Answer is: pH of ammonium hydroxide is 11.13.
Chemical reaction of ammonium hydroxide in water: NH₄OH → NH₄⁺ +
OH⁻<span>.
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Kb(NH₄OH) = 1,8·10⁻⁵<span>.
c</span>₀(NH₄OH<span>) = 0.1 M.
c(NH</span>₄⁺) = c(OH⁻<span>) = x.
c(</span>NH₄OH<span>) = 0.1 M- x.
Kb = c(NH</span>₄⁺) · c(OH⁻) / c(NH₄OH<span>).
0,000018 = x² / 0.1 mol/L - x</span>.
Solve quadratic equation: x = c(OH⁻) = 0.00133 M.
pOH = -logc(OH⁻).
pOH = 2.87.
pH = 14 - 2.87.
pH = 11.13.
The answers are :
1 - F
2- T
The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;
N₂ + 3F₂ → 2NF₃
The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence,
moles of N₂ / moles of F₂ = 1 / 2
moles of N₂ / 25 mol = 0.5
moles of N₂ = 0.5 x 25 mol = 12.5 mol
Hence N₂ moles needed = 12.5 mol
At STP (273 K and 1 atm) 1 mol of gas = 22.4 L
Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
= 280 L
