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ZanzabumX [31]
3 years ago
11

Complete the two acid dissociation reactions for the ethylenediammonium ion and select the correct symbol for the equilibrium co

nstant for each reaction.
Chemistry
1 answer:
murzikaleks [220]3 years ago
5 0

Answer:

Step 1:

N

H

+

3

C

H

2

C

H

2

N

H

+

3

(

a

q

)

⇌

A.

K

a

1

B.

K

a

2

C.

K

b

1

D.

K

b

2

Step 2:

N

H

2

C

H

2

C

H

2

N

H

+

3

(

a

q

)

⇌

A.

K

a

1

B.

K

a

2

C.

K

b

1

D.

K

b

2

Acid-

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If a weak acid, HA, is 3% dissociated in a 0.25 M
masha68 [24]

Answer:

Ka = 2.32 \times 10^{-4}

pH = 2.12

Explanation:

Calculation of Ka:

% Dissociation = 3% = 0.03

Concentration of solution = 0.25 M

HA dissociated as:

       HA \rightarrow H^+ + A^{-}

      C(1 - 0.03)    C×0.03   C×0.03

HA] after dissociation = 0.25×0.97 = 0.2425 M

[H^+]= 0.25\times0.03 = 0.0075 M

[A^{-}]= 0.25 \times 0.03 = 0.0075 M

Ka= \frac{[H^+][A^{-}]}{[HA]}

Ka = \frac{(0.0075)^2}{0.2475} =2.32 \times 10^{-4}

Calculation of pH of the solution

pH = -log [H^+]

H^+[\tex] = 0.0075 M

pH = -log 0.0075 = 2.12

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3 years ago
What is the total number of protons in the nucleus of an atom of potassium-42?
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