double indicator titration calculation

Question

25 cm3 of a solution containing sodium hydroxide and sodium carbonate was titrated with 0.05 M hydrochloric acid and phenolphthalein was used as the indicator. After 18.50 cm3 of tha acid was added, phenolphthalein turned colourless. Methyl orange was then added. As the titration continued, a ferther 10.00 cm3 of acid was needed to turn methyl orange to red colour. Calculate the concentration of sodium hydroxide and sodium carbonate in the original solution.

Answer 1

Assume that there are x mol of NaOH and y mol of Na2CO3 in 25 cm3 of the solution.

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At the end point when phenolphthalein indicator is used, NaOH is completely neutralized, and Na2CO3 is changed to NaHCO3.
NaOH + HCl → NaCl + H2O ...... (*)
Na2CO3 + HCl → NaHCO3 + NaCl ...... (**)

In (*), mole ratio NaOH : HCl = 1 : 1
No. of moles of NaOH used = x mol
Hence, no moles of HCl used = x mol

In (**), mole ratio Na2CO3 : HCl : NaHCO3 = 1 : 1 : 1
No. of moles of Na2CO3 used = y mol
Hence, no. of moles of HCl used = y mol
and, no of moles of NaHCO3 formed = y mol

Total number of moles of HCl used :
x + y = 0.05 x (18.5/1000)
x + y = 9.25 x 10-4 ...... (1)

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At the end point of further titration using methyl orange as indicator, all NaHCO3 left in the solution is changed to CO2.
NaHCO3 + HCl → NaCl + H2O + CO2

Mole ratio NaHCO3 : HCl = 1 : 1
From above, no. of moles of NaHCO3 reacted = y mol
Hence, no. of moles of HCl used = y mol

Since 10 cm3 of 0.05 M HCl is used,
y = 0.05 x (10/1000)
y = 5 x 10-4 (mol)

Substitute y into (1)
x = 4.25 x 10-4 (mol)

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Concentration of NaOH
= mol / vol
= (4.25 x 10-4) / (25/1000)
= 0.017 M

Concentration of Na2CO3
= mol / vol
= (5 x 10-4) / (25/1000)
= 0.02 M

Answer 2

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