<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
Answer:
A.It is the same for every sample of a single substance.
Explanation:
Answer:
293.1 mL.
Explanation:
- Boyle's law states that: at a constant temperature the pressure of a given mass of an ideal gas is inversely proportional to its volume.
- It can be expressed as: <em>P₁V₁ = P₂V₂,</em>
P₁ = 546.0 mm Hg, V₁ = 350.0 mL.
P₂ = 652.0 mm Hg, V₂ = ??? mL.
<em>∴ V₂ = (P₁V₁)/(P₂)</em> = (546.0 mm Hg)(350.0 mL) / (652.0 mm Hg) = <em>293.1 mL.</em>