Question

A chemist must prepare 850.0 mL of potassium hydroxide solution with a pH of 12.20 at 25°C. He will do this in three steps: .
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Answer #1

Solution:

To calculate the amount of KOH needed, we will first calculate the [OH-].

As we know that,

pH + pOH = 14

Here pH = 12.20

Therefore,

12.20 + pOH = 14

pOH = 14 - 12.2 = 1.8

Now, pOH of the solution is given by the concentration of hydroxide anions

pOH = -log [OH-]

To find out the concentration of hydroxide anions, rewrite the above equation as,

log [OH-] = -pOH

10log [OH-] = 10 -pOH

This will get you,

[OH-] = 10 -pOH

[OH-] = 10 -1.8

[OH-] = 0.01584 M

Therefore 0.01584 moles of potassium hydroxide is present in 1000 mL. To calculate number of moles for 850 mL we will use following formula,

(0.01584 x 850) / 1000 = 0.0134 moles

Using these number of moles we can calculate the wt of potassium hydroxide,

number of moles = Weight in g / Molecular Weight

Here number of moles and molecular weight are 0.0134 and 56.10, respectively.

Therefore,

0.0134 = Weight in g / 56.1

Weight in g = 0.0134 x 56.1 = 0.751 g

Therefore, 0.751 g of potassium hydroxide will be needed to make 850 mL of solution of pH =12.2

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