Question

# A buffer solution is prepared by taking 72.3g CH3NH3Cl and 315 mL of 5.85 M CH3NH2...

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.

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

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