Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solution at equilibrium. Ka(HF) = 7.2×10-4; Ka(HOC6H5) = 1.6×10-10. pH = [OC6H5-] = M

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Question:

Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solution at equilibrium. Ka(HF) = 7.2×10-4; Ka(HOC6H5) = 1.6×10-10. pH = [OC6H5-] = M

Answer:[tex]\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}[/tex]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
[tex]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$^{-1}$}[/tex]
2. Calculate the pH
[tex]\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}[/tex]3. Calculate [C₆H₅O⁻]
C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺
2.7 x 0.0441
[tex]K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}[/tex]

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