Introduction and Degree of Hydrolysis
Trending Questions
Q. % hydrolysis of 0.1 M CH3COONH4 when Ka=Kb=1.8×10−5 is -
- 0.55
- 7.63
- 0.55×10−2
- 7.63×10−3
Q. Calculate [OH−] of 0.01 M solution of ammonium hydroxide solution. The ionization constant (Kb) for NH4OH=1.8×10−5.
- 1.80×10−4 mol L−1
- 1.80×10−3 mol L−1
- 4.24×10−4 mol L−1
- 4.24×10−2 mol L−1
Q.
The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10−5 mol l−1. Its
solubility product number will be
4 × 10−10
4 × 10−15
1 × 10−15
1 × 10−10
Q. Calculate the hydrolysis constant of a salt containing NO−2 ion. If Ka for HNO2=4.5×10−10.
- 2.2×10−5
- 1.1×10−3
- 2.8×10−6
- 20.5×10−5
Q.
If the solubility product Ksp of a sparingly soluble salt MX2 at 25∘C is
1.0 × 10−11, the solubility of the salt in mole/litre at this temperature will
be
2.46 × 1014
1.36 × 10−4
2.60 × 10−7
1.20 × 10−10
Q. If the solubility of sparingly soluble salt PbCl2 is 4.41 g L−1 at 25o C. Find its solubility product at the given temperature.
(Take Molecular mass PbCl2≈278 g mol−1)
(Take Molecular mass PbCl2≈278 g mol−1)
- 2.15×10−1 mol3 L−3
- 1.64×10−5 mol3 L−3
- 4.64×10−8 mol3 L−3
- 3.67×10−10 mol3 L−3
Q. If pKb for the fluoride ion at 25∘C is 10.83
The ionization constant of hydrofluoric acid in water at this temperature will be:
(Given:10−10.83=1.48×10−11)
The ionization constant of hydrofluoric acid in water at this temperature will be:
(Given:10−10.83=1.48×10−11)
- 1.74×10−5
- 1.52×10−3
- 6.76×10−4
- 8.38×10−2
Q. 20 mL of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid to give 70 mL of the solution. Calculate the pH of the solution.
The dissociation constant for CH3COOH is 1.8×10−5.
The dissociation constant for CH3COOH is 1.8×10−5.
- 3.56
- 4.56
- 5.56
- 6.56
Q. Calculate the pH of a 0.1 M K3PO4 solution at 25∘C. The third dissociation constant Ka3 of orthophosphoric acid is 1.3×10−12. Assume that hydrolysis proceeds only in the first step.
Take log(4.6)=0.66
Take log(4.6)=0.66
- 11.28
- 10.24
- 9.34
- 12.34
Q. Calculate the pH of a 0.1 M K3PO4 solution at 25∘C. The third dissociation constant Ka3 of orthophosphoric acid is 1.3×10−12. Assume that hydrolysis proceeds only in the first step.
Take log(4.6)=0.66
Take log(4.6)=0.66
- 11.28
- 10.24
- 12.34
- 9.34
Q. What is the concentration of CO2−3 in a 0.1 M Na2CO3 solution after the hydrolysis reactions have come to an equilibrium? (Ka(HCO−3)=4.7×10−11)
- 2×104
- 6.3×10−4
- 2×10−4
- 9.54×10−2
Q. Which of the following expressions for the percentage ionization of a monoacidic base (BOH) in an aqueous solution at appreciable concentration is not correct?
- 100×√Kbc
- 1×1001+10(pKb−pOH)
- Kw[H+]Kb+Kw
- Kb×100Kb+[OH−]
Q. Match List-I with List-II.
List-IList-II(P)Fe(NO3)3(aq)(1)Only cationic hydrolysis(Q)KClO4(aq)(2)Only anionic hydrolysis(R)RCOONa(aq)(3)Both cationic as anionic hydrolysis(S)NH4CN(aq)(4)No hydrolysis
List-IList-II(P)Fe(NO3)3(aq)(1)Only cationic hydrolysis(Q)KClO4(aq)(2)Only anionic hydrolysis(R)RCOONa(aq)(3)Both cationic as anionic hydrolysis(S)NH4CN(aq)(4)No hydrolysis
- PQRS1234
- PQRS1423
- PQRS2143
- PQRS4321
Q.
