Respuesta :
Explanation:
The given data is as follows.
Concentration of given amphetamine = 205mg/L
= 0.205 g/L (as 1000 mg = 1 g)
Molar mass of amphetamine = 135.2062 g/mol
Therefore, calculate the number of moles of amphetamine as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{0.205 g/L}{135.2062 g/mol}[/tex]
= [tex]1.516 \times 10^{-3}[/tex] mol/l
or, = 0.001516 mol/l
As it is given that [tex]pK_{b} = 4.2[/tex]
and, [tex]pK_{b} = -log k_{b}[/tex]
antilog (4.2) = [tex]K_{b}[/tex]
[tex]K_{b}[/tex] = [tex]6.3 \times 10^{-5}[/tex]
The reaction equation along with ICE table is as follows.
[tex]H_{2}O + B \rightarrow BH^{+} + OH^{-}[/tex]
Initially: 0.001516 0 0
Change: -x +x +x
Equilibrium: (0.001516 - x) x x
[tex]K_{b}[/tex] = [tex]\frac{x^{2}}{(0.001516 - x)}[/tex]
[tex]6.3 \times 10^{-5} = \frac{x^{2}}{(0.001516 - x)}[/tex]
x = 0.00029
This means that [tex][OH^{-}][/tex] = 0.00029
As, pOH = [tex]-log[OH^{-}][/tex]
= -log(0.00029)
= 3.53
Also, it is known that pH = 14 - pOH
= 14 - 3.53
= 10.47
Thus, we can conclude that the pH of a solution containing an amphetamine concentration of 250 mg/L is 10.47.
