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Alika [10]
3 years ago
11

Trichloroacetic acid (CCl3CO2H) is a corrosive acid that is used to precipitate proteins. The pH of a 0.050-M solution of trichl

oroacetic acid is the same as the pH of a 0.040-M HClO4 solution. Calculate Ka for trichloroacetic acid.
Chemistry
1 answer:
Nataly [62]3 years ago
6 0

Answer:

Ka = 0.16

Explanation:

HClO₄ is a strong acid and it will dissociate completely to give hydrogen ion and the conjugate base.

The pH of HClO₄ will be calculated as:

pH = -log[H⁺]

pH = -log[0.04] = 1.4

The trichloroacetic acid is a weak acid and it will not dissociate completely.

The dissociation will be:

                CCl_{3}COOH---> CCl_{3}COO^{-} + H^{+}

Initial                     0.05                       0                         0

Change                -x                          +x                           +x

Equilibrium       0.05-x                     x                              x

Ka = \frac{[H^{+}][ClO_{4}^{-}]}{[HClO_{4}]}

As pH of the acid = 1.4 it means the concentration of hydrogen ion is 0.04.

Therefore

x = 0.04

Putting values

Ka = \frac{(0.04)(0.04)}{0.05-0.04}=0.16

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Answer:

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Explanation:

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d_{A}=\frac{m_{A}}{V_{A}}

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d_{B}=\frac{m_{B}}{V_{B}}

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And all the three cubes have the same mass, so:

m_{A}=m_{B}=m_{C}

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d_{A}*V_{A}=d_{B}*V_{B} (Eq.1)

d_{A}*V_{A}=d_{C}*V_{C} (Eq.2)

Solving for d_{1} in Eq.1:

d_{A}=d_{B}\frac{V_{B}}{V_{A}}

Replacing values for the volume:

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d_{A}=2.7\frac{g}{mL}*0.64

d_{A}=1.728\frac{g}{mL}

that is the density of the magnesium.

Solving for d_{C} in Eq.2:

d_{C}=d_{A}\frac{V_{A}}{V_{C}}

d_{C}=d_{A}\frac{25.9mL}{4.29mL}

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d_{C}=1.728\frac{g}{mL}*6.04

d_{C}=10.4\frac{g}{mL}

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Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.

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