Introduction and Degree of Hydrolysis
Trending Questions
pH value of a solution, whose hydronium ion concentration is 6.2 × 10−9
mol/l, is
6.21
7.21
7.75
8.21
A sample of 100 ml of 0.10 M acid HA (Ka=1×10−5) is titrated with standard 0.10 M KOH.How many ml of KOH will have to be added when the pH in the titration flask will be 5.00 ?
0
10
100
50
Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation. (pKa of formic acid = 3.8 and pKb of ammonia = 4.8)
7.5
6
6.5
7.2
- 7.45×10−5
- 7.45×10−3
- 7.45×10−1
- 5.5×10−2
A solution of NaCN is
Alkaline in nature
Neutral in nature
Amphoteric in nature
Acidic in nature
Hydrolysis is the reverse of:
ionization
neutralization
acidification
saponification
In an attempted determination of the solubility product constant of TI2S the solubility of this compound in pure CO2− free water was determined as 6.3 × 10−6 mol/L. Assume that the dissolved sulfidehydrolyzes almost completely to HS− and that the further hydrolysis to H2S can be neglected. What is the computed Ksp ?
3.5 × 10-25
What would be the hydrogen ion concentration in a solution containing 0.04 mole of acetic acid and 0.05 mole of sodium acetate in 500 ml of the solution?
Dissociation constant for acetic acid is 1.8 x 10−5.
The [ H+] is:
1.44 × 10−5
14.5 × 10−9
3.88 × 10−2
none of the above
The solution of a salt of a weak acid and weak base will have pH -
(Kb = 10−6 ; Ka = 10−4)
8
7
6
4
One or both the ions of a salt react with water to produce acidic, basic or neutral solution. This process is called:
neutralisation
ionisation
saponification
hydrolysis
Given : Ka1 and Ka2 for H2CO3 are 4.5×10−7 and 4.7×10−11 respectively at 25 oC.
- 5.66
- 8.33
- 10.57
- 6.33
- 0.01
- 0.1
- 0.001
- None of the above
- 7.82
- 6.18
- 9.40
- 8.40
[Given: Kb(NH4OH)=10−5 and log2=0.301]
- 2.65
- 4.35
- 4.65
- 5.35
Ionization constants of weak acid HA and weak base BOH are 3.0×10−7 each at 298 K. The percent degree of hydrolysis of their salt at the dilution of 10 L is :
40
50
75
25
An aqueous solution of Copper (II) sulphate pentahydrate turns blue litmus red. It is due to the
presence of Cu2+ ions
presence of SO42– ions
hydrolysis of Cu2+ ions
hydrolysis of SO42– ions
- 2
- 4
- 1
- 3
- 0.55
- 7.63
- 0.55×10−2
- 7.63×10−3
- 0.001
- 0.01
- 0.02
- 0.1
Given : antilog(−10.83)=1.479×10−11
- 1.74×10−5
- 3.52×10−3
- 6.76×10−4
- 5.38×10−2
- 7.2
- 12.3
- 8.5
- 9.75
- h1>h2>h3>h4
- h1=h2=h3=h4
- h4>h3>h2>h1
- None of these
Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation. (pKa of formic acid = 3.8 and pKb of ammonia = 4.8)
7.2
7.5
6
6.5
NH+4 (aq)+H2O (l)hydrolysis⇌NH4OH (aq)+H+ (aq)
Which of the following describes correct relation ?
- Kh=KwKb
- Kh=Kw×Kb
- Kh=KbKw
- None of the above
A solution of NaCN is
Acidic in nature
Alkaline in nature
Neutral in nature
Amphoteric in nature
- 2.8×10−6
- 20.5×10−5
- 2.2×10−5
- 1.1×10−3
(i) CO(g)+H2O(g)⇌CO2(g)+H2(g);K1(ii) CH4(g)+H2O(g)⇌CO(g)+3H2(g);K2(iii) CH4(g)+2H2O(g)⇌CO2(g)+4H2(g);K3
which of the following is correct?
- K3=K1/K2
- K3=K21/K3−2
- K3=K1K2
- K3=K1√K2
- 7.45×10−5
- 7.45×10−3
- 7.45×10−1
- 5.5×10−2