R the titration of 70.0 mL of 0.300 M NH 3 with 0.500 M HCl at 25 ∘ C, determine the relative pH at each of these points. Determine the relative pH before the addition of any HCl .

pH = 11.36

Explanation:

Before the addition of any HCl, you have 0.300M NH₃ in equilibrium with water:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

When system is in equilibrium:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

Concentrations in equilibrium are:

[NH₃] = 0.300M - X

[NH₄⁺] = X

[OH⁻] = X

1.8x10⁻⁵ = [X] [X] / [0.300M - X]

5.4x10⁻⁶ - 1.8x10⁻⁵X = X²

X² + 1.8x10⁻⁵X - 5.4x10⁻⁶ = 0

Solving for X:

X = -0.00233 → False answer, there is no negative concentrations

X = 0.00231 → Right answer.

As [OH⁻] = X; [OH⁻] = 0.00231

As pOH = -log [OH⁻] = 2.636

14 = pOH + pH

14 - pOH = pH

pH = 11.36

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )