Answer:
pH =11.62: [tex][H_{2}NCH_{2}NCH_{2}NH_{2}][/tex] = 0.204 M;
[tex]H_{2}NCH_{2}CH_{2}NH^{+} _{3}[/tex] = [tex]4.20*10^{-3} M[/tex]
[tex]^{+}H_{3}NCH_{2}CH_{2}NH^{+} _{3}[/tex] = [tex]7.04*10^{-8} M[/tex]
Explanation:
To determine the pH, we have to calculate the final equilibrium concentration for the hydrogen ion by working through both dissociation equilibria of the system. Therefore:
[tex]H_{2}NCH_{2}NCH_{2}NH_{2(aq)} + H_{2}O_{(l)}[/tex] ⇔[tex]H_{2}NCH_{2}CH_{2}NH^{+} _{3(aq)} + OH_{(aq)} ^{-}[/tex] [tex]k_{b2}[/tex] equation (1)
[tex]H_{2}NCH_{2}CH_{2}NH^{+} _{3(aq)} + H_{2}O_{(l)}[/tex]⇔[tex]^{+}H_{3}NCH_{2}CH_{2}NH^{+} _{3(aq)} + OH^{-}_{(aq)}[/tex] [tex]k_{b1}[/tex] equation (2)
We know that:
[tex]k_{a} = 10^{-pka}[/tex]
[tex]k_{b}=\frac{k_{w} }{k_{a} }[/tex]
[tex]k_{b2}[/tex] =[tex]8.47*10^{-5}[/tex]
[tex]k_{b1}[/tex] = [tex]7.04*10^{-8}[/tex]
Using equation 1 to calculate the value of [tex]k_{b2}[/tex]:
If the equilibrium concentration of [tex]OH^{-}[/tex] = x
Then the equilibrium concentration of [tex]H_{2}NCH_{2}NCH_{2}NH_{2(aq)}[/tex] = 0.208-x
The equilibrium concentration of [tex]H_{2}NCH_{2}CH_{2}NH^{+} _{3(aq)}[/tex] = x
Thus: [tex]k_{b2}[/tex] = x*x/0.208 (if we consider x to be extremely small)
x^2 = [tex]8.47*10^{-5}[/tex]*0.208 =[tex]1.76*10^{-5}[/tex]
x = [tex]4.20*10^{-3} M[/tex] = [tex]H_{2}NCH_{2}CH_{2}NH^{+} _{3(aq)}[/tex]
The equilibrium concentration of [tex][H_{2}NCH_{2}NCH_{2}NH_{2(aq)}][/tex] = 0.208 - [tex]4.20*10^{-3} M[/tex] = 0.204 M
[tex]k_{b1}[/tex] = [tex]^{+}H_{3}NCH_{2}CH_{2}NH^{+} _{3}[/tex]*OH^{-}[/tex]/[tex]H_{2}NCH_{2}CH_{2}NH^{+} _{3} [/tex]
[tex]7.04*10^{-8}[/tex] = x*[tex]4.20*10^{-3} M[/tex] /[tex]4.20*10^{-3} M[/tex] =x
Thus the equilibrium concentration of [tex]^{+}H_{3}NCH_{2}CH_{2}NH^{+} _{3}[/tex] = [tex]7.04*10^{-8} M[/tex]
pOH = -log[OH^-] = -log [tex]4.20*10^{-3} [/tex] = 2.38
pH = pKw - pOH = 14 -2.38 =11.62
