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 .

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

    ------ 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

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons