Class 11th
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New answer posted
6 months agoContributor-Level 10
(i) To calculate [HS−] in absence of HCl:
Let, [HS−] = x M.
H2? S? H+ + HS−
The initial concentrations of H2? S, H+ and HS− are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-x M, x M and x M respectively.
Ka? = [H2? S] [H+] [HS−]?
9.1*10−8 = x * x / (0.1−x)?
In the denominator, 0.1-x can be approximated to 0.1 as x is very small.
9.1*10−8=x*x / (0.1)?
x=9.54*10−5M= [HS−]
(ii) To calculate [HS−] in presence of HCl:
Let [HS−]= y M.
H2? S? H++HS−
The initial concentrations of H2? S, H+ and HS− a
New answer posted
6 months agoContributor-Level 10
C6H5OH? C6H5O- + H+
| C6H5OH | C6H5O- | H+ |
Initial | 0.05 M | 0 | 0 |
After dissociation | 0.05 –x | x | x |
Ka = x2 / (0.05 - x) = 1.0 x 10-10
=> x2 / 0.05 = 1.0 x 10-10
=> x2 = 5 x 10-12
=> x= 2.2 x 10-6 M
In presence of 0.01 C6H5Na, suppose y is the amount of phenol dissociated, then at equilibrium
[C6H5OH] = 0.05 – y ≈ 0.05,
[C6H5O-] = 0.01 + y ≈ 0.01 M, [H+] = yM
Therefore, Ka = (0.01) (y) / (0.05) = 1.0 x 10-10
=> y = 5 x 10-10
and α = y/c = (5 x 10-10) / (5 x 10-2)
= 10-8
New answer posted
6 months agoContributor-Level 10
For F–, Kb =Kw/Ka= 10-14/ (6.8 x 10-4)
= 1.47 x 10-11 = 1.5 x 10-11 .
For HCOO-, Kb = 10-14/ (1.8 x 10-4)
= 5.6 x 10-11
For CN–, Kb= 10-14/ (4.8 X 10-9)
= 2.08 x 10-6
New answer posted
6 months agoContributor-Level 10
pH = – log [H+] or log [H+] = – pH = – 3.76 = 4.24
Therefore, [H+] = Antilog 4.24 = 1.738 x 10-4 = 1.74 x 10-4 M
New answer posted
6 months agoContributor-Level 10
pH = – log [H+] = – log (3.8 x 10-3) = – log 3.8 + 3 = 3 – 0.5798 = 2.4202 = 2.42
New answer posted
6 months agoContributor-Level 10
(a) OH– ions can donate an electron pair and act as Lewis base.
(b) F– ions can donate an electron pair and actas Lewis base.
(c) H+ ions can accept an electron pair and act as Lewis acid.
(d) BCl3 can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.
New question posted
6 months agoTaking an Exam? Selecting a College?
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