Henderson Hassleback Equation
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[Given : pKa1 and pKa2 of H2CO3 are 6.37 and 10.32, respectively; log2=0.30]
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 x 105L
3.6 x 108 L
8 x 104 L
2.4 x 103 L
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
12.8
8.62
The pH of an aqueous solution at 25∘C having 0.1 M NaOH and 0.3 M to acetic acid (pKa = 4.76) would be nearly
4.25
5.05
4.45
4.75
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 2 : 1 molar ratio
All the three gives nearly similar result
Acid chloride and ammonia are treated in 1 : 2 molar ratio
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
pH of 0.5 M Ba(CN)2 solution (pKb of CN−=9.30) is :
8.35
3.35
9.35
9.50
A weak base BOH is titrated with a strong acid HA. When 10ml of HA is added, the pH is found to be 9.00 and when 25 ml is added, pH is 8.00. The volume of the acid required to reach the equivalence point
50 ml
35 ml
40 ml
30 ml
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 ?
75 ml and 75 ml
50 mL and 100 mL
100 ml and 50 ml
60 ml and 90 ml
0.1 mol of methyl amine (Kb = 5 × 10−4) is mixed with 0.08 mol of HCl and the solution is diluted to 1 litre. The hydrogen ion concentration of the resulting solution is:
[H+] = 3.68 × 10−11 M
[H+] = 1.79 × 10−11 M
[H+] = 6.53 × 10−11 M
[H+] = 8 × 10−11 M