JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 18)
0.4 g mixture of NaOH, Na2CO3 and some inert impurities was first titrated with $${N \over {10}}$$ 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)
Answer
4
Explanation
1st end point reaction :
$$NaOH + HCl\buildrel {} \over \longrightarrow NaCl + {H_2}O$$
$$nf = 1$$
$$NaC{O_3} + HCl\buildrel {} \over \longrightarrow NaHC{O_3}$$
$$nf = 1$$
Eq of HCl used $$ = {n_{NaOH}} \times 1 + {n_{N{a_2}C{O_3}}} \times 1$$
$$17.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaOH}} + {n_{N{a_2}C{O_3}}}$$
2nd end point :
$$NaHC{O_3} + HCl\buildrel {} \over \longrightarrow {H_2}C{O_3}$$
$$1.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaHC{O_3}}} \times 1 = {n_{NaHC{O_3}}}$$
0.15 mmol = $${n_{N{a_2}C{O_3}}}$$
$$0.15 = {n_{N{a_2}C{O_3}}}$$
$${w_{N{a_2}C{O_3}}} = {{0.15 \times 106 \times {{10}^{ - 3}}} \over {0.4}} \times 100 \times 10$$
= 3.975%
$$ \simeq $$ 4%
$$NaOH + HCl\buildrel {} \over \longrightarrow NaCl + {H_2}O$$
$$nf = 1$$
$$NaC{O_3} + HCl\buildrel {} \over \longrightarrow NaHC{O_3}$$
$$nf = 1$$
Eq of HCl used $$ = {n_{NaOH}} \times 1 + {n_{N{a_2}C{O_3}}} \times 1$$
$$17.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaOH}} + {n_{N{a_2}C{O_3}}}$$
2nd end point :
$$NaHC{O_3} + HCl\buildrel {} \over \longrightarrow {H_2}C{O_3}$$
$$1.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaHC{O_3}}} \times 1 = {n_{NaHC{O_3}}}$$
0.15 mmol = $${n_{N{a_2}C{O_3}}}$$
$$0.15 = {n_{N{a_2}C{O_3}}}$$
$${w_{N{a_2}C{O_3}}} = {{0.15 \times 106 \times {{10}^{ - 3}}} \over {0.4}} \times 100 \times 10$$
= 3.975%
$$ \simeq $$ 4%
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