A100.0 ml sample of 0.10 m nh3 is titrated with 0.10 m hno3. determine the ph of the solution after the addition of 150.0 ml of hno3. the kb of nh3 is 1.8 × 10-5.

The solution is as follows:
The reaction is written in the attached picture.

Mol NH₃: 0.10 mol/L * 100 mL * 1 L/1000 mL = 0.01 mol
Mol HNO₃: 0.10 mol/L * 150 mL * 1 L/1000mL = 0.015 mol
Mol NH₄NO₃ produced: 0.01 mol NH₄NO₃
Mol HNO₃ left = 0.015 - 0.01 = 0.005 mol

Hydrolyzing NH₄⁺ and applying ICE approach

            NH₄⁺ --> H⁺ + NO₃⁻
I            0.01       0        0
C           -x          +x      +x
E        0.01-x       x        x

Kh = Kw/Kb = [H⁺][NO₃⁻]/[NH₄⁺]
10⁻¹⁴/1.8×10⁻⁵ = [x][x]/[0.01-x]
Solving for x,
x = [H⁺] = 2.357×10⁻⁶ mol

The formula for pH is
pH = -log [H⁺]
Aside from 2.357×10⁻⁶ mol, let's add the H⁺ from the remaining HNO₃ which is 0.005.
Therefore,
pH = -log[2.357×10⁻⁶ mol + 0.005 mol]
pH = 2.3

Elias is almost correct. he just got it backwards.the correct answer is that it demonstrates elastic potential energy transforming into kinetic energy.