The given concentration of boric acid = 0.0500 M
Required volume of the solution = 2 L
Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.
Calculating the moles of 0.0500 M boric acid present in 2 L solution:
Converting moles of boric acid to mass:
Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.
Answer:
Kb = 6.22x10⁻⁷
Explanation:
Triethanolamine, C₆H₁₅O₃N, is in equilibrium with water:
C₆H₁₅O₃N(aq) + H₂O(l) ⇄ C₆H₁₅O₃NH⁺(aq) + OH⁻(aq)
Kb is defined from concentrations in equilibrium, thus:
Kb = [C₆H₁₅O₃NH⁺] [OH⁻] / [C₆H₁₅O₃N]
The equilibrium concentration of these compounds could be written as:
[C₆H₁₅O₃N] = 0.486M - X
[C₆H₁₅O₃NH⁺] = X
[OH⁻] = X
pH is -log [H⁺], thus, [H⁺] = 10^-pH = 1.820x10⁻¹¹M
Also, Kw = [OH⁻] ₓ [H⁺];
1x10⁻¹⁴ = [OH⁻] ₓ [H⁺]
1x10⁻¹⁴ = [OH⁻] ₓ [1.820x10⁻¹¹M]
5.495x10⁻⁴M = [OH⁻], that means <em>X = 5.495x10⁻⁴M</em>
Replacing in Kb formula:
Kb = [5.495x10⁻⁴M] [5.495x10⁻⁴M] / [0.486M-5.495x10⁻⁴M]
<em>Kb = 6.22x10⁻⁷</em>
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