Answer:
Explanation:
NH₄NO₃ = NH₄⁺ +NO₃⁻
heat released by water = msΔ T
m is mass , s is specific heat and ΔT is fall in temperature
= 50 x 4.18 x ( 22 - 16.5 ) ( mass of 50 mL is 50 g )
= 1149.5 J .
This heat will be absorbed by the reaction above .
q for the reaction = + 1149.5 J
2 )
molecular weight of NH₄NO₃ = 80
No of moles reacted = 5/80 = 1 / 16 moles.
3 )
5 g absorbs 1149.5 J
80 g absorbs 1149.5 x 16 J
= 18392 J
= 18.392 kJ.
= + 18.392 kJ
ΔH = 18.392 kJ / mol
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
