NH3 3.0×10^6 L water pH 9.81 density 0.771 g/cm3

NH3 3.0×10^6 L water pH 9.81 density 0.771 g/cm3

 

 

Suppose that ammonia, applied to a field as a fertilizer,

is washed into a farm pond containing 3.0×10^6 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]

 

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pOH = 14 – 9.81 = 4.19

 

[OH^-] = 10^(-4.19) M = x

 

 

 

NH3 이온화 반응식

NH3 + H2O ⇌ NH4^+(aq) + OH^-(aq)

 

Kb = (x) (x) / (C–x)

( 참고 https://ywpop.tistory.com/4294 )

 

 

 

C = [x^2 / Kb] + x

= [(10^(-4.19))^2 / (1.8×10^(-5))] + (10^(-4.19))

= 0.000296 M NH3

 

 

 

(0.000296 mol/L) (3.0×10^6 L) = 888 mol NH3

 

 

 

(888 mol) (17 g/mol) (1 cm3 / 0.771 g) (1 L / 1000 cm3)

= (888) (17) (1 / 0.771) (1 / 1000)

= 19.6 L NH3

 

 

 

답: 19.6 L