Question

A mixture of KOH and na2co3 required 20 mL of N/20 HCL solution when titrated using phenolphthalein as indicator. But the same amount of solution required 30 mL of the same acid solution when titrated using methyl orange indicator. Calculate the amount of KOH in the mixture.

Answer 1

Answer:

Explanation:

when titrated with phenolpthalein

equivalent of koh + equivalent of na2co3 = 1

moles of koh + moles of na2co3 = 0.001

x/56 + y/106 = 0.001 (eq 1)

when titrated with methyl orange

equivalent of koh + equivalent of na2co3 = 1.5

x/56 + 2y/106 = 0.0015 eq (2)

subtract eq 1 from eq 2

y/106 = 0.0005

y=0.053 gm

x/56 + 0.053/106 = 0.001

x/56 + 1/2000 = 0.001

x/56 = 0.001-0.0005

x/56 = 0.0005

x = 0.028 gm

amount of koh = 0.028 gm