In the titration of 50.0 mL of HCl of unknown concentration, the phenolphthalein indicator present in the colorless solution tur
1 answer:
You might be interested in
Answer:
The weight percent in the sample is 17,16%
Explanation:
The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:
2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻
I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻
The moles in the end point of S₂O₃⁻ are:
0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.
As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:
4,228x10⁻⁴ moles of S₂O₃⁻× = <em>2,114x10⁻⁴ moles of I₃⁻</em>
As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:
2,114x10⁻⁴ moles of I₃⁻× = <em>4,228x10⁻⁴ moles of Ce(IV)</em>.
These moles are:
4,228x10⁻⁴ moles of Ce(IV)× = 0,05924 g of Ce(IV)
As was taken an aliquot of 25,00mL from the solution of 250,0mL:
0,05924 g of Ce(IV)× =0,5924g of Ce(IV) in the sample
As the sample has 3,452g, the weight percent is:
0,5924g of Ce(IV) / 3,452g × 100 = <em>17,16 wt%</em>
I hope it helps!
Answer: Equilibrium constant is 0.70.
Explanation:
Initial moles of = 0.35 mole
Volume of container = 1 L
Initial concentration of
Initial moles of = 0.40 mole
Volume of container = 1 L
Initial concentration of
equilibrium concentration of [/tex]
The given balanced equilibrium reaction is,
Initial conc. 0.35 M 0.40M 0 0
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
we are given : (0.35-x)= 0.18
x = 0.17
Now put all the given values in this expression, we get :
Thus the value of the equilibrium constant is 0.70.