Henderson Hassleback Equation
Trending Questions
- 5.21
- 4.34
- 4.76
- 2.52
The pH of a mixture of and is approximately
- 4.5
- 2.5
- 9.5
- 7
(Given Ka1=10−3, Ka2=10−8, Ka3=10−13)
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
- 7.5
- 3.4
- 3.75
- 5.6
- Increasing the concentration of both weak acid and salt by two moles
- Increasing the concentration of both the acid and salt by 10 times
- Diluting the solution by 10 times
- Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times
- Colour transition range for Methyl Orange is nearly (6−8.6)
- Colour transition range for Methyl Orange is nearly (3.2−4.4)
- Colour transition range for Methyl Orange is nearly (8.3−10)
- Colour transition range for Methyl Orange is nearly (6.2−7.4)
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
pH of 0.5 M Ba(CN)2 solution (pKb of CN−=9.30) is :
8.35
3.35
9.35
9.50
(Given : Ka1=1×10−3, Ka2=1×10−8 and Ka3=1×10−13)
pH of 0.5 M Ba(CN)2 solution (pKb of CN−=9.30) is :
8.35
3.35
9.35
9.50
- 3.74 to 4.74
- 4.74 to 5.74
- 3.74 to 5.74
- 1.0 to -1.0
- 8.4
- 4.3
- 5.7
- 7.5
- 4.5
- 2.5
- 9.5
- 7
What is the threshold below which, if the pH of rain water drops, it is called ‘acid rain’?
5.4
5.6
6.5
6.6
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 ?
11
35
23
12
- 6.8
- 7.24
- 8.4
- 7
- 9.3
- 7.3
- 10.3
- 8.3
They yield of acid amide in the reaction, RCOCl+NH3→RCONH2 , is maximum when
Acid chloride and ammonia are treated in equimolar ratio
Acid chloride and ammonia are treated in 1 : 2 molar ratio
Acid chloride and ammonia are treated in 2 : 1 molar ratio
All the three gives nearly similar result
- 6.02×1019
- 6.02×1024
- 6.02×10−19
- 6.02×10−13
- Increasing the concentration of both weak acid and salt by two moles
- Increasing the concentration of both the acid and salt by 10 times
- Diluting the solution by 10 times
- Increasing the concentration of the salt by 10 times and decreasing concentration of the acid by 10 times
log (1.96)=0.292, log (0.1)=−1
- 10:1
- 8:1
- 4:1
- 1:10
- The concentration of proton donor equals the concentration of proton acceptor
- Concentration of proton donor become zero
- Concentration of proton acceptor become zero
- The concentration of proton donor become log 1/10th of concentration of proton acceptor
- 4 mL
- 8 mL
- 2 mL
- 10 mL
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
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
Solution A : Contains 1 mol CH3COONa+0.5 mol HCl
Solution B : Contains 1 mol CH3COONa+0.5 mol CH3COOH
If Ka(CH3COOH)=10−5 then ratio of [H+ (aq.)] of solution A to that of solution B is,
(Given: log 1=0 and log 2=0.3, log(5×10−6=−5.3)
- 1:2
- 2:1
- 1:4
- 4:1
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