Answer:
Explanation:
The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.
1 .Calculate the hydronium ion concentration
We can use an ICE table to organize the calculations.
HF + H₂O ⇌ H₃O⁺ + F⁻
I/mol·L⁻¹: 2.7 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 2.7 - x x x
![K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Ba%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BH%7D_%7B3%7D%5Ctext%7BO%7D%5E%7B%2B%7D%5D%20%5Ctext%7BF%7D%5E%7B-%7D%5D%7D%20%7B%5Ctext%7B%5BHF%5D%7D%7D%20%3D%207.2%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B2.7%20-%20x%7D%20%3D%207.2%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5C%5Ctext%7BCheck%20for%20negligibility%20of%20%7Dx%5C%5C%5C%5C%5Cdfrac%7B2.7%7D%7B7.2%20%5Ctimes%2010%5E%7B-4%7D%7D%20%3D%204000%20%3E%20400%5C%5C%5C%5C%5Ctherefore%20x%20%5Cll%202.7%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B2.7%7D%20%3D%207.2%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%202.7%20%5Ctimes%207.2%20%5Ctimes%2010%5E%7B-4%7D%20%3D%201.94%20%5Ctimes%2010%5E%7B-3%7D%5C%5Cx%20%3D%200.0441%5C%5C%5Ctext%7B%5BH%24_%7B3%7D%24O%24%5E%7B%2B%7D%24%5D%7D%3D%20%5Ctext%7Bx%20mol%24%5Ccdot%24L%24%5E%7B-1%7D%24%7D%20%3D%20%5Ctext%7B0.0441%20mol%24%5Ccdot%24L%24%5E%7B-1%7D%24%7D)
2. Calculate the pH
![\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20-%5Clog%7B%5Crm%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%20%3D%20-%5Clog%7B0.0441%7D%20%3D%20%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.36%7D%7D)
3. Calculate [C₆H₅O⁻]
C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺
2.7 x 0.0441
Answer:
In an ionic bonds, the metal loses electrons to become a positively charged cation, In which the nonmetal accepts those electrons to become a negatively charged anion.
Explanation:
Sulfur trioxide (SO3) is a chemical compound that is a significant pollutant in gaseous form as it is involved in the production of acid rain.
Industrially, sulfur trioxide is an important precursor to sulfuric acid and is formed from the reaction between sulfur dioxide (SO2) and oxygen gas (O2) as shown in the chemical equation below.
