The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
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Ka (mol/L ) = <span>0.00002340</span>
Answer:
The calculated concentration of sodium thiosulphate solution will be less than the actual value.
Explanation:
When IO3^2- solution is added to KI solution, I2 gas is released ,then sulphuric acid is now added to facilitate reduction. In order to prevent the escape of iodine (I2) gas ,the solution must immediately be titrated with thiosulphate.
If the solution is not immediately titrated with thiosulphate, the concentration of iodine available in the system decreases. When this occurs, it will also cause a decrease in the amount of iodine available to react with thiosulphate thus decreasing the concentration of thiosulphate obtained from calculation
Answer:
A positive ions is always smaller than the corresponding atom.
A negative ion is always larger than the corresponding atom.
Explanation:
The reason for this is that, when a positive ion is formed, a full shell is usually removed with its electrons thereby reducing the size of the electron cloud and decreasing the size of the electron cloud.
A negative ion is formed by addition of more electrons to the electron cloud hence it spreads out. Interelectronic repulsion accounts for the larger size of the negative ion.
The amount of chemical energy is equal to the amount of heat and light energy.
This is due to the principle of conservation of energy, which states that the total energy of a system remains constant.
