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allochka39001 [22]
3 years ago
11

Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.

Chemistry
1 answer:
kiruha [24]3 years ago
3 0

Answer:

Explanation:

For pH of a buffer solution , the formula is

pH = pKa + log [ Base ] / [ conjugate acid ]

=   pKa + log [ NH₃ ] / [ NH₄⁺ ]

Ka = Kw / Kb

Kb for NH₄OH = 1.8 x 10⁻⁵

Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵

= 5.6 x 10⁻¹⁰

pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2

= 10 - log 5.6

= 9.25

Effect of addition of HCl

H⁺ of HCl will react with NH₃ to produce NH₄⁺

25 mL of .1 HCl = 2.5 mM of HCl

25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺

65 mL of .2 M NH₃ = 13 mM of NH₃

65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺

NH₃ + H⁺ = NH₄⁺

NH₄⁺ formed = 2.5 + 13  mM

15.5 mM of NH₄⁺

NH₃ = 13 mM

Concentration of NH₃ = 13 / 90

Concentration of NH₄⁺ = 15.5 / 90

pH of final buffer mixture

= 9.6 + log 13 / 15.5

= 9.25 - .076

= 9.174

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