Answer:
ph = 4.73
Explanation:
Determine the pH of a 0.62 M NH4NO3 solution at 25C
The Kb for NH3 is 1.76 x 10^-5
NH4+ is the conjugate acid of the base NH3
Ka for NH4+ = K water / K NH3 conjugate base
Ka = 1 × 10⁻¹⁴ / 1.76 10⁻⁵
= 5.68 × 10⁻¹⁰
which is he Ka for
NH4 + H2O --> NH4OH + H⁺
Ka = [NH4OH] [H+] / [NH4)+]
5.68 × 10⁻¹⁰ = [X] [X] / [0.62]
X² = 3.52 × 10⁻¹⁰
X = [H+] = 1.88 × 10⁻⁵
p of that H+ = 4.73
ph = 4.73
The pH of ammonium nitrate solution has been 4.728.
Ammonium nitrate has been a weak base. The pH of the solution has been given by:
[tex]\text {pH}=\dfrac{1}{2}\;[pKw-pKb-logC][/tex]
Where, the self ionization equilibrium of water, [tex]pKw=14[/tex]
The base dissociation constant, [tex]Kb=1.76\;\times\;10^-^5[/tex]
The value of pKb has been given as:
[tex]pKb=-log\;Kb\\pKb=-log(1.76\;\times\;10^-5)\\pKb=-4.75[/tex]
The concentration of the solution, [tex]c=0.62 \rm m[/tex]
Substituting the values for pH of the solution:
[tex]\rm pH=\dfrac{1}{2}\;[14-(-4.75)-log \;0.62] \\pH=\dfrac{1}{2}\;[14+4.75+0.207)\\pH=4.728[/tex]
The pH of ammonium nitrate solution has been 4.728.
For more information about pH, refer to the link:
https://brainly.com/question/491373
