Math Problem Statement
Solution
Let's break down the problem step by step:
-
Given Reaction: Chlorine gas reacts with water to form hypochlorous acid (HClO) and hydrochloric acid (HCl).
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Buffer Solution:
- The solution contains 0.170 moles of HClO and 0.170 moles of HCl.
- 0.2 moles of NaOH is added to this solution. NaOH reacts with HCl, as NaOH neutralizes the acid (HCl).
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Reaction Between NaOH and HCl: Since NaOH reacts completely with HCl, the moles of NaOH (0.2 moles) will neutralize an equal amount of HCl (0.170 moles of HCl). After the reaction: All 0.170 moles of HCl are consumed, leaving no HCl, and 0.030 moles of NaOH remain unreacted.
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Buffer Composition After Reaction:
- After neutralization, all the remaining NaOH will react with HClO (since HClO is a weak acid), converting it to its conjugate base (ClO⁻).
- So, 0.030 moles of HClO will react with 0.030 moles of NaOH to form 0.030 moles of ClO⁻, and 0.170 moles of HClO are reduced by 0.030 moles.
Final buffer composition:
- Moles of HClO = .
- Moles of ClO⁻ = 0.030 moles.
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Using the Henderson-Hasselbalch Equation: [ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) ] Where:
- = concentration of
- = concentration of
Let's calculate each part.The pH of the buffer solution is approximately 6.87.
Would you like a detailed explanation or have any further questions?
Here are 5 related questions you can explore:
- How does the buffer capacity change with different amounts of NaOH?
- What happens to the pH if the volume of the solution increases?
- How is the pKa related to the strength of an acid?
- Why does the addition of NaOH affect the ratio of HClO to ClO⁻?
- What would happen if more NaOH were added beyond the neutralization point?
Tip: In buffer solutions, the closer the amounts of the weak acid and its conjugate base, the better the buffer resists changes in pH.
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Math Problem Analysis
Mathematical Concepts
Buffer Solution Chemistry
Henderson-Hasselbalch Equation
Logarithmic Calculations
Formulas
Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA])
pKa = -log(Ka)
Theorems
Henderson-Hasselbalch Equation
Suitable Grade Level
Undergraduate Chemistry
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