Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
<h3>Answer:</h3>
2.55 × 10²² Na Atoms
<h3>Solution:</h3>
Data Given:
M.Mass of Na = 23 g.mol⁻¹
Mass of Na = 973 mg = 0.973 g
# of Na Atoms = ??
Step 1: Calculate Moles of Na as:
Moles = Mass ÷ M.Mass
Moles = 0.973 g ÷ 23 g.mol⁻¹
Moles = 0.0423 mol
Step 2: Calculate No, of Na Atoms as:
As 1 mole of sodium atoms counts 6.022 × 10²³ and equals exactly to the mass of 23 g. So, we can write,
Moles = No. of Na Atoms ÷ 6.022 × 10²³ Na Atoms.mol⁻¹
Solving for No. of Na Atoms,
No. of Na Atoms = Moles × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 0.0423 mol × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 2.55 × 10²² Na Atoms
<h3>Conclusion: </h3>
2.55 × 10²² sodium atoms are required to reach a total mass of 973 mg in a substance of pure sodium.
