<u>Answer:</u>
<em>A. 10.25</em>
<em></em>
<u>Explanation:</u>
Pkb =4.77
So pka = 14 - pka = 9.23
Initial 0.50M 0 0
Change -x +x +x
Equilibrium 0.50M-x +x +x
(-x is neglected) so we get
![pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E3%20O%5E%2B%5D%5C%5C%5C%5CpH%3D-log%5B1.72%5Ctimes10%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D4.76)
pOH = 14 - pH
= 14 - 4.76
pOH = 9.24 is the answer
Option A - 10.25 is the answer which is close to 9.24
Because copper has a higher electrical and thermal conductivity than iron and is not susceptible to oxidation as easily.
Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
The atmosphere is a mixture of gasses.
Lord dragon you are wrong.
The answer is A: Incandescent light bulbs are almost 100 percent efficient. Because b-d do not make any sense.
