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
Answer:
1255.4L
Explanation:
Given parameters:
P₁ = 928kpa
T₁ = 129°C
V₁ = 569L
P₂ = 319kpa
T₂ = 32°C
Unknown:
V₂ = ?
Solution:
The combined gas law application to this problem can help us solve it. It is mathematically expressed as;
P, V and T are pressure, volume and temperature
where 1 and 2 are initial and final states.
Now,
take the units to the appropriate ones;
kpa to atm, °C to K
P₂ = 319kpa in atm gives 3.15atm
P₁ = 928kpa gives 9.16atm
T₂ = 32°C gives 273 + 32 = 305K
T₁ = 129°C gives 129 + 273 = 402K
Input the values in the equation and solve for V₂;
V₂ = 1255.4L
From the calculation, the pH of the solution after dilution is 3.
<h3>What is the pH?</h3>
The pH is the hydrogen ion concentration of the solution. Now we know that;
C1 = 0.010 m
V1 = 10.0 ml
V2 = 10.0 ml + 100.0 ml = 110 ml
C2 = ?
C1V1 = C2V2
C2 = C1V1 /V2
C2 = 0.010 m * 10.0 ml / 110 ml
C2 = 0.00091 M
pH = -log[0.00091 M]
pH = 3
Learn more about pH:brainly.com/question/15289741
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The original options for this question were cleavage, luster and hardness. The answer would be cleavage.
