (aq)+H2O2 (aq)
This method is now used for t
In which of the following solvents is silver chloride most soluble?
(i) 0.1 mol dm–3 AgNO3 solution
(ii) 0.1 mol dm–3 HCl solution
(iii) H2O
(iv) Aqueous ammonia
In which of the following solvents is silver chloride most soluble?
(i) 0.1 mol dm–3 AgNO3 solution
(ii) 0.1 mol dm–3 HCl solution
(iii) H2O
(iv) Aqueous ammonia
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: option (iv)
Aqueous ammonia will absorb chloride ions and thus the equilibria will shift in forward direction and solubility of silver chloride will increase.
How is heavy water prepared? Compare its physical properties with those of ordinary water.
How is heavy water prepared? Compare its physical properties with those of ordinary water.
This is a short answer type question as classified in NCERT Exemplar
Heavy water can be made by electrolyzing water repeatedly or as a by-product in the fertiliser industry. It's a deuterium compound that's used to make other deuterium compounds.
Physical Properties of H2O and D2O
Property | H2O | D2O |
Molecular mass (g mol-1) | 18.015 | 20.0276 |
Melting point/K | 273.0 | 276.8 |
Boiling point/K | 373.0 | 374.4 |
Enthalpy of formation/kJ mol- | -285.9 | -294.6 |
Enthalpy of Vaporisation (373 K)/kJ mol-1 | 40.66 | 41.61 |
Enthalpy of fusion/kJ mol-1 | 6.01 | — |
Temp of max. density/K | 276.98 | 284.2 |
Density (298 K)/g cm-3 | 1.0000 | 1.1059 |
Viscosity/centipoise | 0.8903 | 1.107 |
Dielectric constant/C2/N.m2 | 78.39 | 78.06 |
Which of the following will produce a buffer solution when mixed in equal volumes?
(i) 0.1 mol dm–3 NH4OH and 0.1 mol dm–3 HCl
(ii) 0.05 mol dm–3 NH4OH and 0.1 mol dm–3 HCl
(iii) 0.1 mol dm–3 NH4OH and 0.05 mol dm–3 HCl
(iv) 0.1 mol dm–3 CH4COONa and 0.1 mol dm–3 NaOH
Which of the following will produce a buffer solution when mixed in equal volumes?
(i) 0.1 mol dm–3 NH4OH and 0.1 mol dm–3 HCl
(ii) 0.05 mol dm–3 NH4OH and 0.1 mol dm–3 HCl
(iii) 0.1 mol dm–3 NH4OH and 0.05 mol dm–3 HCl
(iv) 0.1 mol dm–3 CH4COONa and 0.1 mol dm–3 NaOH
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: option (iii)
When 0.1 mol dm-3 NH4OH and 0.05 mol dm-3 HCl react total amount of HCl reacts with NH4OH to form NH4Cl and some NH4OH will be left unreacted. Thus, the resultan
Acidity of BF3 can be explained on the basis of which of the following concepts?
(i) Arrhenius concept
(ii) Bronsted Lowry concept
(iii) Lewis concept
(iv) Bronsted Lowry as well as Lewis concept
Acidity of BF3 can be explained on the basis of which of the following concepts?
(i) Arrhenius concept
(ii) Bronsted Lowry concept
(iii) Lewis concept
(iv) Bronsted Lowry as well as Lewis concept
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: option (iii)
BF3 is an electron deficient species that is why it is a Lewis acid. (all Lewis acid can accept a pair of electrons.
Molecular hydrides are classified as electron deficient, electron precise and electron rich compounds. Explain each type with two examples.
Molecular hydrides are classified as electron deficient, electron precise and electron rich compounds. Explain each type with two examples.
This is a short answer type question as classified in NCERT Exemplar
(a) Electron deficient hydrides are those that lack sufficient number of electrons to form typical covalent bonds. Group 13 hydrides, for example (BH3, AlH3, etc.).
(b) Electron precise: Electron precise hydrides are those that hav
<p><strong>Acidity of BF<sub>3</sub> can be explained on the basis of which of the following concepts?</strong></p>
<p><strong>(i)</strong> Arrhenius concept</p>
<p><strong>(ii)</strong> Bronsted Lowry concept</p>
<p><strong>(iii)</strong> Lewis concept</p>
<p><strong>(iv)</strong> Bronsted Lowry as well as Lewis concept</p>
Ka1 , Ka2 and Ka3 are the respective ionisation constants for the following reactions.
H2S H+ + HS
HS– H+ + S2
H2S 2H+ + S2
The correct relationship between is Ka1 , Ka2 and Ka3
- Ka3= Ka1 × Ka2
- Ka3= Ka1 + Ka2
- Ka3= Ka1 – Ka2
- Ka3= Ka1 / Ka2
Ka1 , Ka2 and Ka3 are the respective ionisation constants for the following reactions.
H2S H+ + HS
HS– H+ + S2
H2S 2H+ + S2
The correct relationship between is Ka1 , Ka2 and Ka3
- Ka3= Ka1 × Ka2
- Ka3= Ka1 + Ka2
- Ka3= Ka1 – Ka2
- Ka3= Ka1 / Ka2
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option(i)
H2S?H+ +HS− Ka1
HS−?H++S2− Ka2
H2S?2H++S2− Ka3
Ka1=
Ka2=
Ka3=Ka1.Ka2