3-ethyl-2.4-dimethyl-octanoic acid
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
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
16.4 grams is the mass of solute in a 500 mL solution of 0.200 M
.
sodium phosphate
Explanation:
Given data about sodium phosphate
atomic mass of Na3PO4 = 164 grams/mole
volume of the solution = 500 ml or 0.5 litres
molarity of sodium phosphate solution = 0.200 M
The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:
The formula is
molarity = 
putting the values in the equation, we get
molarity x volume = number of moles
0.200 X 0.5= number of moles
number of moles = 0.1 moles
Atomic mass x number of moles = mass
putting the values in the above equation
164 x 0.1 = 16.4 grams
16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.