Question

Given that ammonia (NH3) acts as a weak base in water with a Kb of 1.8 x 10-5 at 298.0 K, calculate the pH at 298.0 K of a solution prepared by mixing 100.0 mL of a 0.050 M aqueous solution of ammonia with 20.0 mL of a 1.00 M aqueous solution of nitric acid. The final volume of the solution is 120.0 mL.

Answer 1

pH = 0.903.

Step-by-step explanation:

Hello there!

In this case, according to the given information about the problem, we can notice there is a reaction between nitric acid and ammonia, whereas just the H ions from the former are relevant:

`NH_3(aq)+H^+(aq)\rightleftharpoons NH_4^+(aq)`

Thus, we first calculate the consumed moles of ammonia, it order to evaluate the moles after the reaction:

`n_(H^+)=0.020L*1.00mol/L=0.020mol\\\\n_(NH_3)=0.100L*0.050mol=0.005mol`

It means there is a leftover of moles of hydrogen ions of 0.015mol; then, we calculate the molarity of the resulting solution as follows:

`[H^+]=(0.015mol)/(0.120L)=0.125M`

Finally, we use the formula for the calculation of the pH to obtain:

`pH=-log(0.125)\\\\pH=0.903`

Which means this solution is highly acidic.

Regards!