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
The pH value is mathematically given as
pH= -6.332.
Question Parameters:
the pH of a 0.20 M NH3/0.20 M NH4Cl buffer
the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Generally, the equation for the Chemical Reaction is mathematically given as
HCl + NH3 --> NH4^+ + Cl^-
Therefore
pH= pka + log(13/14).
pH= -6.3 + log 0.93.
pH= -6.3+ (-0.032).
pH= -6.332.
Read more about Chemical Reaction
https://brainly.com/question/11231920
