Respuesta :
Answer:
[tex]Ka=0.000346[/tex]
[tex]Kb=0.000593[/tex]
Explanation:
We have to start with the calculation of the concentration of the acid, using the molar mass:
Molas mass of [tex]C_9H_8O_4[/tex]= 180.16 g/mol
[tex]2*g\frac{1~mol}{180.16~g}=~0.011~mol[/tex]
[tex]M=\frac{0.011~mol}{0.6~L}=0.019~M[/tex]
The, with the ionization equation for aspirin, using the ICE table, we can calculate the Ka expression.
HA <=> A^- + H^+
I 0.019 Zero Zero
C -X +X +X
E 0.019-X X X
[tex]Ka=\frac{[X][X]}{[0.019-X]}=\frac{[X]^2}{[0.019-X]}[/tex]
Now, we can calculate X using the pH value:
[tex]pH=-Log[H^+][/tex]
[tex][H^+]=10^-^p^H[/tex]
[tex][H^+]=10^-^2^.^6^2=0.00240[/tex]
With this value we can calculate the Ka, so:
[tex]Ka=\frac{[0.00240]^2}{[0.019-0.00240]}=0.000346[/tex].
For the ethylamine we have to follow the same procedure:
B^- + H2O <=> BH + OH^-
I 0.1 / Zero Zero
C -X / +X +X
E 0.1-X / X X
[tex]Kb=\frac{[X][X]}{[0.1-X]}=\frac{[X]^2}{[0.1-X]}[/tex]
Now, we can calculate X using the pH value:
[tex]14=pH+pOH[/tex]
[tex]pOH=14-pH[/tex]
[tex]pOH=-Log[OH^-][/tex]
[tex][OH^-]=10^-^p^O^H[/tex]
[tex][OH^-]=10^-^2^.^1^3=0.00741[/tex]
With this value we can calculate the Kb, so:
[tex]Kb=\frac{[0.00741]^2}{[0.1-0.00741]}=0.000593[/tex]
