determine the pH of the solution.

Question

A solution is made by dissolving 15.7 g of CH₃NH₃NO₃ in 500.0 mL of water. Using Ka(CH₃NH₃⁺) = 2.3 × 10⁻¹¹, determine the pH of the solution.

Answer 1

CH₃NH₃NO₃ is the salt formed by neutralizing the strong acid HNO₃ (nitric acid) with the weak base CH₃NH₂(methyl amine). As a result, that salt is acidic and can hydrolyze water. Ignore the spectator ion NO₃^- and write the equation:

CH₃NH₃⁺ + H₂O <--> CH₃NH₂ + H₃O⁺

The equlibrium expression is:

Ka = [CH₃NH₂] [H3O⁺] / [CH₃NH₃⁺]

The initial concentration of CH₃NH₃⁺ is 0.334 M (15.7 g / 94.0 g/mol) / 0.5000 L) and Ka is 2.3 x 10^-11.

2.3 x 10^-11 = [x] [x] / [0.334 - x]

Since the value of x is negligible in comparison to 0.334, you can skip doing the quadratic and just divide instead by 0.334. So x = 2.77 x 10^-6. That means [H₃O⁺] = 2.77 x 10^-6. Taking the negative log of this number gives us a pH of 5.557.