Question

Your standard iron solution is 0.1511 M Fe(II), your dichromate solution is 0.0181 M dichromate, and it took 26.84 mL of your standard iron solution to titrate the excess dichromate in your unknown. What is the chemical oxygen demand of the sample, in units of mg O2/L? (Report your answer with 4 sig figs)

Answer 1

Answer:

$COD=2030\frac{mg}{L}$

Explanation:

Hello,

In this case, considering the given information the redox reaction is:

$Fe^{+2}+Cr_2O_7^{-2}\rightarrow Fe^{+3}+Cr^{+3}$

Which properly balanced turns out:

$Fe^{+2}\rightarrow Fe^{+3}+1e^-\\Cr^{+6}_2O_7^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow\\6Fe^{+2}+Cr_2O_7^{-2}+14H^+\rightarrow 6Fe^{+3}+2Cr^{+3}+7H_2O$

In such a way, one sees a 6 to 1 molar relationship between the standard iron (II) solution and the dichromate, therefore, by using the following equation it is possible to determine the oxygen as shown below:

$n_{Fe^{+2}}=n_{Cr_2O_7^{-2}}$

Thus, the moles of iron (II) solution are:

$n_{Fe^{+2}}=0.1511\frac{molFe^{+2}}{L_{sln}}*0.02684L_{sln}=0.00406molFe^{+2}$

Moreover, the moles of dichormate result:

$n_{Cr_2O_7^{-2}}=0.00406molFe^{+2}*\frac{1molCr_2O_7^{-2}}{6molFe^{+2}}=6.76x10^{-4}}molCr_2O_7^{-2}$

Thereby, the volume of the sample is:

$V_{sample}=\frac{n_{Cr_2O_7^{-2}}}{M_{Cr_2O_7^{-2}}}=\frac{6.76x10^{-4}molCr_2O_7^{-2}}{0.0181\frac{molCr_2O_7^{-2}}{L_{sln}} } =0.0373L_{sample}$

Finally, the chemical oxygen demand result:

$COD=6.76x10^{-4}molCr_2O_7^{-2}*\frac{7molO}{1molCr_2O_7^{-2}}*\frac{16gO}{1molO}*\frac{1000mgO}{1gO}*\frac{1}{0.0373L_{sln}} \\COD=2030\frac{mg}{L}$

Best regards.