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REY [17]
3 years ago
14

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 solu

tion at equilibrium. Ka(HF) = 7.2×10-4; Ka(HOC6H5) = 1.6×10-10. pH = [OC6H5-] = M
Chemistry
1 answer:
borishaifa [10]3 years ago
6 0

Answer:

\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}

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$^{-1}$}

2. Calculate the pH

\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

     2.7                         x        0.0441

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 $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}

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It is true. a) 0.25 mol

Explanation:

<em>Hello </em><em>there?</em>

To begin solving this problem, you have to write down the chemical equation and make sure it is well balanced.

The chemical equation is;

3Mg(s) + N2(g) => Mg3N2(s)

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<em>I </em><em>hope</em><em> </em><em>this </em><em>helps</em><em> </em><em>you </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Ha</em><em>v</em><em>e </em><em>a </em><em>nice </em><em>studies.</em><em> </em><em />

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You will often see high clouds like cirrus, cirrostratus, and middle clouds like altostratus ahead of a warm front. These clouds form in the warm air that is high above the cool air. As the front passes over an area, the clouds become lower, and rain is likely. There can be thunderstorms around the warm front if the air is unstable.

On weather maps, the surface location of a warm front is represented by a solid red line with red, filled-in semicircles along it, like in the map on the right (B). The semicircles indicate the direction that the front is moving. They are on the side of the line where the front is moving. Notice on the map that temperatures at ground level are cooler in front of the front than behind it.

3 0
3 years ago
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Answer:

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Explanation:

Given the following mathematical expression;

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First of all, we would cross-multiply;

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Opening the bracket, we have;

2A = ah + bh

Rearranging the mathematical expression, we have;

ah = 2A - bh

a = \frac {2A - bh}{h}

6 0
2 years ago
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