Aerobic cellular respiration requires an adequate supply of
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Answer:
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- <u><em>pOH = 0.36</em></u>
Explanation:
Both <em>potassium hydroxide</em> and <em>lithium hydroxide </em>solutions are strong bases, so you assume 100% dissociation.
<u>1. Potassium hydroxide solution, KOH</u>
- Volume, V = 304 mL = 0.304 liter
- Molarity, M = 0.36 M
- number of moles, n = M × V = 0.36M × 0.304 liter = 0.10944 mol
- 1 mole of KOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.10944
<u>2. LIthium hydroxide, LiOH</u>
- Volume, V = 341 mL = 0.341 liter
- Molarity, M = 0.51 M
- number of moles, n = M × V = 0.341 liter × 0.51 M = 0.17391 mol
- 1mole of LiOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.17391
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<u>3. Resulting solution</u>
- Number of moles of OH⁻ ions = 0.10944 mol + 0.17391 mol = 0.28335 mol
- Volume of solution = 0.304 liter + 0.341 liter = 0.645 liter
- Molar concentration = 0.28335 mol / 0.645 liter = 0.4393 M
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<u>4. </u><em><u>pOH</u></em>
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