Answer to Question #234591 in General Chemistry for Olioko

Question

A typical protein contains 16wt.% Nitrogen. A 0.500ml aliquot of protein solution was digested and the liberated NH_(3) was diluted into 10.00ml of 0.02140mol HCL. Unreacted HCL require 3.26ml of 0.0198mol of NaOH for complete titration. Find the concentration of protein in the original sample.

Answer 1

If we consider that all the reactions occurred without any interference, we have to calculate how much HCl was left at the end to then know how much HCl reacted with the NH₃ produced.

$3.26\,{mL}\,sol.\,{NaOH}(\dfrac{0.0198\,mol\,{NaOH}}{1000\,mL\,sol.\,{NaOH}})(\dfrac{1\,mol\,{HCl}}{1\,mol\,{NaOH}})(\dfrac{1000\,mL\,sol.\,{HCl}}{0.02140\,mol\,{HCl}}) \\ =3.016\,{mL}\,sol.\,{HCl}$

Now that we know the volume of HCl equivalent to the remaining we can confirm how much HCl reacted before as $V_{rx}=V_i-V_{f}=(10.00-3.016)\,mL=6.984\,mL$

With that last piece of information we can calculate the mass of nitrogen contained in the sample and with that the concentration of protein in the original sample:

$6.984\,{mL}\,sol.\,{HCl}(\dfrac{0.02140\,mol\,{HCl}}{1000\,mL\,sol.\,{HCl}})(\dfrac{1\,mol\,{NH_3}}{1\,mol\,{HCl}})(\dfrac{14.01\,g\,{N}}{1\,mol\,{NH_3}})(\dfrac{100\,g\,{protein}}{16\,g\,{N}}) \\ =0.01309 \text{ g of protein.}$

$C=\dfrac{0.01309 \text{ g of protein}}{0.500\,{mL}}(1000\,mL/L)=26.18\,g/L$

In conclusion, the concentration of protein on the sample is about 26.18 g/L