The pKaoof acetylsalicyclic acid (aspirin) is 3.5. The pH of gastric juice in the human stomach is about 2 to 3 and the pH in small intestine is about 8. Aspirin will be
unionized in the small intestine and in the stomach
completely ionized in the small intenstine and in the stomach
ionized in the stomach and almost unionized in the small intestine
ionized in the small intenstine and almost unionized in the stomach
Q. Calcium lactate is a salt of weak acid and represented as Ca(LaC)2. A saturated solution of Ca(LaC)2 contains 0.13 mole of salt in 0.50 litre solution. The pOH of solution is 5. Assuming complete dissociation of salt, calculate Ka of lactic acid.
- 5.2×10−5
- 8.6×10−5
- 7.8×10−6
- 4×10−4
Q. Calculate [H+] from a CH3COOH in a solution which is 10−1 M in HCl and 10−3 M in CH3COOH; [Ka=2×10−5].
- 2×10−4
- 4×10−7
- 2×10−7
- None of the above
Q. Calculate the solubility of a sparingly soluble compund A2X3 at 25∘C.
The solubility product Ksp of A2X3 is 1.08×10−23 at the given temperature.
The solubility product Ksp of A2X3 is 1.08×10−23 at the given temperature.
- 1×10−7 mol L−1
- 1.08×10−7 mol L−1
- 1×10−5 mol L−1
- 1.08×10−3 mol L−1
Q.
Definition (produces hydroxyl ions on dissolution in water)
(b) Ca(OH)2 🡪
Q. If pKb for the fluoride ion at 25∘C is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is
Given : antilog(−10.83)=1.479×10−11
Given : antilog(−10.83)=1.479×10−11
- 1.74×10−5
- 3.52×10−3
- 6.76×10−4
- 5.38×10−2
Q. Solubility product Ksp of a sparingly soluble salt Zn(CN)2 is 8×10−12. Its solubility value in (mol L−1) is:
(Take 3√2=1.26)
(Take 3√2=1.26)
- 1.26×10−5
- 1.26×10−6
- 1.26×10−4
- 1.26×10−12
Q.
Which of the following salt solutions (0.1 M each) will have the lowest pH values?
CaCO3
Ca(OH)2
CaCl2
CH3COONa
Q. What is the pH of 0.01 M NaHCO3 ?
Given : Ka1 and Ka2 for H2CO3 are 4.5×10−7 and 4.7×10−11 respectively at 25 oC.
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
Q. K1 and K2 for oxalic acid C2H2O4 are 6.5×10−2 and 6.1×10−5 respectively. What will be [OH−] in a 0.01 M solution of sodium oxalate?
- 9.6×10−16
- 1.4×10−1
- 1.26×10−6
- 3.3×10−9
Q. Calculate the percentage of hydrolysis in 0.003 M aqueous solution of NaOCN. Ka for HOCN=3.33×10−4.
- 0.001
- 0.01
- 0.02
- 0.1
Q. % hydrolysis of 0.1 M CH3COONH4 when Ka=Kb=1.8×10−5 is -
- 0.55
- 7.63
- 7.63×10−3
- 0.55×10−2
Q. HX is a weak acid (Ka=10−5). It forms a salt NaX (0.1M) on reacting with caustic soda. What is the percentage of degree of hydrolysis (h) of NaX ?
- 0.01
- 0.1
- 0.001
- None of the above
Q.
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
Q. The compound whose 0.1 M solution at 25∘C is basic is :
- NH4Cl
- CH3COONa
- All of the above.
- (NH4)2SO4
Q. Hydrolysis constant of NH+4 is 5.55×10−10. Calculate the ionisation constant of NH+4.
- 5.55×10−10
- 5.55×10−5
- 1.88×10−10
- 1.88×10−4
Q. Calcium lactate is a salt of weak acid and represented as Ca(LaC)2. A saturated solution of Ca(LaC)2 contains 0.13 mol of salt in 0.50 L solution. The pOH of solution is 5. Assuming complete dissociation of salt, calculate Ka of lactic acid.
- 8.6×10−5
- 5.2×10−5
- 7.8×10−6
- 4×10−4
