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3 years ago

What’s the ionic equation for lithium + water

Chemistry

1 answer:

disa [49]3 years ago

7 0

Answer:

2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)

Explanation:

An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.

In this reaction, solid lithium reacts with liquid water to form soluble lithium hydroxide and gaseous hydrogen .

1. Molecular equation

2Li(s) + 2H₂O(ℓ) ⟶ 2LiOH(aq) + H₂(g)

2. Ionic equation

Lithium hydroxide is a soluble ionic compound, so we write it as hydrated ions.

2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)

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g 304 mL of a 0.36 M potassium hydroxide solution is added to 341 mL of a 0.51 M lithium hydroxide solution. Calculate the pOH o

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Answer:

pOH = 0.36

Explanation:

Both potassium hydroxide and lithium hydroxide solutions are strong bases, so you assume 100% dissociation.

1. Potassium hydroxide solution, KOH

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

2. LIthium hydroxide, LiOH

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

3. Resulting solution

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

4. pOH

$pOH=-\log [OH^-]=-\log (0.439)=0.36$ ← answer

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