What is the Arrhenius definition of a base?
Why is NH₃ considered to be a base according to Brønsted-Lowry theory? (6)
Calculate the pH value of
(i) a 0.50 M solution of hydrochloric acid,
(ii) a 0.50 M solution of ethanoic acid - Leaving Cert Chemistry - Question b - 2018
Question b
What is the Arrhenius definition of a base?
Why is NH₃ considered to be a base according to Brønsted-Lowry theory? (6)
Calculate the pH value of
(i) a 0.50 M ... show full transcript
Worked Solution & Example Answer:What is the Arrhenius definition of a base?
Why is NH₃ considered to be a base according to Brønsted-Lowry theory? (6)
Calculate the pH value of
(i) a 0.50 M solution of hydrochloric acid,
(ii) a 0.50 M solution of ethanoic acid - Leaving Cert Chemistry - Question b - 2018
Step 1
What is the Arrhenius definition of a base?
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Answer
The Arrhenius definition of a base is a substance that dissociates into OH⁻ (hydroxide ions) in water, producing an aqueous solution. This means that an Arrhenius base increases the concentration of hydroxide ions in an aqueous solution.
Step 2
Why is NH₃ considered to be a base according to Brønsted-Lowry theory?
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Answer
According to Brønsted-Lowry theory, NH₃ (ammonia) is considered a base because it accepts a proton (H⁺). This characteristic defines it as a proton acceptor, which is the central aspect of Brønsted-Lowry acid-base theory.
Step 3
Calculate the pH value of (i) a 0.50 M solution of hydrochloric acid.
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Answer
For hydrochloric acid (HCl), which is a strong acid, the pH can be calculated using the following equation:
pH=−extlog[H+]
Since HCl dissociates completely in solution,
[H+]=0.50extM
So,
= 0.30$$
Step 4
Calculate the pH value of (ii) a 0.50 M solution of ethanoic acid.
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Answer
To calculate the pH of a 0.50 M solution of ethanoic acid (CH₃COOH), we use the acid dissociation constant (Kₐ):
K_a = rac{[H^+][A^-]}{[HA]}
Here, let the concentration of dissociated
H⁺ be denoted as x:
Following the expression:
yielding the equation:
K_a = 1.8 imes 10^{-5} = rac{x*x}{0.50 - x}
For simplicity, we assume x is very small compared to 0.50, hence:
1.8 imes 10^{-5} = x^2 \
\ x = ext{sqrt}(1.8 imes 10^{-5} imes 0.50) \
ightarrow [H^+] = 0.012 ext{ M}$$
Thus, we can now calculate the pH:
$$pH = - ext{log}(0.012) \
= 1.92$$
Step 5
Which of the two acid solutions was neutralised?
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Answer
The acid solution that was neutralised can be identified from its pH curve. As sodium hydroxide (NaOH) is added, a steep rise in pH indicates that a strong acid was neutralised. If the pH rapidly increased before reaching a neutral point, it suggests that hydrochloric acid (HCl) has been neutralised.
Step 6
State two pieces of evidence from the pH curve in support of your answer.
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Answer
The rapid increase in pH at the equivalence point demonstrates the strong acidic nature of HCl being neutralised by NaOH.
The pH remains relatively stable at 7 after neutralisation, indicating a complete reaction between the acid and base.
Step 7
What is the essential property of an indicator used to detect the end point in a titration between this acid and NaOH?
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The essential property of an indicator used in this titration is that it must change color at a pH close to 7, which is the expected endpoint for strong acid and strong base titrations. Thus, the indicator should have a transition range (pH) that coincides with the steep vertical region of the pH curve.
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