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
6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution
therefore, 250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g
# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles
1 mole of CuCl2 contain 1 mole of Cu2+ ion
Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2
Molar mass of CuCl2 = 134.452 g/mole
The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams
<span>If the aqueous solution is 34% Licl then it is 100 - 34% water = 66%
From the calculation we've found out that it is 66% water. Then we need to find the weight from a 250 g solution.
66/100 * 250 = 165g
Hence it is 165g</span>
Answer:
Its passive
Explanation:
Facilitated diffusion is a type of passive transport that allows substances to cross membranes with the assistance of special transport proteins.
