Question

What is the pH of a buffer solution that contains 0.55 M methylamine, CH 3NH2, and 0.29 M methylammonium chloride, CH3NH3Cl?A

Answer 1

Answer:

pH = 10.93

Explanation:

To find the pH we need to use the Henderson - Hasselbalch equation:

$pH = pKa + log(\frac{[A^{-}]}{[HA]})$

Where:

[A⁻]: is the conjugate base of the acid = [CH₃NH₂] = 0.55 M

[HA]: is the acid = [CH₃NH₃Cl] = 0.29 M

Since the pkb of methylamine is 3.35, the pka is:

$pk_{a} + pk_{b} = 14$

$pk_{a} = 14 - pk_{b} = 14 - 3.35 = 10.65$  

Now, the pH is:

$pH = pKa + log(\frac{[A^{-}]}{[HA]}) = 10.65 + log(\frac{0.55}{0.29}) = 10.93$

Therefore, the pH of the buffer solution is 10.93.

I hope it helps you!