Question

Calculate the pH at the equivalence point for the titration of 0.230 M methylamine (CH3NH2) with 0.230 M HCl. The Kb of methylamine is 5.0× 10–4.

What is the pKa of the indicator?

What is the color of this indicator in a solution with pH = 6?

Answer 1

The reaction of titration of methylamine with HCl:

CH₃NH₂ + HCl <-----> CH₃NH₃⁺ + Cl^-

The equivalence point is a point where the moles of hydrochloric acid added is equal to moles of methylamine in solution.

Ka = Kw / Kb

= 1 x 10⁻¹⁴ / 5.0 x 10⁻⁴

= 2.0 x 10⁻¹¹

After equivalence point :

                CH₃NH₃⁺ + H₂O ---> H₃O⁺ + CH₃NH₂

at end       0.230 - x                      x    x

Ka = x _x x / 0.230 - x = 2.0 x 10⁻¹¹

x ² -4.60 x 10⁻¹² + 2.0 x 10⁻¹¹ x = 0

x = 2.3 x 10⁻⁶

So, concentration of H₃O⁺ is 2.3 x10^-6 M.

Therefore, pH at equivalence point is:

pH = - log [H3O+] = - log (2.3 x10^-6) = 5.63

pH = 5.63

the equilibrium HIn H⁺ + In^- lies in exact middle.

applying the equilibrium law to this situation then:

Ka = [H⁺] [In^-] / [HIn]

And when the equilibrium lies then:

[In^-] = [HIn]

we can cancel that out to give the equation:

Ka = [H⁺] [In^-] / [HIn]

Ka = H⁺

therefore:

pKa = pH

The consequence of this answer is that the indicator will change color when the pH is the same value as it is pKa value. Hence, indicators have different regions of opertion. As the change in pH usually large at equivalence point this meams that the provided pH change takes place through the pKaof the indicator then it can be used for titration.

Equivalence point is when the moles of a standard solution (titrant) equal the moles of a solution of unknown concentration (analyte).

As the pH increases, the intensity of the color of HIn decreases and the equilibrium is pushed to the right. Therefore, the intensity of the color of In^- increases. But if we go to pH = 6 will show distnt colour if use methyl red as indicator we get yellow color.