Question

Suppose that ammonia, applied to a field as a fertilizer, is washed into a farm pond containing 3.0 × 106 L of water. If the pH of this pond is found to be 9.81, what volume of liquid ammonia found its way into the pond? [Given: Kb(NH3) = 1.8 × 10–5; the density of liquid ammonia is 0.771 g/cm3]

Answer 1

Answer:

19,7 L of ammonia.

Explanation:

The reaction of NH₃ with water is:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ With kb = 1,8x10⁻⁵

OH⁻ concentration is:

[OH⁻] = $10^{-14+pH}$ = 6,46x10⁻⁵ M

That is the same than [NH₄⁺]

Also, Kb is:

kb = $\frac{[OH^-][NH_{4}^+]}{[NH_3}}$

Replacing:

[NH₃] = 2,32x10⁻⁴ M

Total volume is 3,0x10⁶ L, thus:

2,32x10⁻⁴ M + 6,46x10⁻⁵ M = $\frac{NH_{3} + NH_{4}^+ moles}{3,0x10^{6} L }$

= 889,8 moles of NH₃ + NH₄⁺

These moles were initially just of liquid ammonia, thus, volume of ammonia added into the pond is:

889,8 mol NH₃ ₓ $\frac{17,031 g}{1mol} \frac{1 mL}{0,771 g}$ = 19,7 L of ammonia

I hope it helps!