Henderson Hassleback Equation
Trending Questions
- 4.87
- 5.8
- 2.4
- 9.2
To 1L solution containing 0.1 mol each of NH3 and NH4Cl, 0.05 mol of NaOH is added. The change in pH will be (pKb for NH3=4.74)
-0.48
0.48
0.30
-0.30
- 3.74 to 4.74
- 4.74 to 5.74
- 3.74 to 5.74
- 1.0 to -1.0
- 1.8×10−6
- 1.8×10−5
- 9×10−6
- 9×10−5
Calculate the pH of a solution of 0.10 M acetic acid after 50.0 mL of 0.10 M acetic acid solution is treated with 25.0 mL of 0.10 M NaOH. (Ka of acetic acid = 1.8×10−5)
5.6
4.74
8.62
12.8
- 60 mL
- 40 mL
- 30 mL
- 50 mL
To what volume one litre of 0.1M acetic acid solution diluted so that the pH of the solution becomes double of the initial ? ( Ka=1.8×10−5)
5.5×105L
8×104L
2.4×103L
3.6×108L
H2S, SiCl4, BeF2, CO2−3 andHCOOH
Calculate the pH of a solution of 0.10 M acetic acid after 50.0 mL of 0.10 M acetic acid solution is treated with 25.0 mL of 0.10 M NaOH. (Ka of acetic acid = 1.8×10−5)
5.6
4.74
8.62
12.8
The base imidazole has Kb of 1.0×10−7. What volumes of 0.02 M HCl and 0.02 M imidazole should be mixed to make 150 mL of a buffer of pH 7 ?
100 ml and 50 ml
50 mL and 100 mL
75 ml and 75 ml
60 ml and 90 ml
- 60 mL
- 40 mL
- 30 mL
- 50 mL
- 14×10−7 M
- 14×10−5 M
- 1.4×10−5 M
- 1.4×10−7 M
Electrolyte | Λ0(S cm2 mol−1) |
KCl | 149.9 |
NaCl | 126.5 |
HCl | 426.2 |
KNO3 | 145.0 |
NaOAc | 91.0 |
- Λ0HNO3=421.3 S cm2mol−1
Λ0HOAc=390.7 S cm2mol−1 - Λ0HNO3=517.2 S cm2mol−1
Λ0HOAc=149.8 S cm2mol−1 - Λ0HNO3=390.7 S cm2mol−1
Λ0HOAc=517.2 S cm2mol−1 - Λ0HNO3=149.8 S cm2mol−1
Λ0HOAc=444.7 S cm2mol−1
(Given Ka1=10−3, Ka2=10−8, Ka3=10−13)
- 6.8
- 7.24
- 8.4
- 7
- 1.8×10−5
- 8.22×10−6
- 1.8×10−6
- 8.2×10−5
The base imidazole has a Kb of 9.8×10−8 at 25∘ C.
In what ratio of volumes should 0.02 M HCl and 0.02 M imidazole be mixed to make 100 ml of a buffer at pH 7 ?
35
23
12
11
To what volume one litre of 0.1M acetic acid solution diluted so that the pH of the solution becomes double of the initial ? ( Ka=1.8×10−5)
5.5×105L
8×104L
2.4×103L
3.6×108L