0.4g mixture of NaOH, Na₂CO₃ and some inert impurities was first titrated with $\frac{N}{10} H C l$ using phenolphthalein as an indicator, 17.5mL of HCl was required at the end point . After this methyl orange was added and titrated. 1.5mL of same HCl was required for the next end point. The weight percentage of Na₂CO₃ in the mixture is----------. (Rounded – off to the nearest integer)

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alok kumar singh | Contributor-Level 10

2 months ago

NaOH + Na₂CO₃

(1) When Hph is added

m.e NaOH + $\frac{1}{2} m . e N a_{2} C O_{3} = m . e . H C l = 17.5 \times \frac{1}{10} = 1.75$ (m.e = milli equivalents)

(2) When MeOH is added after Hph

$\frac{1}{2} m e N a_{2} C O_{3} = m . e H C l = 1.5 \times \frac{1}{10} = 0.15$

$\begin{array}{l} ∴ m . e N a O H = 1.75 - 0.15 = 1.6 \\ m . e N a_{2} C O_{3} = 0.15 \times 2 = 0.3 \end{array}$

$\frac{W_{N a_{2} C O_{3}}}{E_{N a_{2} C O_{3}}} \times 1000 = 0.3$

$W e i g h t % o f N a_{2} C O_{3} = (\frac{\frac{0.3 \times 53}{1000}}{0.4}) \times 100 = \frac{0.3 \times 53}{10 \times 0.4} = \frac{15.9}{4} = 3.975 % \approx 4 % .$

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