The base-dissociation constant of ethylamine (c2h5nh2) is 6.4 × 10-4 at 25.0 °c. the [h ] in a solution of ethylamine is m.

[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:  

Where Kb is the base ionization constant  

LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)  

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:  

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 =  

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

an equilibrium constant link  

link  

Calculate the value of the equilibrium constant, Kc  

link  

Concentration of hi at equilibrium  

link