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] = [tex]\frac{0.190}{2}[/tex]
= 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
[tex]K_a =\frac{[CH_3NH_2][H^+]}{[CH_3NH^+_3]}[/tex]
[tex]K_a = \frac{[x][x]}{[0.095-x]}[/tex]
[tex]K_a = \frac{[x^2]}{[0.095-x]}[/tex] ------ equation (1)
We all know that:
[tex]K_a = \frac{K_w}{K_b}[/tex]
and we are given [tex]K_b[/tex] = [tex]5.0*10^{-4[/tex]; and [tex]K_w = 10^{-14[/tex] ;
Then;
[tex]K_a= \frac{10^{-14}}{5*10^{-4}}[/tex]
[tex]K_a = 2*10^{-11}[/tex]
We can now re-write the equation (1) to be :
[tex]2*10^{-11} = \frac{x^2}{0.095-x}[/tex]
[tex]x^2 = 2*10^{-11} (0.095 -x)[/tex]
[tex]x^2 = 1.9*10^{-12} - 2*10^{-11}x[/tex]
[tex]x^2 + (2*10^{-11})x - (1.9*10^{-12}) = 0[/tex]
where a = 1; b = 2×10⁻¹¹; c = 1.9×10⁻¹²
Using the quadratic formula; [tex]\frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex] ; we have
= [tex]\frac{-(2*10^{-11})+\sqrt{(-(2*10^{-11})^2-4(1)(1.9*10^{-12})} }{2*1}[/tex] OR [tex]\frac{-(2*10^{-11})-\sqrt{(-(2*10^{-11})^2-4(1)(1.9*10^{-12})} }{2*1}[/tex]
x = [tex]1.38*10^{-6}M[/tex]
[CH₃NH₂] = [H⁺] = [tex]1.38*10^{-6}M[/tex]
∴ pH = -log [H⁺]
pH = -log [tex]1.38*10^{-6}M[/tex]
pH = 5.86
Hence, the pH at the equivalence point for the titration = 5.86
