Will concentration of sodium hydroxide affect the result of chicken feather hydrolyzate if i don't standardized it?

Question

The title of the paper: Isolation and characterization of keratin from chicken feathers https://doi.org/10.1063/5.0002792

Here is the method: Keratin was isolated by dissolving 5g of chicken feather in 95 ml sodium hydroxide 1N solutions at 70°C for 3 hrs. Glacial acetic acid was added to neutralize the solution (pH 7) to obtain a solution of 5% w/v concentration. The chicken feather is then heated at 70°C in a water bath to form a gel. Then heated in an oven at 70°C until dry.

My overall question whether the reaction between acetic acid and the chicken feather with 1N sodium hydroxide be affected if I didn't standardized the sodium hydroxide?

And the second one is that .. how did the author ended up with 5% w/v concentration? Does this mean that the precipitated chicken hydrolyzate constitute only 5g of weight and 95 ml of acetic acid is added? I am confuse with how the author obtain a 5% w/v concentration.

Answer 1

The lye is the agent to push the hydrolysis of the peptide forward. The initial concentration of the lye may affect the rate of this digestion (temperature, pressure, time of residence of the reagents in the reactor are equally factors to be considered), as well if the digestion is exhaustive. As shown by the reaction scheme, the reaction yields water; water thus dilutes the lye and the concentration of lye decreases.

As stated in your quotation, glacial acetic acid is added after $\pu{3 h}$ of reaction. This is to quench any excess of lye which did not react (yet). Using glacial acetic acid instead of just «acetic acid» as from the groceries has the advantage of its high concentration ($\pu{100\%}$ HOAc), thus the introduced change of volume caused by the addition of this reagent is much smaller.

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Back on an envelope: $\pu{95E-3 L} \times \pu{1 mol/L} = \pu{95E-3 mol}$ of $\ce{NaOH}$ engaged. With data from Wikipedia's entry about acetic acid ($M = \pu{60 g/mol}$, and $d = \pu{1.049 g cm−3}$), one needs at maximum $\pu{5.7 g}$ (or $\pu{5.5 mL}$) of glacial acetic acid to neutralize it if there were no digestion. (However, because of the digestion of the peptide consuming NaOH, the amount of HOAc needed for the neutralization likely is lower.)

Then, assuming the density of the the digest were $\pu{1 g cm-3}$, you add $\pu{95 mL}$ (from NaOH) and $\pu{5 mL}$ (from HOAc) to yield $\approx \pu{100 mL}$ (in total) containing $\pu{5 g}$ of digest. Here you are, $\approx \pu{5\%wt}$ (m/v).