Salt of Weak Acid and Strong Base
Trending Questions
Q. What is the pH when 25.0 mL of 0.2 M of CH3COOH has been titrated with 35.0 mL of 0.1 M NaOH ?
pKa(CH3COOH)=4.752
log 2.333=0.368
pKa(CH3COOH)=4.752
log 2.333=0.368
- 7
- 3.45
- 5.12
- 6.52
Q. Ka for butyric acid is 2.0×10−5. Calculate pH and hydroxyl ion concentration in 0.2 M aqueous solution of sodium butyrate.
- pH=5;[OH]−=10−5M
- pH=9;[OH]−=10−9M
- pH=9;[OH]−=10−5M
- pH=5;[OH]−=10−9M
Q.
Which of the following statement(s) is/are true? |
- The pH of 10−8M HCl is 8
- The conjugate base of H2PO−4 is HPO2−4
- If in an aqueous solution [H+]strong acid<10−6 M
[H+] of water cannot be neglected.
- If in an aqueous solution [OH−]strong base<10−6 then in [OH−] of water can be neglected.
Q. Which of the following statement(s) is/are correct for an equimolar aqueous solution of HCl and NH4OH?
- NH+4 ions will undergo hydrolysis to form NH4OH
- Solution will be basic in nature.
- Solution will be acidic in nature.
- Hydrolysis does not occur
Q. A solution consists of CH3COOH(pKa=4.74) and CH3COONa. If the pH of the solution reaches 6.74, the percentage dissociation of CH3COOH is:
- 50%
- 1%
- 99%
- None of these
Q. Calculate pH of 0.002 N NH4OH having 2 % dissociation.
- 7.6
- 8.6
- 9.6
- 10.6
Q.
The degree of hydrolysis and pH of 0.1 M sodium acetate solution are:
Hydrolysis constant of sodium acetate is 5.6 × 10−10.
h = 7.5 × 10−7 M, pH = 7.9
h = 8.5 × 10−6, pH = 6.9
h = 8.5 × 10−7, pH = 7.9
h = 7.5 × 10−6, pH = 7.9
Q. The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of 3% in its 0.1 M solution at 25∘ C. Given that the ionic product of water is 10−14 at this temperature, what is the dissociation constant of the acid ?
- 1×10−10
- 1×10−9
- 3.33×10−9
- 3.33×10−10
Q. 40 mL of 0.5 M CH3COOH is titrated with of 0.5 M NaOH. Find the pH of the solution when 39 mL NaOH has been added.
Given: Ka(CH3COOH)=1.9×10−5
Given: Ka(CH3COOH)=1.9×10−5
- 3.4
- 6.3
- 4.6
- 5.3
Q. Which of the following is correct for a salt of weak acid and strong base?
Here C is the concentration of the salt.
Here C is the concentration of the salt.
- pH=7+12(pKa+logC)
- pH=7+12(pKb+logC)
- Kh=KwKa
- Kh×Kw=Ka
Q. Given
NO−2+H2O⇋HNO2+OH−, Kb=2.22×10−11
Thus, the degree of hydrolysis of 0.04 M NaNO2 solution is:
NO−2+H2O⇋HNO2+OH−, Kb=2.22×10−11
Thus, the degree of hydrolysis of 0.04 M NaNO2 solution is:
- 1.14×10−2
- 2.36×10−5
- 0.11
- 4.86×10−3
Q. Choose the correct statements for an aqueous solution of KOCl
- The solution will be basic in nature
- The solution will be acidic in nature.
- Cationic (K+) hydrolysis occurs.
- Anionic (OCl−) hydrolysis occurs
Q. Calculate [OH−] of 0.01 M solution of ammonium hydroxide solution. The ionization constant (Kb) for NH4OH=1.8×10−5.
- 4.24×10−4 mol L−1
- 4.24×10−2 mol L−1
- 1.80×10−4 mol L−1
- 1.80×10−3 mol L−1
Q. Which of the following concentrations of NH+4 will be sufficient to prevent the precipitation of Mg(OH)2 from a solution which is 0.01 M MgCl2 and 0.1 M NH3(aq)? Given that the Ksp of Mg(OH)2=2.5×10−11 and Kb for NH3(aq)=2×10−5.
- 0.01 M
- 0.02 M
- 0.03 M
- 0.04 M
Q. 20 mL of 0.2 M CH3COONa solution is titrated against 0.1 M HCl. Calculate the pH of solution when 10 mL HCl has been added.
