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QJEEM21Chbec8: JEE Main 2021 Chemistry - Class 11 JEEM21Chbec8 Solution: 0.4 g mixture of NaOH, Na 2 CO 3 and some inert impurities was…

0.4 g mixture of NaOH, Na2CO3 and some inert impurities was first titrated with N10{N \over {10}} HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL 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)

Step-by-Step Verified Solution

1st end point reaction :

NaOH+HCl⟶NaCl+H2O

nf=1nf = 1

NaCO3+HCl⟶NaHCO3

nf=1nf = 1

Eq of HCl used =nNaOH×1+nNa2CO3×1= {n_{NaOH}} \times 1 + {n_{N{a_2}C{O_3}}} \times 1

17.5×110×10−3=nNaOH+nNa2CO317.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaOH}} + {n_{N{a_2}C{O_3}}}

2nd end point :

NaHCO3+HCl⟶H2CO3

1.5×110×10−3=nNaHCO3×1=nNaHCO31.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaHC{O_3}}} \times 1 = {n_{NaHC{O_3}}}

0.15 mmol = nNa2CO3{n_{N{a_2}C{O_3}}}

0.15=nNa2CO30.15 = {n_{N{a_2}C{O_3}}}

wNa2CO3=0.15×106×10−30.4×100×10{w_{N{a_2}C{O_3}}} = {{0.15 \times 106 \times {{10}^{ - 3}}} \over {0.4}} \times 100 \times 10

= 3.975%

≃\simeq 4%

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