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Define (i) acid, (ii) conjugate acid, according to the Brønsted-Lowry theory - Leaving Cert Chemistry - Question 8 - 2010

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Define (i) acid, (ii) conjugate acid, according to the Brønsted-Lowry theory. In acting as an acid-base indicator methyl orange behaves like a weak acid. Letting H... show full transcript

Worked Solution & Example Answer:Define (i) acid, (ii) conjugate acid, according to the Brønsted-Lowry theory - Leaving Cert Chemistry - Question 8 - 2010

Step 1

(i) acid, (ii) conjugate acid

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Answer

A Brønsted-Lowry acid is defined as a proton (H⁺) donor. This means that an acid is any species that can donate a proton to another species during a chemical reaction.

A conjugate acid is the species formed when a Brønsted-Lowry base accepts a proton. When the base gains a proton, it becomes the conjugate acid of that base.

Step 2

Distinguish between a strong acid and a weak acid

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Answer

A strong acid completely dissociates in aqueous solution, meaning that it donates all its available protons to the solution, resulting in a high concentration of H⁺ ions. An example of a strong acid is hydrochloric acid (HCl). In contrast, a weak acid only partially dissociates in solution, leading to a lower concentration of H⁺ ions. A typical example of a weak acid is acetic acid. The key difference lies in their degree of ionization in water.

Step 3

What is the conjugate base of HX?

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Answer

The conjugate base of HX can be determined by removing a proton from it, which results in X⁻. Therefore, if HX represents methyl orange, the conjugate base is X⁻.

Step 4

State and explain the colour observed

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Answer

(i) When a few drops of methyl orange are added to a 0.1 M solution of HCl, the solution will appear red. This occurs because HCl is a strong acid, and the resulting high concentration of H⁺ ions favors the undissociated form (HX) of the indicator, which is red.

(ii) When the same number of drops is added to a 0.1 M solution of NaOH, the solution will appear yellow. NaOH is a strong base, and it causes the methyl orange to dissociate into its conjugate base (X⁻), which gives a yellow coloration.

Step 5

Calculate the pH

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Answer

(i) To calculate the pH of a 0.1 M NaOH solution, we first find the concentration of OH⁻ ions. For a strong base like NaOH, this is equal to its concentration: [OH⁻] = 0.1 M.
Using the formula:

extpOH=extlog([OH])=extlog(0.1)=1 ext{pOH} = - ext{log}([OH⁻]) = - ext{log}(0.1) = 1
Calculating pH from pOH:
extpH=14extpOH=141=13ext{pH} = 14 - ext{pOH} = 14 - 1 = 13

(ii) For the 0.004 M solution of methyl orange with a Ka of 3.5 x 10⁻⁴, we can utilize the equation:

extpH=extlog[extH+] ext{pH} = - ext{log}[ ext{H}⁺]
Where the concentration of H⁺ in equilibrium can be found using the relation:
K_a = rac{[H^+][X^-]}{[HX]}
For a weak acid, we can assume that [H⁺] is approximately equal to [X⁻]. Thus,
K_a = rac{x^2}{0.004 - x} ext{ (where x is negligible)}
We find
ext[H+]=extsqrt(Ka0.004)=extsqrt(3.5imes1040.004)ext{[H}⁺] = ext{sqrt}(K_a * 0.004) = ext{sqrt}(3.5 imes 10^{-4} * 0.004)
ext{Solving gives,} ext{pH} = 2.9.

Step 6

Draw a clearly labelled diagram of the pH curve

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Answer

The pH curve for the titration of HCl with NaOH is characterized by a steep slope at the equivalence point.
The x-axis represents the volume of NaOH added, while the y-axis represents the pH.

  1. Label the pH scale from 0 to 14 and indicate the expected shape of the curve starting from low pH, rising steeply around the equivalence point, and leveling out as it approaches neutral pH.

  2. Mark the initial volumes: 25 cm³ of HCl and the volume at which the equivalence point occurs, which should correspond to the neutralization of the acid and base, around 50 cm³ of NaOH added.

  3. Explain that the curve shows a rapid pH change in the range below 7 during the addition of NaOH until it approaches 7, where most indicators will change color due to the steep slope of the curve.

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