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Calculate the ph at the equivalence point for these titrations: (а) 0.10 М НСІ versus 0.10 М NH3, (b)010M CH,COOH versus 0.10m Nаoh
Accepted Answer
Here's how to calculate the pH at the equivalence point for these titrations:
(a) 0.10 M HCl versus 0.10 M NH3
Reaction: HCl (strong acid) reacts with NH3 (weak base) to form NH4Cl (salt of a weak base).
At the equivalence point: All the NH3 has reacted with HCl, leaving only NH4+ ions in solution. NH4+ is the conjugate acid of NH3, and it will undergo hydrolysis to produce H+ ions, making the solution acidic.
Calculation:
1. Write the hydrolysis reaction: NH4+ + H2O ⇌ NH3 + H3O+
2. Set up an ICE table:
| | NH4+ | NH3 | H3O+ |
|-------------|-------|-------|------|
| Initial | 0.10 | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 0.10-x| x | x |
3. Write the Ka expression: Ka = [NH3][H3O+]/[NH4+]
4. Substitute values and solve for x (which is equal to [H3O+]): Since Ka for NH4+ is 5.6 x 10^-10, we can assume x is small compared to 0.10. Therefore, 5.6 x 10^-10 ≈ (x^2)/0.10. Solving for x, we get [H3O+] = 7.48 x 10^-6 M.
5. Calculate pH: pH = -log[H3O+] = -log(7.48 x 10^-6) ≈ 5.13.
(b) 0.10 M CH3COOH versus 0.10 M NaOH
Reaction: CH3COOH (weak acid) reacts with NaOH (strong base) to form CH3COONa (salt of a weak acid).
At the equivalence point: All the CH3COOH has reacted with NaOH, leaving only CH3COO- ions in solution. CH3COO- is the conjugate base of CH3COOH and will undergo hydrolysis to produce OH- ions, making the solution basic.
Calculation:
1. Write the hydrolysis reaction: CH3COO- + H2O ⇌ CH3COOH + OH-
2. Set up an ICE table:
| | CH3COO- | CH3COOH | OH- |
|-------------|---------|-----------|-----|
| Initial | 0.10 | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 0.10-x | x | x |
3. Write the Kb expression: Kb = [CH3COOH][OH-]/[CH3COO-]
4. Substitute values and solve for x (which is equal to [OH-]): Since Kb for CH3COO- is 5.6 x 10^-10, we can assume x is small compared to 0.10. Therefore, 5.6 x 10^-10 ≈ (x^2)/0.10. Solving for x, we get [OH-] = 7.48 x 10^-6 M.
5. Calculate pOH: pOH = -log[OH-] = -log(7.48 x 10^-6) ≈ 5.13.
6. Calculate pH: pH + pOH = 14, therefore, pH = 14 - 5.13 ≈ 8.87.
In summary:
For (a), the pH at the equivalence point is approximately 5.13, indicating an acidic solution.
For (b), the pH at the equivalence point is approximately 8.87, indicating a basic solution.
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