Question

The determination of milk acidity is one of the most widely used measures to control the raw material milk by ndústria. The test is used to rank the milk and as a guide to control the manufacturing of products such as cheese.

The titratable acidity is expressed in percentage ((mass / vol) of lactic acid represented by the following structural formula:

In one assay, an aliquot of 10 ml milk sample and was placed in a flask and added two drops of phenolphthalein. Titration of the sample consumed NaOH solution 1,68mL-1 0.100 mol L -1.

The acidity of the milk in the sample, expressed as percentage (weight / volume) is

Data: Pasta molars: H = 1 g mol -1; C = 12 g mol -1; O = 16 g mol -1.

   a) 0.076%.

   b) 0.15%.

   c) 0.30%.

   d) 1.5%.

   e) 7.6%.

Answer 1

The mass weight for lactic acid is: 90 g/mol (C₃H₆O₃)

First you need to transform the 1.68 of NaOH into moles:

$1.68mL\frac{1L}{1000mL}\times \frac{0.1moles NaOH}{1L}=0.000168moles NaoH$

Then as you can see in the structure there is just one acid group, it means that the NaOH and the lactic acid react in a ration 1:1, so you now can calculate the mass of lactic acid:

$0.000168moles NaoH\times \frac{1mol LacAc}{1mol NaOh}\times \frac{90gLacAc}{1molLacAc}=0.01512gLacAc$

If you divide the mass by the aliquot, and then multiply times 100 you get the percentage:

$\frac{0.01512gLacAc}{10mL}*100=0.15$

So the answer is B)