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 Henderson-Hasselbach equation:

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.

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

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

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3 years ago

Chloroform is a liquid once used for anesthetic. What is the volume of 5.0 g of chloroform? The density of chloroform 1.49 g/mL

mario62 [17]

Answer:

3.3557047 mL

Explanation:

The density can be found using the following formula:

$d=\frac{m}{v}$

Let's rearrange the formula to find the volume, $v$ .

$d*v=\frac{m}{v}*v$

$d*v=m$

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$v=\frac{m}{d}$

The volume can be found by dividing the mass by the density. The mass of the chloroform is 5 grams and the density is 1.49 grams per milliliter. Therefore,

$m= 5g \\d= 1.49 g/mL$