Question

PH of solution after adding 0.1 mole of solid NaOH to a liter​

Answer 1

Answer:

K

b

=2×10

−5

(NH

3

)

NH

3

→NH

4

+OH

−

Cα

2

=Ka

α=

c

Ka

=

0.1

2×10

−5

=

2

×10

−3

[OH

−

]=0.1×

2

×10

−3

=

2

×10

−4

NaOH→Na

+

+OH

−

[OH

−

]=0.1M

P

H

=14−log0.1

=10

P

H

due to NH

3

=14−log

2

×10

−4

=14(

2

1

log

2

+log10

−4

)

=14−(−4+0.15)=10.15

(A) degree of dissociation approaches to zero

α→0

(B) change in P

H

be 13−10.15=2.85

(C)(Na

+

)=0.1M NH

3

≃0.1M

[OH

−

]=0.1+2 ×10 −4

≃0.1

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