Answer:
The NaCl concentration will be 0.03 M.
Explanation:
Given data:
Initial volume = V₁ = 56.98 mL (56.98/1000 = 0.05698 L)
Initial concentration = M₁= 0.5894 M
Final volume = V₂= 1.20 L
Final concentration = M₂= ?
Solution:
By diluting the solution volume of solution will increase while number of moles of solute remain the same.
Formula:
Initial concentration × Initial volume = Final concentration × Final volume
M₁V₁ = M₂V₂
M₂ = M₁V₁ / V₂
M₂ = 0.5894 M × 0.05698 L / 1.20 L
M₂ = 0.0336 M /1.20
M₂ = 0.03 M
Removing chloride ion is the reaction away
Answer:
Explanation:
1)<u><em> Ionization equilibrium equation: given</em></u>
- H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
2) <em><u>Ionization equilibrium constant, at 25°C, Kw: given</u></em>
<u>3) Stoichiometric mole ratio:</u>
As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:
- [H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M
- ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M
<u><em>4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C </em></u><em><u>and you need to calculate what the [H₃O⁺(aq)] is.</u></em>
Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:
Then you can substitute the known values and solve for the unknown:
- 1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M
- ⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M
As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.