Respuesta :
Ethylamine is a weak base. concentration of weak base, [OH⁻]= ([tex]K_{b} C_{b}[/tex])[tex]^{\frac{1}{2}}[/tex], where [tex]K_{b}[/tex] is the dissociation constant of weak base and [tex]C_{b}[/tex] is the concentration of weak base. Ethyl ammonium ion is acid which will dissociate into H⁺ ion and ethyl amine. [tex]P^{k}a_{}[/tex] of ethylammonium ion is given as 10.70. Concentration is given as 0.1 m.[tex][H^{+} ]= (K_{a} C_{a})^{\frac{1}{2}}, where K_{a}[/tex] is dissociation constant of acid. [tex]P^{k}a_{}[/tex] = -log [tex]K_{a}[/tex]=10.70
So, [tex]K_{a}[/tex]= [tex]10^{-10.70}[/tex]=1.99 X [tex]10^{-11}[/tex]. [tex][H^{+} ][/tex]=([tex]1.99 X 10^{-11}[/tex]X0.1)=1.99X[tex]10^{-12}[/tex]. [tex]P^{H}[/tex]= -log[tex][H^{+} ][/tex]= - log [tex]1.99X10^{-12}[/tex]=11.7
Answer : The pH of the solution is, 11.8
Solution : Given,
Concentration (c) = 0.1 M
Base dissociation constant = [tex]pK_a=10.70[/tex]
Now we have to calculate the value of [tex]pK_b[/tex].
As we know that,
[tex]pK_b=14-pK_a=14-10.70=3.3[/tex]
Now we have to calculate the value of [tex]K_b[/tex].
The expression used for the calculation of [tex]pK_b[/tex] is,
[tex]pK_b=-\log (K_b)[/tex]
Now put the value of [tex]pK_b[/tex] in this expression, we get:
[tex]3.3=-\log (K_b)[/tex]
[tex]K_b=5.0\times 10^{-4}[/tex]
The given equilibrium reaction is,
[tex]CH_3CH_2NH_2+H_2O\rightleftharpoons CH_3CH_2NH_3^++OH^-[/tex]
initially conc. 0.1 0 0
At eqm. (0.1-x) x x
Formula used :
[tex]k_b=\frac{[CH_3CH_2NH_3^+][OH^-]}{[CH_3CH_2NH_2]}[/tex]
Now put all the given values in this formula ,we get:
[tex]5.0\times 10^{-4}=\frac{(x)(x)}{(0.1-x)}[/tex]
By solving the terms, we get:
[tex]x=6.8\timees 10^{-3}[/tex]
Now we have to calculate the concentration of hydroxide ion.
[tex][OH^-]=x=6.8\times 10^{-3}M[/tex]
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (6.8\times 10^{-3})[/tex]
[tex]pOH=2.2[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-2.2\\\\pH=11.8[/tex]
Therefore, the pH of the solution is, 11.8
