By using ICE table:
CH3NH3+ + H2O → CH3NH4 2+ + OH-
initial 0.175 0 0
change -X +X +X
Equ (0.175-X) X X
when: Ka = Kw / Kb
= (1 x 10^-14) / (4.4 x 10^-4) = 2.3 x 10^-11
when Ka = [CH3NH42+][OH-] / [CH3NH3+]
by substitution:
2.3 x 10^-11 = X^2 / (0.175 - X ) by solving for X
∴ X = 2 x 10^-6
∴[OH-] = 2 x 10^-6
∴POH = -㏒[OH-]
= -㏒(2 x 10^-6)
= 5.7
when PH + POH = 14
∴PH = 14 - 5.7 = 8.3
<span>Chlorine has 7 valence electrons, but in the diagram, there are 2 chlorine atom making there a total of 14 valence electrons. There are 2 valence electrons in between the two atoms. There are 6 valence electrons around the outside of each individual chlorine atom. The electrons are arranged in groups of two around the outside of each atom.</span>
Answer:
pH = 14
Explanation:
pOH = -log[OH-] = -log1,0 = 0
pH + pOH = 14
pH = 14 - pOH = 14 - 0 = 14
Im not really sure but my best guess is d 52.8%