2nd of same first question Fifteen (15.00) mL of a bleach bath (sp.gr 1.02) requ

Question

2nd of same first question

Fifteen (15.00) mL of a bleach bath (sp.gr 1.02) required 12.00 mL of a 0.100 N H2SO4 to fully titrate the total alkali to the equivalence point. Assuming all the alkali in the bath is NaOH, calculate:

a. The normality of the alkali in the bleach bath;

b. The concentration of NaOH in grams/liter;

c. The concentration of NaOH in percent weight to volume (%w/v);

d. The concentration of NaOH in percent weight to weight (%w/w);

e. The degrees Twaddell;

f. Considering that a jet dye machine will be used to prepare and then dye the fabric and that this machine operates at a 8:1 liquor ratio with a working capacity of 300 Kg fabric, what quantity of 30% (w/w) concentrated NaOH you would use to prepare the bath in this machine;

g. The volume of .05 N Hall, which would have been required to perform the same titration.

h. The pH of the bath

Explanation / Answer

a) Equivalents of bleach bath = equivalents of H2SO4 = 12*0.1/1000 = N*15/1000

Normality = 1.2/15 N = 0.08 N

B) 15 ml of alkali has 0.08 N NaOH . n factor of NaOH =1, so Molarity will be 0.08 M
total moles of NaOH in 15 ml = 15*0;08M = 1.2 milli moles = 1.2*40/1000 gm = 0.048 gms
Concentration will be in grams / litre = 0.048*1000/15 = 48/15 gms/litre = 3.2 gm/litre
Now 15 ml of solution has 0.08 moles of NaOH

C )

Percentage weight by volume = weight of solue in grams*100/ volume of solution in ml
= 0.048*100/15
= 0.32 %

Percentage weight by weight = weight of solue in grams*100/ weight of solution in gm

Weight of solution = volume * specific gravity = 1.02*15 = 15.3 gm

= 0.048*100/15.3 = 4.8/15.3 = 0.313 %

e

degree Twaddle (

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