Question

0.4 gm of solution of mixture of NaOH and Na2​CO3​ and inert impurities was first titrated with phenolphthalein and N/10 HCl 17.5ml of HCl was required at the endpoint. After this methyl orange was added and 2.5ml of same HCl was again required for next endpoint. Find out percentage of NaOH and Na2​CO3​ in the mixture.


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Answer 1

Explanation:

as given in the question NaOH and Na2Co3 is titrated with N/10 HCl

for NaOH

equivalent of NaOH = equivalent of HCl

equivalent of HCl = normality × volume (L)

= 0.1 × 17.5 / 1000 = 1.75 × 10-³

from here equivalent of NaOH = 1.75 × 10-³

wt of NaOH = equivalent × equivalent wt of NaOH

= 40 × 1.75 × 10-³ = 0.07 gram

now calculate % of NaOH

= 0.07/0.4 × 100 = 70/4 = 17.5 %

similarly for Na2Co3

equivalent of HCl = equivalent of Na2Co3

equivalent of Na2Co3 = 0.1 × 2.5/1000

= 0.25 ×10-³

wt of Na2Co3 = equivalent × equivalent wt of Na2Co3

= 0.25 × 10-³ × 106 = 0.026 gm

%of Na2Co3 = 0.026/0.4 × 100 = 26/4 = 6.75 %

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