<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
<span><span>Argon,</span><span>Carbon dioxide,</span><span>Neon,</span><span>Helium, and </span><span>Methane</span></span>
Answer:
<em>Hydrogen.</em>
Explanation:
You've probably seen "
" which is the formula for water. It means that there's 2 hydrogen atoms, and one oxygen atom, in one molecule of water.
<em>Hope this helps! Feel free to mark me Brainliest if you feel this helped. :)</em>
Answer:
pH = 12.20
Explanation:
Ca(OH)2 is a strong base, so it dissociates completely. A 0.08 M solution of Ca(OH)2 is 0.16 M OH-, since every mole of Ca(OH)2 has 2 OH-.
Calculate pOH using [OH-] = 0.16 M
pOH = -log(0.16) = 0.80
pH = 14 - pOH = 14 - 0.80 = 12.20
