Equilibrium

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New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

pH=3.44

We know that,

pH=−log [H+]

∴ [H+]=3.63*104

Then, Kb= (3.63*104)2 / 0.02 (? concentration =0.02M)

⇒Kb=6.6*106

Now, Kb=Kw / Ka

⇒Ka=Kw / Kb=1014 / 6.6*106=1.51*109

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Since NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2);

NO2+ H2O ↔ HNO2 + OH

Then, Kb= [HNO2] [OH] / [NO2]

⇒ Kw/ Ka = (10−14) / (4.5*10−4)

= 0.22*10−10

Now, if x moles of the salt undergo hydrolysis, and then the concentration of various species present in the solution will be:

  [NO2]= 0.04−x; 0.04

[HNO2]= x

[OH]= x

Kb= x2 / 0.04

= 0.22*10−10

 x2= 0.0088*10−10

x= 0.093*10−5

∴ [OH]= 0.093*10−5 M

[H3O+]=10−14 / 0.093*10−5=10.75*10−9 M

⇒ pH= −log (10.75*10−9)=7.96

Now, degree of hydrolysis

= x / 0.04= (0.093*105)/ 0.04

= 2.325*105

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

c=0.1M 

pH= −log [H+]

=> 2.34 = −log [H+]

So, −log [H+]= 2.34

=> [H+]= 4.5*103

Also,

[H+]= cα

=>4.5*10−3= 0.1*α

=> α=4.5*10−2= 0.045

Then,  

Ka = α2/c = (45*103)2/ 0.1

=202.5*106

=2.02*10−4

New answer posted

5 months ago

0 Follower 17 Views

V
Vishal Baghel

Contributor-Level 10

Let the degree of ionization of propanoic acid be α. Then, representing propionic acid as HA, we have:

HA         +        H2O ⇔ H3O+    +    A

(.05−0.0α)≈0.5                         0.05α      0.05α

Ka= [H3O+] [A] / [HA]

    = (0.05α) (0.05α) / 0.05

    = 0.05α2

=> α = (Ka /0.05)1/2

          &nbs

...more

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Solubility of Sr (OH)2=19.23g/L

The molecular weight of Sr (OH)2  is 87.6 + 2 (17)=121.6

Then, concentration of Sr (OH)2=19.23 /121.63M=0.1581M

Sr (OH)2 (aq)→Sr2+ (aq)+2 (OH) (aq)

∴ [Sr2+]=0.1581M

  [OH]=2*0.1581M=0.3126M

Now, Kw= [OH] [H+]

=> [H+] = 1014 / 0.3126

=> [H+]=3.2*1014

      ∴pH= 13.495

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

[KOH]= [K+]= [OH]= (0.561*1000) / (56*200)? =0.050M
[H+]=Kw / [OH]? =10−14 / 0.05? =2.0*10−13
pH=−log [H+]=−log (2.0*10−13)

=12.7

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH=−log [H+]

Hence,   [H+] = 10−pH
Milk:  [H+] = 10−6.8 = 1.58*10−7M
Black coffee:  [H+] = 10−5.0 =1*10−5M
Tomato juice:  [H+] = 10−4.2 =6.31*10−5M
Lemon juice:  [H+]=10−2.2 = 6.31*10−3M
Egg white:  [H+]=10−7.8=1.58*10−8M

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

a. Human muscle fluid 6.83

pH=6.83

pH=−log [H+]

∴6.83=−log [H+]

[H+]= 1.48 x 107M

b. Human stomach fluid, 1.2:

pH=1.2

1.2=−log [H+]

∴ [H+] = 0.063 M = 6.3 x 10-2 M

c. Human blood, 7.38:

pH=7.38=−log [H+]

∴ [H+]= 4.17 x 108M

d. Human saliva, 6.4:

pH=6.4

6.4=−log [H+]

[H+]= 3.98 x 10−7 M

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Kb= 5.4*104

c= 0.02M

Then, α= (Kb /c)1/2

α= (5.4*104 / 2 x 10-2)1/2 =0.1643

(CH3)2NH+H2O ↔ (CH3)2NH+2+OH-

[ (CH3)2NH] = 0.02 – x ≈ 0.02

[ (CH3)2NH+2] = x

[OH-] = 0.1 + x

≈ 0.1

Now, Kb= [ (CH3)2NH+2] [OH−]/ [ (CH3)2NH] = (x * 0.1) / (0.025).

x = 1.08 x 10-4

% of dimethylamine ionised = (1.08 x 10-4) x (100 / 0.02) = 0.54%

New answer posted

5 months ago

0 Follower 41 Views

V
Vishal Baghel

Contributor-Level 10

pKa? =? logKa= 4.74

Ka? = 10? pKa =10?4.74 = 1.8*10?5
Let x be the degree of dissociation. The concentration of acetic acid solution, C = 0.05 M
The degree of dissociation,  

x= (Ka / C)1/2? = (1.8*10?5 / 0.05)1/2 ? = 0.019

(a) The solution is also 0.01 M in HCl.
Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration

[H+] = 0.01 + x
The acetate ion concentration is equal to the hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.
[CH3?

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