Photo AI

Define an acid in terms of the Brønsted-Lowry theory - Leaving Cert Chemistry - Question b - 2012

Question icon

Question b

Define-an-acid-in-terms-of-the-Brønsted-Lowry-theory-Leaving Cert Chemistry-Question b-2012.png

Define an acid in terms of the Brønsted-Lowry theory. What is a conjugate pair? A certain water soluble acid-base indicator represented by HIn is a weak acid which ... show full transcript

Worked Solution & Example Answer:Define an acid in terms of the Brønsted-Lowry theory - Leaving Cert Chemistry - Question b - 2012

Step 1

Define an acid in terms of the Brønsted-Lowry theory.

96%

114 rated

Answer

In the Brønsted-Lowry theory, an acid is defined as a proton (hydrogen ion, H⁺) donor. This means that an acid releases H⁺ ions into the solution.

Step 2

What is a conjugate pair?

99%

104 rated

Answer

A conjugate pair consists of an acid and its corresponding base that differ by a proton. When an acid donates a proton, it becomes its conjugate base, and vice versa.

Step 3

State and explain the colour observed when a few drops of a solution of the indicator are added to a 0.5 M NaOH solution.

96%

101 rated

Answer

The colour observed when a few drops of the indicator solution are added to a 0.5 M NaOH solution is purple. This occurs because hydroxide ions (OH⁻) from the NaOH remove H⁺ ions from HIn, causing the equilibrium of the dissociation reaction to shift to the right, where In⁻ predominates, resulting in the purple colour.

Step 4

Calculate the pH of the 0.5 M NaOH solution.

98%

120 rated

Answer

To calculate the pH of the 0.5 M NaOH solution, we first calculate the pOH:

pOH=extlog[OH]=extlog(0.5) =0.301pOH = - ext{log}[OH^-] = - ext{log}(0.5) \ = 0.301

Then, we find the pH using:

pH=14pOH=140.301=13.699pH = 14 - pOH = 14 - 0.301 = 13.699

Step 5

Calculate the pH of a 0.1 M solution of the indicator.

97%

117 rated

Answer

To calculate the pH of a 0.1 M solution of the indicator:

First, we find the concentration of H⁺ ions using the formula for Ka:

Ka=[H+][In]/[HIn]=2.0×103Ka = [H^+][In^-] / [HIn] = 2.0 × 10^{-3}

At equilibrium, let us assume [H⁺] = [In⁻] and hence:

Ka = rac{[H^+]^2}{0.1 - [H^+]}

With [H⁺] much smaller than 0.1:

ightarrow rac{[H^+]^2}{0.1} = 2.0 × 10^{-3}$$ Thus, $$[H^+] = ext{sqrt}(2.0 × 10^{-3} × 0.1) = ext{sqrt}(2.0 × 10^{-4}) = 0.01414$$ Now we calculate the pH: $$pH = - ext{log}[H^+] = - ext{log}(0.01414) \ ≈ 1.85$$

Join the Leaving Cert students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;