Answer to Question #139068 in Inorganic Chemistry for chi

Question

A sample of caustic soda weighing 1.675 g was found to contain 97.25% of total alkali calculated as NaOH and of which is 2.48% was Na2CO3. What volume of 1.15 N acid could have been consumed in the Polypropylene and MO titrations as ordinarily performed?

MW: NaOH = 40.0, Na2CO3 = 106.0

Answer 1

(Assuming that 1.15 N actually means 1.15 M HCl acid, Polypropylene was meant to be Phenolphthalein indicator and MO is Methyl Orange indicator.)

Mass of NaOH $=\dfrac{97.25}{100}x1.475 = 1.4344375g$

Mass of Na2CO3$\dfrac{2.48}{100}x1.475 = 0.03658g$

Moles of NaOH $=\dfrac{1.4344375}{40} = 0.03586mol$

Moles of Na2CO3$=\dfrac{0.03658}{106} = 3.383x10^-4mol$

Moles of acid required to react with all the NaOH

Mole ratio=1:1

Hence moles of acid $=\dfrac{1}{1}x0.03586 = 0.03586 mol$

Moles of acid required to react with all Na2CO3

Mole ratio = 2:1

Hence moles of acid$=\dfrac{1}{2}x3.383x10^-4 = 1.692x10^-4 mol$

Total moles of acid used = 0.03586 + 1.692x10^-4 = 0.03602915mol.

Volume of acid$=\dfrac{moles x 1000}{molarity} = \dfrac{0.0360291x1000}{1.15}$

= 31.33 cm³ for Phenolphthalein

For MO$=\dfrac{31.33}{100} = 0.3133$ cm³