Respuesta :
Answer:
The answer to the question is;
The pH of a solution containing an amphetamine (with pKb of 4.2) concentration of 205 mg/l is 10.4.
Explanation:
To solve the question we note that the
The mass of amphetamine = 205 mg/l
The molar mass of amphetamine = 135.2062 g/mol
Number of moles of amphetamine = (205 mg)/(135.2062 g/mol) = 1.5×10⁻³ moles
That is the initial concentration of amphetamine = 1.5×10⁻³ M
We have [tex]pK_b = -log K_b[/tex] therefore [tex]K_b[/tex] = [tex]10^{-pK_b} = 10^{-4.2}[/tex] = 6.3×10⁻⁵
Then [tex]K_b=\frac{[C_9H_{14}N][OH^-]}{[C_9H_{13}N]}[/tex] = 6.3×10⁻⁵
Which gives 6.3×10⁻⁵ = [tex]\frac{x^{2} }{0.001516-x}[/tex]
Solving, we get [tex](0.001516-x)*6.3*10^{-5} = x^{2}[/tex]
[tex](0.001516-x)*6.3*10^{-5} - x^{2} = 0[/tex]
From which by factorizing gives (x+3.42×10⁻⁴)×(x-2.79×10⁻⁴) = 0
That is x = -3.42×10⁻⁴ or 2.79×10⁻⁴ which gives x= 2.79×10⁻⁴ since we are dealing with concentration
However x = [OH⁻] = 2.79×10⁻⁴
pOH = -log[OH⁻] = -log(2.79×10⁻⁴) = 3.5538
pH = 14 - pOH = 14 - 3.5538 = 10.446 ≈ 10.4
the pH of a solution containing an amphetamine concentration of 205 mg/l = 10.4.
