1. Tardigrade
  2. Question
  3. Chemistry
  4. A solution containing Na2CO3 and NaOH requires 300 ml of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further 25 ml of 0.2 N HCI is required. The amount of NaOH present in solution is (NaOH = 40, Na2CO3 = 10)

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Q. A solution containing $Na_2CO_3$ and $NaOH$ requires $300\, ml$ of $0.1\, N\, HCl$ using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further $25\, ml$ of $0.2 \,N\, HCI$ is required. The amount of $NaOH$ present in solution is $\left(NaOH = 40, Na_2CO_3 = 10\right)$

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Solution:

$\left(I\right)$ Phenolphthalein indicate partial neutralisation of
$Na_2O_3 \to NaHCO_3$
$Meq. of Na_2O_3+ Meq. of NaOH = Meq. of HCl$
$\frac{W}{E}\times1000+\frac{W}{E}\times1000=NV$
(Suppose $Na_2O_3 = a g, NaOH = b g)$
$\frac{a}{106}\times1000+\frac{b}{40}\times1000=300\times0.1 .......\left(1\right)$
$\left(II\right)$ Methyl orange indicate complete neutralisation
$HCl \,\,\,\,\, HCl$
$N_1V_1 = N_1V_1, 25 \times 0.2 = 0.1 \times V_2 so V_2 = 50 ml$ excess
$\therefore \frac{a}{50}\times1000+\frac{b}{40}\times1000=350\times0.1 .......\left(2\right)$
From (1) and (2) b=1 gm.