Question

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

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.