Question

If the dissociation constant of 5×10^-4m aqueous solution of diethylamine is 2.5×10^-5 its ph value is

Answer 1

Answer:

Explanation:

We assess the reaction.....

H

N

(

C

2

H

5

)

2

(

a

q

)

+

H

2

O

(

l

)

⇌

H

2

+

N

(

C

2

H

5

)

2

+

H

O

−

And

K

b

=

[

H

2

N

+

(

C

2

H

5

)

2

]

[

H

O

−

]

[

H

N

(

C

2

H

5

)

2

]

Now initially

[

H

N

(

C

2

H

5

)

2

]

=

0.05

⋅

m

o

l

⋅

L

−

1

....and we are given that the

p

H

=

12

, and thus

p

O

H

=

14

−

12

=

2

, and thus

[

H

O

−

]

=

10

−

2

⋅

m

o

l

⋅

L

−

1

And thus

[

H

O

−

]

=

10

−

2

⋅

m

o

l

⋅

L

−

1

; and so

H

2

+

N

(

C

2

H

5

)

2

=

10

−

2

⋅

m

o

l

⋅

L

−

1

, and

[

H

N

(

C

2

H

5

)

]

=

(

0.05

−

0.01

)

⋅

m

o

l

⋅

L

−

1

=

0.04

⋅

m

o

l

⋅

L

−

1

.

Answer 2

pH of solution is 8.4

Explanation:

$(C_2H_5)_2NH+H_2O\rightarrow (C_2H_5)_2NH_2^++OH^-$

 cM              0             0

$c-c\alpha$                      $c\alpha$          $c\alpha$  

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= $5\times 10^{-4}M$ and $\alpha$ = ?

$K_a=2.5\times 10^{-5}$

Putting in the values we get:

$2.5\times 10^{-5}=\frac{(5\times 10^{-4}\times \alpha)^2}{(5\times 10^{-4}-5\times 10^{-4}\times \alpha)}$

$(\alpha)=0.005$

$OH^-]=c\times \alpha$

$[OH^-]=5\times 10^{-4}\times 0.005=2.5\times 10^{-6}$

Also $pOH=-log[OH^-]$

$pOH=-log[2.5\times 10^{-6}]=5.6$

$pH=14-5.6=8.4$

Thus pH of solution is 8.4

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