To calculate the acid dissociation constant (Kₐ) for the weak acid and the base dissociation constant (Kᵦ) for the weak base, we use the following general formulas:

For Weak Acids:

[ K_a = \frac{[H_3O^+][\text{conjugate base}]}{[\text{acid}]} ]

For Weak Bases:

[ K_b = \frac{[OH^-][\text{conjugate acid}]}{[\text{base}]} ]

Now let's apply these formulas to each part.

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(a) For the Weak Acid (HCH₃CO₂):

• Given:
  • $[H_3O^+] = 1.56 \times 10^{-3} \, \text{M}$
  • $[CH₃CO₂^-] = 1.56 \times 10^{-3} \, \text{M}$
  • $[HCH₃CO₂] = 0.143 \, \text{M}$

Using the Kₐ formula: [ K_a = \frac{[H_3O^+][\text{CH₃CO₂}^-]}{[\text{HCH₃CO₂}]} ] $K_a = \frac{(1.56 \times 10^{-3})(1.56 \times 10^{-3})}{0.143}$ $K_a = \frac{2.4336 \times 10^{-6}}{0.143}$ $K_a = 1.70 \times 10^{-5}$

So, for the weak acid HCH₃CO₂, $K_a = 1.70 \times 10^{-5}$.

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(b) For the Weak Base (CN⁻):

• Given:
  • $[OH^-] = 1.93 \times 10^{-3} \, \text{M}$
  • $[HCN] = 1.93 \times 10^{-3} \, \text{M}$
  • $[CN^-] = 0.228 \, \text{M}$

Using the Kᵦ formula: [ K_b = \frac{[OH^-][\text{HCN}]}{[\text{CN}^-]} ] $K_b = \frac{(1.93 \times 10^{-3})(1.93 \times 10^{-3})}{0.228}$ $K_b = \frac{3.7249 \times 10^{-6}}{0.228}$ $K_b = 1.63 \times 10^{-5}$

So, for the weak base CN⁻, $K_b = 1.63 \times 10^{-5}$.

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Summary:

• $K_a$ for HCH₃CO₂ = $1.70 \times 10^{-5}$
• $K_b$ for CN⁻ = $1.63 \times 10^{-5}$

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Would you like more details on the steps, or do you have any specific questions?

Related Questions:

1. How is the equilibrium constant related to the strength of acids and bases?
2. Can you explain how the autoionization of water affects Kₐ and Kᵦ?
3. What is the relationship between Kₐ and Kᵦ for conjugate acid-base pairs?
4. How would changing the initial concentration of the acid or base affect the calculated Kₐ or Kᵦ?
5. What are the implications of having a small Kₐ or Kᵦ value?

Tip:

Remember that Kₐ and Kᵦ are temperature-dependent, so they can vary if the temperature changes from the standard 25°C.