Answer:
pH = 10.49
Explanation:
The reaction is :
C₉H₁₃N(aq) + H₂O(l) ⇄ C₉H₁₃NH⁺(aq) + OH⁻(aq) (1)
The dissociation constant (Kb) of the above reaction is:
This Kb can be caculated using the pkb:
To find the pH of the reaction (1) first we need to convert the concentration of the amphetamine in mg/L to mol/L as follows:
Now, from the reaction (1) we have:
C₉H₁₃N(aq) + H₂O(l) ⇄ C₉H₁₃NH⁺(aq) + OH⁻(aq) (1)
<em>At equilibrium:</em> 1.85x10⁻³- x x x
(2)
Solving equation (2) for x, we have:
x = 0.0003116 = [OH⁻] = [C₉H₁₃NH⁺]
Therefore, the pH of reaction (1) is:
I hope it helps you!