Answer :
Answer:
The pH of a solution containing an amphetamine (with pKb of 4.2) concentration of 250 mg/l is 10.5
Explanation:
To solve the question we note that the
The mass of amphetamine = 250 mg/l
The molar mass of amphetamine = 135.2062 g/mol
Number of moles of amphetamine = (250 mg)/(135.2062 g/mol) = 1.85×10⁻³ moles
That is the initial concentration of amphetamine = 1.85×10⁻³ M
We have [tex]pK_b = -log K_b[/tex]
therefore
[tex]K_b = 10^{-pK_b} = 10^{-4.2} = 6.3 \times 10^-^5[/tex]
[tex]K_b=\frac{[C_9H_{14}N][OH^-]}{[C_9H_{13}N]}\\= 6.3 \times 10^-^5[/tex]
Which gives 6.3×10⁻⁵ [tex]= \frac{x^{2} }{0.00185-x}[/tex]
[tex]we get \\(0.00185-x)*6.3*10^{-5} = x^{2}(0.00185-x)*6.3*10^{-5} - x^{2} = 0[/tex]
x = 0.0003116 = [OH⁻] = [C₉H₁₃NH⁺]
However x = [OH⁻] = 3.116×10⁻⁴
pOH = -log[OH⁻] = -log(3.116×10⁻⁴) = 3.506
pH = 14 - pOH
= 14 - 3.506
= 10.49
≈ 10.5
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:
[tex] K_{b} = \frac{[C_{9}H_{13}NH^{+}][OH^{-}]}{[C_{9}H_{13}N]} [/tex]
This Kb can be caculated using the pkb:
[tex]p_{K_{b}} = -log(K_{b}) \rightarrow K_{b} = 10^{-p_{K_{b}}} = 10^{- 4.2}= 6.31 \cdot 10^{-5}[/tex]
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:
[tex] [C_{9}H_{13}N] = \frac{250 mg}{L} \cdot \frac{1 g}{1000 mg} \cdot \frac{1 mol}{135.206 g} = 1.85 \cdot 10^{-3} mol/L [/tex]
Now, from the reaction (1) we have:
C₉H₁₃N(aq) + H₂O(l) ⇄ C₉H₁₃NH⁺(aq) + OH⁻(aq) (1)
At equilibrium: 1.85x10⁻³- x x x
[tex] K_{b} = \frac{[C_{9}H_{13}NH^{+}][OH^{-}]}{[C_{9}H_{13}N]} [/tex]
[tex] 6.31 \cdot 10^{-5} = \frac{x*x}{1.85 \cdot 10^{-3} - x} [/tex]
[tex]x^{2} - 6.31 \cdot 10^{-5}*(1.85 \cdot 10^{-3} - x) = 0[/tex] (2)
Solving equation (2) for x, we have:
x = 0.0003116 = [OH⁻] = [C₉H₁₃NH⁺]
Therefore, the pH of reaction (1) is:
[tex] pOH = -log([OH^{-}]) = -log(0.0003116) = 3.51 [/tex]
[tex] pH = 14 - pOH = 14 - 3.51 = 10.49 [/tex]
I hope it helps you!