A buffer is prepared by adding 139 mL of 0.39 M NH3 to 169 mL of 0.19 M NH4NO3. What is the pH of the final solution? (Assume the value of Kw is 1.0 ✕ 10-14. Enter an unrounded value.)

pH = 9.48

Explanation:

We have first to realize that NH₃ is a weak base:

NH₃ + H₂O ⇔ NH₄⁺ + OH⁻ Kb = 1.8 x 10⁻⁵

and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.

Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:

pOH = pKb + log ([ conjugate acid ] / [ weak base ]

mol NH₃ = 0.139 L x 0.39 M = 0.054 mol

mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol

Now we have all the information required to calculate the pOH (Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)

pOH = - log (1.8 x 10⁻⁵) + log (0.032/0.054) = 4.52

pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48

The solution is basic which agrees with NH₃ being a weak base.