A substance is tested and has a pH of 1.2. How would you classify it?
2 answers:
You might be interested in
Answer:
Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹
Explanation:
First convert Concentration from ppm inM or mol/l
⇒ Molar mass of KMnO₄ = 158.03 g
⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l
⇒ Molarity = = 2.83 x 10⁻⁵ molar
Absorbance (A) = - log(T) ( T = % transmittance)
= - log(0.859)
= 0.06
According to Lambert Beer's law
ε =
or, ε =
or, ε = 2120.14 cm⁻¹M⁻¹
Where
ε = Molar absorptivity
A = absorbance
C = Molar concentration of KMnO₄ solution
l = length
Answer: Concentration of = 0.328 M
Concentration of = 0.328 M
Concentration of = 0.532 M
Explanation:
Moles of and
= 0.430 mole
Volume of solution = 0.500 L
Initial concentration of and
=
The given balanced equilibrium reaction is,
Initial conc. 0.860M 0.860M 0
At eqm. conc. (0.860-x) M (0.860-x) M (x) M
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 0.532 M
Thus, the concentrations of at equilibrium are :
Concentration of = (0.860-x) M =(0.860-0.532) M = 0.328 M
Concentration of = (0.860-x) M = (0.860-0.532) M = 0.328 M
Concentration of = x M = 0.532 M