Question

A chemist must prepare 800ml of potassium hydroxide solution with a pH 13 of at 25. He will do this in three steps:

Fill a volumetric flask about halfway with distilled water.
Weigh out a small amount of solid potassium hydroxide and add it to the flask.
Fill the flask to the mark with distilled water.

Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to significant digits.

Answer 1

Answer:

4.48 grams of potassium hydroxide that the chemist must be weighing out.

Explanation:

The pH of the KOH solution = 13

pH + pOH = 14

pOH = 14 - pH = 14 - 13 = 1

$pOH=-\log[OH^-]$

$1=-\log[OH^-]$

$[OH^-]=0.1 M$

$KOH(aq)\rightarrow K^+(aq)+OH^-(aq)$

1 mole of hydroxide ions are obtained from 1 mole of KOH. Then 0.1 mole of hydroxide ions will be obtained from :

$\frac{1}{1}\times 0.1 M=0.1 M$ of KOH

$[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}$

Volume of KOH solution = 800 mL = 0.800 L ( 1 mL = 0.001 L)

$0.1 M=\frac{\text{Moles of KOH}}{0.800 L}$

Moles of KOH = 0.1 M × 0.800 L = 0.08 mol

Mass of 0.08 moles of KOH :

0.08 mol × 56 g/mol = 4.48 g

4.48 grams of potassium hydroxide that the chemist must be weighing out.