What is the concentration of the hydroxide ion given that the concentration of the hydronium ion?
1 answer:
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Answer:
24.9 L Ar
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Aqueous Solutions</u>
- States of Matter
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 40.0 g Ar
[Solve] L Ar
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Ar - 39.95 g/mol
[STP] 22.4 L = 1 mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
24.9235 L Ar ≈ 24.9 L Ar
Answer: 0.0023
Explanation:
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1) First, you must write the balanced equilibrium equation, which will let you to determine the mole ratios of the different species in solution:
HF ⇆ H⁺ + F⁻
Therefore, the mole ratio is: 1 HF: 1 H⁺ : 1 F⁻
2) Second, write the equilibrium constant of the acid, Ka, which will permit you to state the relationship of the concentrations in equilibrium:
Ka = [H⁺] [F⁻] / [HF]
[HF] is the searched concentration of the hydrofluoric acid
From the mole ratio of the balanced chemical equation [H⁺] = [F⁻]
Hence:
Ka = [H⁺] [H⁺] / [HF] = [H⁺]² / [HF]
3) From the pH value you can calculate [H⁺], using the definition:
- pH= - log [H⁺]
- Substitute: 2.90 = - log [H⁺]
Clear [H⁺] using logarithm properties:
4) Now you can substitute [H⁺] and the value of Ka in the equation for Ka:
Ka = [H⁺]² / [HF] ⇒ [HF] = [H⁺]² / Ka = (0.00126)² / (6.8×10⁻⁴) = 0.00233.
Since the value of Ka has two significant figures, you must report the answer with two significan figures, i.e. 0.0023