A chromate πΆππ4 β2 solution has an absorbance of 1.534 when the cell length is 1.5 cm. If the molar absorptivity of chromate ion is 5.21 Γ 103 πβ1 ππβ1 , calculate the concentration of chromate ion.
I don't know your problem here but it appears to me to be a plug and chug problem.
A = abc
A = 1.534
a = 5210
b = 1.5
c = ?
Substitute and solve for c. The units for c will be mols/L since a is the molar absorptivity constant.
thank you dr bob, i get the answer.
To calculate the concentration of chromate ion, you can use the Beer-Lambert Law, which relates the absorbance of a solution to its concentration.
The Beer-Lambert Law is expressed as: A = Ι * c * β
Where:
A is the absorbance of the solution
Ι is the molar absorptivity (also known as the molar absorptivity coefficient)
c is the concentration of the solute
β is the path length of the cell (in this case, 1.5 cm)
In your case, the given information is:
A = 1.534 (absorbance)
Ι = 5.21 Γ 10^3 π^β1 ππ^β1 (molar absorptivity)
β = 1.5 cm (cell length)
To calculate the concentration (c), rearrange the Beer-Lambert Law equation to isolate c:
c = A / (Ι * β)
Substituting the given values:
c = 1.534 / (5.21 Γ 10^3 * 1.5)
Simplifying the equation:
c = 1.534 / 7.815 π
Therefore, the concentration of chromate ion is approximately 0.1962 M.