НОХ 2005, II кръг (областен), 10–12 клас — Задача 1

• 1. 0.5 mol/L
• 2. $3.33 \times 10^{-12}$ mol/L
• 3. 11.477
• 5. sp³ и в двата случая

1. $c_0(\mathrm{NH_3}) = \dfrac{4480\ \mathrm{mL}}{22400\ \mathrm{mL/mol} \times 0.40\ \mathrm{L}} = 0.5\ \mathrm{mol/L}$ - 4т.

2. $\mathrm{NH_4OH \rightleftharpoons NH_4^+ + OH^-}$, $K_D = \dfrac{c(\mathrm{NH_4^+})\,c(\mathrm{OH^-})}{c(\mathrm{NH_4OH})}$ - 3т.

$c(\mathrm{NH_4^+}) = c(\mathrm{OH^-})$, $c(\mathrm{NH_4OH}) = c_0 - c(\mathrm{OH^-}) \approx c_0$

Тъй като $K_D \ll 1$, то $c(\mathrm{OH^-}) \ll c_0$, $c(\mathrm{NH_4OH}) \approx c_0$ и $K_D = \dfrac{c^2(\mathrm{OH^-})}{c_0}$

$c(\mathrm{OH^-}) = \sqrt{K_D\, c_0} = \sqrt{1.8\cdot 10-5 \times 0.5} = 3\cdot 10^{-3}\ \mathrm{mol/L}$ - 8т.

$c(\mathrm{H^+}) = \dfrac{10^{-14}}{c(\mathrm{OH^-})} = \dfrac{10^{-14}}{3\cdot 10^{-3}} = 3.33\cdot 10^{-12}\ \mathrm{mol/L}$ - 2т

3. $\mathrm{pH} = -\lg(\mathrm{H^+}) = -\lg(3.33\cdot 10^{-12}) = 11.477$ - 3т.

4. - 1т

5. Хибридизация на азот в амоняка – $sp^3$; в амониев катион – $sp^3$ - 4т