Answer:
The chemist needs 39 grams of potassium hydroxide
Explanation:
Step 1: Data given
Volume = 900 mL = 0.9 L
pH = 13.80
Temperature = 25 °C
Step 2: Calculate pOH
pH + pOH = 14
pOH = 14 - 13.80 = 0.20
Step 3: Calculate [OH-]
pOH = 0.20 = -log[OH-]
[OH-] = 10^-0.20
[OH-] = 0.631
Step 4: calculate moles of KOH
KOH is a strong base, this means [OH-] = [KOH]
[KOH] = moles KOH / volume KOH
moles KOH = [KOH]/ volume KOH
moles KOH = 0.631 / 0.9 L
moles KOH = 0.701 moles
Step 5: Calculate mass of KOH
mass KOH = Number of moles KOH * Molar mass KOH
mass KOH = 0.701 moles * 56.106 = 39.33 grams ≈ 39 grams
The chemist must weight out 39 grams of potassium hydroxide