The standardized 0.1M NaOH was then used to titrate a vinegar sample that is ~5%

Question

The standardized 0.1M NaOH was then used to titrate a vinegar sample that is ~5% (wt/wt) with a density of 1.056g/mL. The vinegar sample was diluted before titration by taking 25.00mL by volumetric pipet and diluting it to 250.0mL in a volumetric flask. 20.00mL of the diluted vinegar was titrated until a light pink endpoint with phenolphthalein. 16.93mL of NaOH was used.
-What is the molarity (M/L) of the diluted vinegar?

-What is the molarity of the undiluted vinegar?

-What is the concentration of the vinegar reported as a mass % (grams acetic acid/100g solution *100%). Use the density given above in your calculation.

Explanation / Answer

The 5% (w/w) means 100 gm solution contains 5 gm of acetic acid.

From the titration ,we know

VNaOH    SNaOH =VVinegar    Svinegar

i) Then molarity of diluted vinegar is( Svinegar)=(16.93X0.1/20)=0.08465 M

ii)Then 25ml is taken and diluted to 250 ml. The   molarity of diluted vinegar is( Svinegar)=0.08465 M.

Then,Vvinegar(undiluted)    Svinegar (undiluted)=VVinegar (diluted) Svinegar (diluted)

25XSundiluted=250 X 0.08465

Sundiluted=0.8465 M

c)The 5% (w/w) means 100 gm solution contains 5 gm of vinegar.

Volume of the vinegar=(mass/density)=(5/1.056)=4.734ml

concentration=(1000X weight percentage )/{molar mass of the soluteX(100-weight percentage)}=(1000X5)/{60X(100-5)}=0.877 molal

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