Question

The PH of a 0.02 M solution of ammonia is 10.78 calculate
i) OH- ion concentration.
ii) the degree of dissociaton iii)the dissociaton constant.​

Answer 1

ph =c×alpha or degree of dissociaton

H×oH=10power14

kor dissociation constant =c×alpha2

where c=0.02m

Answer 2

Answer:

• OH- ion concentration is $6.3$ × $10^{-9}$
• The degree of dissociation is 0.0315
• Degree constant is  1.9 × $10^{-5}$

Explanation:

Given: The PH of a 0.02 M solution of ammonia is 10.78

To find:

(I)OH- ion concentration.

ii) the degree of dissociation

iii)the dissociation constant.​

Solution:

We know that

pOH = $- log_{10} [OH^{-} ]$

pH + pOH = 14

pH of ammonia = 10.78

10.78 + pOH = 14

pOH = 14 - 10.78

pOH = 3.22

pOH = $- log_{10} [OH^{-} ]$

⇒  3.22 = $- log_{10} [OH^{-} ]$

⇒  - 3.22 = $- log_{10} [OH^{-} ]$

⇒ $10^{-3.22} = [OH^{-} ]$

$[OH^{-} ] = 6.3$ × $10^{-9}$

OH- ion concentration is $6.3$ × $10^{-9}$

(II) The degree of dissociation:

α = $\frac{[OH^{-} }{C}$

  = $\frac{6.3 * 10^{-4} }{0.02}$

α = 0.0315

The degree of dissociation is 0.0315

(III) The dissociation constant.​

$K_{b} = \frac{c\alpha _{2} }{1 - \alpha }$

    = $\frac{0.02 * (0.0315)^{2} }{1 - 0.0315}$

    = $\frac{0.02 * 0.000992 }{0.968}$

  =  $\frac{0.02 * 9.92 * 10^{-4} }{1}$

 = 1.9 × $10^{-5}$

Degree constant is  1.9 × $10^{-5}$

Final answer:

OH- ion concentration is $6.3$ × $10^{-9}$ and The degree of dissociation is 0.0315 and Degree constant is  1.9 × $10^{-5}$

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