Answer:
pH = 4.56
Explanation:
The strychnine reacts with HCl as follows:
C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻
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For strychnine buffer:
pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]
Initial moles of C₂₁H₂₂N₂O₂ are:
0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂
And of HCl are:
0.05L * (0.200 mol / L) = 0.01 moles HCl
That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:
[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M
This conjugate acid, is in equilibrium with water as follows:
C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺
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<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>
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Ka is defined as:
Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]
In equilibrium, concentrations are:
C₂₁H₂₂N₂O₂ = X
H₃O⁺ = X
C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X
Replacing in Ka expression:
5.556x10⁻⁹ = [X] [X] / [0.1333M - X]
7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²
7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0
Solving for X:
X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations
X = 2.72x10⁻⁵M → Right solution.
As H₃O⁺ = X
H₃O⁺ = 2.72x10⁻⁵M
And pH = -log H₃O⁺
<h3>pH = 4.56</h3>
![pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BCH_3NH_2%5D%7D%7B%5BCH_3NH_3Cl%5D%7D%20%29%3D10%2C65%2Blog%28%20%5Cfrac%7B0%2C18M%7D%7B0%2C73M%7D%20%29%3D10%2C04)