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
The molarity of the solution will be 0.72 m.
The majority of reactions take place in solutions, making it crucial to comprehend how the substance's concentration is expressed in a solution when it is present. The number of chemicals in a solution can be stated in a variety of ways, including.
The symbol for it is M, and it serves as one of the most often used concentration units. Its definition states how many moles of solute there are in a liter of solution.
Given data:
Molarity can be determined by the formula:
where, M is molarity and V is volume.
Put the value of given data in above equation.
57.3 × 0.497 m = M × 39.5 L
M = 0.72 m
Therefore, the molarity of the solution will be 0.72 m
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