Question

Calculate the pH at the equivalence point in the titration of 30.0 mL of 0.135 M methylamine

(K_b = 4.4 × 10⁻⁴⁾ with 0.270 M HCl.

Answer 1

find the volume of HCl used to reach equivalence point

M(CH3NH2)*V(CH3NH2) =M(HCl)*V(HCl)

0.135 M *30.0 mL = 0.27M *V(HCl)

V(HCl) = 15 mL

we have:

Molarity of HCl = 0.27 M

Volume of HCl = 15 mL

Molarity of CH3NH2 = 0.135 M

Volume of CH3NH2 = 30 mL

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 0.27 M * 15 mL = 4.05 mmol

mol of CH3NH2 = Molarity of CH3NH2 * Volume of CH3NH2

mol of CH3NH2 = 0.135 M * 30 mL = 4.05 mmol

We have:

mol of HCl = 4.05 mmol

mol of CH3NH2 = 4.05 mmol

4.05 mmol of both will react to form CH3NH3+ and H2O

CH3NH3+ here is strong acid

CH3NH3+ formed = 4.05 mmol

Volume of Solution = 15 + 30 = 45 mL

Ka of CH3NH3+ = Kw/Kb = 1.0E-14/4.4E-4 = 2.273*10^-11

concentration ofCH3NH3+,c = 4.05 mmol/45 mL = 0.09 M

CH3NH3+ + H2O -----> CH3NH2 + H+

9*10^-2 0 0

9*10^-2-x x x

Ka = [H+][CH3NH2]/[CH3NH3+]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.273*10^-11)*9*10^-2) = 1.43*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.43*10^-6 M

[H+] = x = 1.43*10^-6 M

we have below equation to be used:

pH = -log [H+]

= -log (1.43*10^-6)

= 5.84

Answer: 5.84