[Incomplete question. I found this data on other source]
c= 1.6 × 10^-2 M
c2h5nh2 + H20 ⇄ c2h5nh3+ + OH
co
co-x x x
The final concentration of OH- will be the same as the final concentration of c2h5nh3+ and we will call that x.
The final concentration of c2h5nh2 will be co-x.
You should use the equilibrium equation:
Kb=[c2h5nh3]*[OH]/[c2h5nh2]
We will use the Kb data and the concentration data:
6.4 × 10-4=x*x/(co-x)
6.4 × 10-4=x²/(1.6 × 10-2 M-x)
6.4 × 10-4*(1.6 × 10-2 M-x)=x²
6.4 × 10-4*1.6 × 10-2 M-6.4 × 10-4*x=x²
-x²+6.4 × 10-4*1.6 × 10-2 M-6.4 × 10-4*x=0
-x²-6.4 × 10-4*x+6.4 × 10-4*1.6 × 10-2 M =0
So now we can use the Bhaskara formula to find x:
x=(-b±√(b²-4*a*c)))/(2*a)
a=-1
b=-6.4 × 10-4
c=6.4 × 10-4*1.6 × 10-2
x=(-(-6.4 × 10-4)+√((6.4 × 10-4)²-4*(-1)* 6.4 × 10-4*1.6 × 10-2)))/(2*(-1))
x= -0,00353596 -> nonsense, x can’t be negative!
or x= 0,00289596 -> makes sense
So the correct value is x= 0,00289596
This means [OH-] is 0,00289596M
Now you should use the Ka equation:
Ka=[H+]*[OH-]
At 25°C, Ka is 10-14
10-14=[H+]*00289596
10-14/00289596=[H+]
[H+]=3,45309x10-12
[H⁺] in a solution of ethylamine =3.608.10⁻¹² M
Further explanation
Weak acid/base ionization reaction occurs partially (not ionizing perfectly as in strong acids/base)
The ionization reaction of a weak acid is an equilibrium reaction
HA (aq) ---> H⁺ (aq) + A⁻ (aq)
The equilibrium constant for acid ionization is called the acid ionization constant, which is symbolized by Ka
The values for the weak acid reactions above:
[tex]\rm Ka=\dfrac{[H][A^-]}{[HA]}[/tex]
Where Kb is the base ionization constant
LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)
[tex]\rm Kb=\dfrac{[L][OH^-]}{[LOH]}[/tex]
Kb of Ethylamine (C₂H₅NH₂) : 6.4.10⁻⁴
The ethylamine ionization reactions occur in water as follows:
C₂H₅NH₂ + H₂O ⇒ C₂H₅NH₃⁺ + OH⁻
with a Kb value:
[tex]\rm Kb=\dfrac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]}[/tex]
for example x = number of moles / concentration that reacts
Initial concentration of Ethylamine (C₂H₅NH₂) : 1.2.10⁻²
C₂H₅NH₂ + H₂O ⇒ C₂H₅NH₃⁺ + OH⁻
1.2.10⁻²
x x x x
1.2.10⁻² -x x x x
the value of Kb =
[tex]\rm Kb=\dfrac{[x][x]}{[1.2.10^{-2}-x]}\\\\assumption\:x=so\:small\:then\\\\6.4.10^{-4}=\dfrac{x^2}{1.2.10^{-2}}\\\\x^2=7.68.10^{-6}\\\\x=2.771.10^{-3}[/tex]
x = [OH⁻] = 2.771.10⁻³
Ka x Kb = [H⁺] [OH-]
a water equilibrium constant value (Kw) = 1.10⁻¹⁴ at 25 °C
Ka x Kb = [H +] [OH-] = 1.10⁻¹⁴
1.10⁻¹⁴ = [H⁺] . 2.771.10⁻³
[H⁺]=3.608.10⁻¹²
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