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Define pH. What are the limitations of the pH scale? Calculate the approximate pH of a vinegar solution that contains 4.5 g of ethanoic acid per 100 cm³ - Leaving Cert Chemistry - Question b - 2004
Question b
Define pH. What are the limitations of the pH scale? Calculate the approximate pH of a vinegar solution that contains 4.5 g of ethanoic acid per 100 cm³. The value... show full transcript
Worked Solution & Example Answer:Define pH. What are the limitations of the pH scale? Calculate the approximate pH of a vinegar solution that contains 4.5 g of ethanoic acid per 100 cm³ - Leaving Cert Chemistry - Question b - 2004
Step 1
Define pH.
Answer
The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration [H⁺] in a solution. Mathematically, it is expressed as:
This scale usually ranges from 0 to 14, with lower values indicating higher acidity and higher values indicating alkalinity. pH is a crucial measure in many chemical and biological systems.
Step 2
What are the limitations of the pH scale?
Answer
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Limited Range: The pH scale typically ranges from 0 to 14, which may not be sufficient for extremely acidic or alkaline solutions.
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Temperature Dependence: The pH value can vary with temperature, affecting the reliability in different conditions.
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Concentration Limitations: The scale does not work well at very low concentrations, especially when dealing with very dilute solutions.
Step 3
Calculate the approximate pH of a vinegar solution that contains 4.5 g of ethanoic acid per 100 cm³.
Answer
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First, calculate the number of moles of ethanoic acid (C₂H₄O₂):
Molar mass of ethanoic acid = 60 g/mol
Moles of ethanoic acid = ( \frac{4.5 ext{ g}}{60 ext{ g/mol}} = 0.075 ext{ mol} )
Since the solution volume is 100 cm³ (0.1 L), the concentration [C₂H₄O₂] is:
[ C = \frac{0.075 ext{ mol}}{0.1 ext{ L}} = 0.75 ext{ mol/L} ]
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Now, using the dissociation constant (Kₐ): Kₐ = 1.8 × 10⁻⁵, the relationship is:
From the dissociation of ethanoic acid:
[ K_a = \frac{[H^+][A^-]}{[HA]} \approx \frac{[H^+]^2}{0.75} ]
Solving for [H⁺]:
[ 1.8 \times 10^{-5} = \frac{[H^+]^2}{0.75} ]
[ [H^+]^2 = 1.8 \times 10^{-5} \times 0.75 \approx 1.35 \times 10^{-5} ]
[ [H^+] = \sqrt{1.35 \times 10^{-5}} \approx 3.67 \times 10^{-3} \text{ mol/L} ]
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Finally, calculate the pH:
[ pH = -\log_{10}(3.67 \times 10^{-3}) \approx 2.43 ]
Thus, the approximate pH of the vinegar solution is 2.43.