Spectrophotometry area of chemistry

[thenewbosco]

I am not familiar with this area of chemistry. The theory given in the manual doesn't really help. The only thing it says is that absorption is proportional to concentration. Maybe you can give me some assistance on how to get started and some equations that i can use.

The eqilibrium constant for: $$AuBr_{4}^{-} + 2Au + 2Br^{-} --> 3AuBr_{2}^{-}$$ can be determined by preparing a solution of AuBr$$_{4}^{-}$$ and AuBr$$_{2}^{-}$$ in contact with a piece of gold metal.
In one experiment a solution initially containing $$6.41x10^{-4}$$ mol/L of dissolved gold (both AuBr$$_{4}^{-}$$ and AuBr$$_{2}^{-}$$) in 0.4 M HBr was allowed to attain equilibrium in the presence of gold metal. The absorbance was 0.445 in a 1.00 cm cell at 382nm.

In a separate experiment, the absorbance of a $$8.54x10^{-5}$$M solution of only AuBr$$_{4}^{-}$$ (no AuBr$$_{2}^{-}$$ ) in 0.4 M HBr was determined in a 1.00cm cell to be 0.410 at 382nm. (AuBr$$_{2}^{-}$$ does not absorb at 382nm.)

a) calculate the equilibrium concentrations and b) evaluate the equilibrium constant.

I can get b) if i knew how to do a) can someone help me with this one?


Thanks

 

[thenewbosco]

here is what i have done can someone tell me if it is correct:

Given that $$A={\epsilon}cl$$ (from beers law i found on the internet)

I can use this and experiment two to get the $${\epsilon} = 0.410/8.54^{-5}=4800.94$$

Now from this value of epsilon, can i use it with beer's law again to calculate the concentration of AuBr$$_{4}^{-}$$ in the first experiment with the new A value (.445)?

Then i would have a concentration for AuBr$$_{4}^{-}$$ and i can subtract it from the combined concentration of 6.41E-4 M, to get [AuBr2].

Is this correct?

thanks

 

[chem_tr]

You should use compound-specific extinction coefficients, I mean, $\displaystyle \epsilon_{{AuCl_4}^-}$ and $\displaystyle \epsilon_{{AuCl_2}^-}$

However, it was said that the latter species does not absorb at 382 nm, which means that the extinction coefficient for [AuBr₂]^- will be zero.

The second separate experiment will allow you to know the extinction coefficient for the first species, namely, [AuBr₄]^-. In the first experiment, you can find the final concentration of this species after equilbrium has been reached. The difference will totally be from the second Au⁺-bearing species, as solid Au is not spectrometrically specific.

 

[fantasydreams]

wow thanks
i got that exact same question on my lab manual too! :P