A 1. 30 m solution of the weak acid ha is created. if the ka of the weak acid is 3. 06×10−9, what is the ph of the solution? the

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2 years ago

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equilibrium expression is:____.

1 answer:

babymother [125] · Answer 1 · 2023-12-07T18:03:12+03:00

A 1. 30 m solution of the weak acid ha is created. if the ka of the weak acid is 3. 06×10−9, the pH of the solution is 3.71.

Firstly we will get the equilibrium concentration, we will write the ICE chart for the reaction.

HA ------ (H+) + (A-)

1.30 0 0

1.30-x x x

Now,

the equilibrium expression can be given as

Ka = [H+] [A-]/[HA]

3. 06×10−9 = x^2/(1.30-x)

x = 1.99 × 10^(-4)

Therefore, the concentration of [H+] is 1.99 × 10^(-4).

<h3>Calculation of pH</h3>

pH = -log[H+]

pH = -log(1.99 × 10^(-4))

pH = 4-0.29

pH = 3.71

Thus, we find that the value pH of the solution is 3.71.

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