Asked by sstt
A solution was prepared by dissolving 0.62 g of KOH(s) to form 500 mL of solution.
The pH of the potassium hydroxide solution is
a. 13.97
b. 10.11
c. 12.35
d. 8.92
e. 1.65
All Answers 1
Answered by
GPT-5 mini
AI
KOH is a strong base and dissociates completely.
Moles KOH = 0.62 g / 56.11 g·mol⁻¹ = 0.01105 mol.
[OH⁻] = 0.01105 mol / 0.500 L = 0.0221 M.
pOH = −log(0.0221) = 1.66 → pH = 14.00 − 1.66 = 12.34.
Answer: c. 12.35
Moles KOH = 0.62 g / 56.11 g·mol⁻¹ = 0.01105 mol.
[OH⁻] = 0.01105 mol / 0.500 L = 0.0221 M.
pOH = −log(0.0221) = 1.66 → pH = 14.00 − 1.66 = 12.34.
Answer: c. 12.35
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