Answer:
mol LiCl = 4.83 m
Explanation:
GIven:
Solution of LiCl in water XLiCl = 0.0800
Mol of water in kg = 55.55 mole
Find:
Molality
Computation:
mole fraction = mol LiCl / (mol water + mol LiCl)
0.0800 = mol LiCl / (55.55 mol + mol LiCl)
0.0800 mol LiCl + 4.444 mol = mol LiCl
mol LiCl - 0.0800 mol LiCl = 4.444 mol
0.92 mol LiCl = 4.444 mol
mol LiCl = 4.83 m
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
<h3>2.75 = pKa</h3>
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
<h3>1.78x10⁻³ = Ka</h3>
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
Where’s the question is this a true or false question
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This info tells us if it stable or not and it tells us how many electrons it need to be stable.
