From the equilibrium concentrations given, calculate *K*a for each of the weak acids and *K*b for each of the weak bases.

(a)

HCH3CO2: [H3O+ ] = 1.56 ✕ 10−3 *M*, [CH3CO2− ] = 1.56 ✕ 10−3 *M*, [HCH3CO2] = 0.143 *M*

(b)

CN−: [OH− ] = 1.93 ✕ 10−3 *M*, [HCN] = 1.93 ✕ 10−3 *M*, [CN− ] = 0.228 *M*

To calculate the acid dissociation constant (*K*ₐ) for the weak acid and the base dissociation constant (*K*ᵦ) 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.

---

### (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 *K*ₐ 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**, \( K_a = 1.70 \times 10^{-5} \).

---

### (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 *K*ᵦ 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⁻**, \( K_b = 1.63 \times 10^{-5} \).

---

### Summary:
- \( K_a \) for **HCH₃CO₂** = \( 1.70 \times 10^{-5} \)
- \( K_b \) for **CN⁻** = \( 1.63 \times 10^{-5} \)

---

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 *K*ₐ and *K*ᵦ?
3. What is the relationship between *K*ₐ and *K*ᵦ for conjugate acid-base pairs?
4. How would changing the initial concentration of the acid or base affect the calculated *K*ₐ or *K*ᵦ?
5. What are the implications of having a small *K*ₐ or *K*ᵦ value?

### Tip:
Remember that *K*ₐ and *K*ᵦ are temperature-dependent, so they can vary if the temperature changes from the standard 25°C.

From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.

(a)
HCH3CO2: [H3O+] = 1.56 ✕ 10−3 M, [CH3CO2−] = 1.56 ✕ 10−3 M, [HCH3CO2] = 0.143 M

(b)
CN−: [OH−] = 1.93 ✕ 10−3 M, [HCN] = 1.93 ✕ 10−3 M, [CN−] = 0.228 M

Learn how to calculate the dissociation constants, Ka for acetic acid (HCH3CO2) and Kb for cyanide (CN−), using given equilibrium concentrations. This problem covers key concepts of acid-base equilibrium, dissociation constants, and chemical formulas for advanced chemistry students.

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