Answer:
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Answer: The solution is a SATURATED solution.
Explanation:
Although most substances are soluble in water, some are more soluble than others,that is , their solubilities differ. SOLUBILITY is a means of comparing the extent to which different solutes can dissolve in a particular solvent at a definite temperature.
From the question above, when water was added to the sodium acetate in the flask, SOME of the chemical dissolved into the water, meaning that some remained undissolved. This is because a given volume of water can only dissolve a certain amount of chemical in it at room temperature. If more chemical is added to such a solution, the chemical will remain undissolved. Such a chemical solution is said to be a SATURATED SOLUTION.
A saturated solution of a solute at a particular temperature is on which contains as much solute as it can dissolve at that temperature in the presence of undissolved solute particles.
Unsaturated solution is a type of solution that dissolves all its solutes with no presence of undissolved solute.
Supersaturated solution is one which contains more of the solute than it can normally hold at that temperature. It is an unstable solution which crystallizes out when disturbed.
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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