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
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In order to find molarity, you must first find the number of moles that was dissolved.
Now, Moles = Mass ÷ Molar Mass
⇒ Moles of NaCl = 2.922 g ÷ 58.44 g/mol
= 0.05 moles
∴ the Molarity of the NaCl is 0.05 M [Option 1]
Answer: acid dissociation constant Ka= 2.00×10^-7
Explanation:
For the reaction
HA + H20. ----> H3O+ A-
Initially: C. 0. 0
After : C-Cx. Cx. Cx
Ka= [H3O+][A-]/[HA]
Ka= Cx × Cx/C-Cx
Ka= C²X²/C(1-x)
Ka= Cx²/1-x
Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2
Ka= 0.2(0.001²)/(1-0.001)
Ka= 2.00×10^-7
Therefore the dissociation constant is
2.00×10^-7
