Titanium (IV) and vanadium (V) form colored complexes when treated with hydrogen

Question

Titanium (IV) and vanadium (V) form colored complexes when treated with hydrogen peroxide in 1M sulfuric acid. The titanium complex has an absorption maximum at 415nm, and the vanadium complex has an absorption maximum at 455nm. A 1.00x10-3M solution of the titanium complex exhibits an absorbance of 0.805 at 415nm and 0.465 at 455nm, while a 1.00x10-2M solution of the vanadium complex exhibits absorbances of 0.400 and 0.600 at 415nm and 455nm, respectively. A 1.0000g sample of an alloy containing titanium and vanadium was dissolved, treated with excess hydrogen peroxide, and diluted to a final volume of 100.0mL. The absorbance of the solution was 0.685 at 415nm and 0.513 at 455nm. What were the percentages of titanium and vanadium in the alloy?

Explanation / Answer

According to Beer's law, the absorbance of a mixture is sum of each component represented as,

A = ebC[x] + ebC[y]

x = titanium and y = vanadium

where,

A = absorbance

e = molar absorptivity coefficients

C = concentration in M

b = path length

with b = path length same for all measurements and cancels out

Let us calculate the molar absorptivity coefficients for the two at given Lmax values.

For titanium, Lmax = 415 nm [C = 0.001 M]

0.805 = e x 0.001

e = 805

at Lmax = 455 nm

0.465 = e x 0.001

e = 465

For vanadium, Lmax = 415 nm [C = 0.001 M]

0.400 = e x 0.001

e = 400

At Lmax = 455 nm

0.600 = e x 0.001

e = 600

So we have 2 equations:

A = eC[x] + eC[y] at 415 nm

A = eC[x] + eC[y] at 455 nm

Plug in the molar absorptivity and the mixture absorbances the sample at their respective wavelengths. With [x] being Ti and [y] being V. This will give us 2 equations with 2 unknowns.

0.685 = 805C[x] + 400C[y] ----- 415 nm

0.513 = 465C[x] + 600C[y] ----- 455 nm

Solving the linear equation for C[x] and C[y],

C[x] = 0.000693 M

C[y] = 0.000318 M

So the mixture contains 0.000693 M Titanium and 0.000318 M Vanadium.

the percentage of each would be,

percentage of titanium = 0.000693 x 100 / (0.000693+0.000318) = 68.55 %

percentage of vanadium = 0.000318 x 100 / (0.000693+0.000318) = 31.45 %

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