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Paha777 [63]
3 years ago
8

If the [A-]/[HA] ratio of a base/weak acid mixture changes from 10/1 to 1/10, how much does the pH change:

Chemistry
1 answer:
andreev551 [17]3 years ago
3 0

Answer:

The pH changes by 2.0 if the [A-]/[HA] ratio of a base/weak acid mixture changes from 10/1 to 1/10.

Explanation:

To solve this problem we use the<em> Henderson-Hasselbach equation</em>:

  • pH = pKa + log [A⁻]/[HA]

Let's say we have a weak acid whose pKa is 7.0:

  • pH = 7.0 + log [A⁻]/[HA]

If the [A⁻]/[HA] ratio is 10/1, we're left with:

  • pH = 7.0 + log (10/1)
  • pH = 7.0 + 1
  • pH = 8.0

Now if the ratio is 1/10:

  • pH = 7.0 + log (1/10)
  • pH = 7.0 - 1
  • pH = 6.0

The difference in pH from one case to the other is (8.0-6.0) 2.0.

<em>So the pH changes by 2.0</em> if the [A-]/[HA] ratio of a base/weak acid mixture changes from 10/1 to 1/10.

<u>Keep in mind that no matter the value of pKa, the answer to this question will be the same.</u>

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Y = 57.15 − 0.5 X If Y equals 5 , what is X ?<br><br> Then is Y=3, what is X?
Fiesta28 [93]

When y equals 5, x is 104.3

When y equals 3 then x is 108.3

<em><u>Solution:</u></em>

<em><u>Given expression is:</u></em>

y = 57.15 - 0.5x

<h3><u>If y equals 5 what is x ?</u></h3>

Substitute y = 5 in given expression

5 = 57.15 - 0.5(x)

5 = 57.15 - 0.5x

0.5x = 57.15 - 5

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Divide both sides by 0.5

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<h3><u>If y = 3 what is x ?</u></h3>

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