Question

Calculate the pH at the equivalence point for the titration of 0.190 M 0.190 M methylamine ( CH 3 NH 2 ) (CH3NH2) with 0.190 M HCl . 0.190 M HCl. The K b Kb of methylamine is 5.0 × 10 − 4 .

Answer 1

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] = $\frac{0.190}{2}$

= 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

$K_a =\frac{[CH_3NH_2][H^+]}{[CH_3NH^+_3]}$

$K_a = \frac{[x][x]}{[0.095-x]}$

$K_a = \frac{[x^2]}{[0.095-x]}$     ------ equation (1)

We all know that:

$K_a = \frac{K_w}{K_b}$

and we are given $K_b$ = $5.0*10^{-4$;  and $K_w = 10^{-14$ ;

Then;

$K_a= \frac{10^{-14}}{5*10^{-4}}$

$K_a = 2*10^{-11}$

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

$2*10^{-11} = \frac{x^2}{0.095-x}$

$x^2 = 2*10^{-11} (0.095 -x)$

$x^2 = 1.9*10^{-12} - 2*10^{-11}x$

$x^2 + (2*10^{-11})x - (1.9*10^{-12}) = 0$

where a = 1; b = 2×10⁻¹¹; c = 1.9×10⁻¹²

Using the quadratic formula; $\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; we have

= $\frac{-(2*10^{-11})+\sqrt{(-(2*10^{-11})^2-4(1)(1.9*10^{-12})} }{2*1}$  OR $\frac{-(2*10^{-11})-\sqrt{(-(2*10^{-11})^2-4(1)(1.9*10^{-12})} }{2*1}$

x = $1.38*10^{-6}M$

[CH₃NH₂] = [H⁺] =  $1.38*10^{-6}M$

∴ pH = -log [H⁺]

pH = -log $1.38*10^{-6}M$

pH = 5.86

Hence, the  pH at the equivalence point for the titration  = 5.86