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kobusy [5.1K]
3 years ago
15

What is the ph of an acetic acid solution if 10 drops are titrated with 70 drops of a 0.65 m koh solution? (ka for acetic acid =

1.8 x 10-5)?
Chemistry
1 answer:
laiz [17]3 years ago
7 0
<span>What is the ph of an acetic acid solution if 10 drops are titrated with 70 drops of a 0.65 m koh solution? (ka for acetic acid = 1.8 x 10-5)?

[KOH] = 0.65 M
[OH] = 0.65 M
</span>Dilute your mom
<span>[OH]Dil= 0.65 M * 70/80 = 0.56875 M

pH = 5.4
</span>
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Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

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P at t=20000 is:

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For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

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