Given: pKa(CH3COOH)=4.74
Given: pKa(CH3COOH)=4.74
- 5.22
- 6.44
- 3.55
- 7.64
Q.
Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is___
KCN K2SO4 (NH4)2C2O4 NaCl Zn(NO3)2 FeCl3 K2CO3 NH4NO3
LiCN
Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is
KCN K2SO4 (NH4)2C2O4 NaCl Zn(NO3)2 FeCl3 K2CO3 NH4NO3
LiCN
Q. What amount (in milli mole) of sodium acetate (CH3COONa) must be added to 2 L of 0.1 M acetic acid solution to given a solution of pH = 3 [Given: Ka(CH3COOH)=10−5]
Q. Calcium lactate is a salt of weak organic acid and is represented by Ca(Lac)2. A saturated solution of Ca(Lac)2 contains 0.13 mol of this salt in 0.5 L solution. The pOH of this solution is 5.60. Assuming complete dissociation of the salt. Calculate pKa of lactic acid.
- 2.5
- 5.7
- 4.5
- 3.1
Q. 1.00 L of a buffer that is 0.100 M in HOAc (pKa=4.74) and 0.100 M in NaOAc. What is the pH after the addition of 0.0200 mol NaOH? (log 1.5=0.176)
- 4.92
- 4.74
- 8.74
- 6.74
Q. When 50 mL of 0.1 M NaOH is added to 50 mL of 0.1 M CH3COOH solution, the pH will be:
(use :(pKa)acetic acid=5, log2=0.3, log3=0.5, log5=0.7)
(use :(pKa)acetic acid=5, log2=0.3, log3=0.5, log5=0.7)
- 4.75
- 9.75
- 4.25
- 8.85
Q. A solution consists of CH3COOH(pKa=4.74) and CH3COONa. If the pH of the solution reaches 6.74, the percentage dissociation of CH3COOH is:
- 50%
- 1%
- 99%
- None of these
Q. 50 mL of a 0.1 M CH3COONa(aq) is titrated against a 0.1 M HCl(aq). Calculate the pH of solution when 50 mL HCl was added.
Given: pKa(CH3COOH)=4.74 at 25oC
Given: pKa(CH3COOH)=4.74 at 25oC
- 4.74
- 3.02
- 5.25
- 4.24
Q.
The addition of ammonium chloride to 0.1 M acetic acid will cause:
increase in its pH value
decrease in its pH value
no change in pH value
change in pH which cannot be predicted
Q. For sodium acetate solution in water, the given equilibrium reaction occur:
CH3COO− (aq)+H2O (l)hydrolysis⇌CH3COOH (aq)+OH− (aq)
Which of the following describes the correct relation ?
(where h is the degree of hydrolysis of salt)
CH3COO− (aq)+H2O (l)hydrolysis⇌CH3COOH (aq)+OH− (aq)
Which of the following describes the correct relation ?
(where h is the degree of hydrolysis of salt)
- h=√Kw⋅CKa
- h=√KwKa⋅C
- h=√Ka⋅CKw
- h=√KaKw⋅C
Q. Calculate the pH at the equivalence point in the titration of 25 mL of 0.10 M formic acid with a 0.1 M NaOH solution (given that pKa of formic acid = 3.74).
log 0.1=−1
log 0.1=−1
- 4.74
- 8.74
- 8.37
- 6.06
Q. HX is a weak acid (Ka=10−5). It forms a salt NaX having concentration of 0.1M, on reacting with caustic soda. What is the degree of hydrolysis (h) of the salt?
- 10−4
- 10−2
- 10−6
- 10−5
Q.
The degree of hydrolysis and pH of 0.1 M sodium acetate solution are:
Hydrolysis constant of sodium acetate is 5.6 × 10−10.
h = 7.5 × 10−7 M, pH = 7.9
h = 8.5 × 10−6, pH = 6.9
h = 8.5 × 10−7, pH = 7.9
h = 7.5 × 10−6, pH = 7.9
Q. Calcium lactate is a salt of weak organic acid and is represented by Ca(Lac)2. A saturated solution of Ca(Lac)2 contains 0.13 mol of this salt in 0.5 L solution. The pOH of this solution is 5.60. Assuming complete dissociation of the salt. Calculate pKa of lactic acid.
- 2.5
- 5.7
- 4.5
- 3.1
Q. On shaking H2O2 with acidified potassium dichromate and ether, the ethereal layer becomes:
- Black
- Green
- Red
- Blue
Q. Which of the following property of ammonia explains its basic nature?
- Reaction with metal salts
- All of the above
- Reaction with acids
- Complex formation