Question

A buffer solution is prepared by taking 72.3g CH3NH3Cl and 315 mL of 5.85 M CH3NH2 (Kb = 4.4x10^-4) and diluting to a final volume of 500.0 mL, with a pH of 10.88 resulting.  

50.0 mL of this buffer is titrated with 3.86 M HCl. Determine the pH of the resulting solution after 47.80 mL of HCl is added (at the equivalence point).

The answer is pH = 5.08 but I don't know how to get to it.

Answer 1

Since, after addition of 47.80 mL of HCl you are at equivalence point where you have reacted all of the HCl and converted it to CH3NH3Cl

Initial mmoles CH3NH2 = Molarity x mL of CH3NH2

= (5.85 M)(0.05 L)

= 0.2925 mol of CH3NH2 = mol of CH3NH3Cl

[CH3NH3Cl] at equivalence point = moles CH3NH3Cl/ L of solution

= 0.2925 mol / 0.0978 L

= 2.99 M

Since CH3NH3Cl is made using Strong acid and weak base, the resulting CH3NH3Cl solution will be acidic,

For solution of weak acids

[H+= (Ka x [B]o)^0.5

Where, Ka = Kw / Kb

= (1 x 10^-14) / (4.4x 10^-4)   

= 2.27 x 10^-11

Putting this value in above equation :

[H+] = ((2.27x 10^-11](2.99 M))^0.5

= 8.23 x 10^-6

pH = -log [H+]

= -log (8.23x 10^-6)

= 5.08