Answer:
5.86
Explanation:
The equation of reaction is between methylamine and hydrochloric acid
CH₃NH₂ + HCl -------> CH₃NH⁺₃ + Cl⁻
At equivalence point, there is a mixture of equal volume of acid and base. As such, the concentration of CH₃NH⁺₃ will be half the initial starting concentration of the reactant.
i.e [CH₃NH⁺₃] =
= 0.095 M
Now, the dissociation of CH₃NH⁺₃ yields
CH₃NH⁺₃ ⇄ CH₃NH₂ + H⁺
The ICE Table can be constructed as follows:
CH₃NH⁺₃ ⇄ CH₃NH₂ + H⁺
Initial 0.095 0 0
Change -x x x
Equilibrium (0.095 - x) x x
![K_a =\frac{[CH_3NH_2][H^+]}{[CH_3NH^+_3]}](https://tex.z-dn.net/?f=K_a%20%3D%5Cfrac%7B%5BCH_3NH_2%5D%5BH%5E%2B%5D%7D%7B%5BCH_3NH%5E%2B_3%5D%7D)
![K_a = \frac{[x][x]}{[0.095-x]}](https://tex.z-dn.net/?f=K_a%20%3D%20%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B%5B0.095-x%5D%7D)
------ equation (1)
We all know that:

and we are given
=
; and
;
Then;


We can now re-write the equation (1) to be :




where a = 1; b = 2×10⁻¹¹; c = 1.9×10⁻¹²
Using the quadratic formula;
; we have
=
OR 
x = 
[CH₃NH₂] = [H⁺] = 
∴ pH = -log [H⁺]
pH = -log 
pH = 5.86
Hence, the pH at the equivalence point for the titration = 5.86