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

Learn more

an equilibrium constant link  

link  

Calculate the value of the equilibrium constant, Kc  

link  

Concentration of hi at equilibrium  

link  

D-no effect on the rate

There is no reason to think the results are biased or not conducted properly but you should not assume the results are correct just because they were done in a laboratory so the answer would be c. it is likely to be reliable because the results support the conclusion presented.