(B) Let $x$ be the milliequivalents of $NaOH$ and $y$ be the milliequivalents of $Na_{2}CO_{3}$.At the first end point (phenolphthalein indicator),$Na

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(B) Let $x$ be the milliequivalents of $NaOH$ and $y$ be the milliequivalents of $Na_{2}CO_{3}$.At the first end point (phenolphthalein indicator),$NaOH$ is neutralized and $Na_{2}CO_{3}$ is converted to $NaHCO_{3}$:$x + y = \frac{1}{10} \times 17.5 = 1.75 \text{ meq} \dots (1)$At the second end point (methyl orange indicator),$NaHCO_{3}$ is neutralized:$y = \frac{1}{10} \times 1.5 = 0.15 \text{ meq} \dots (2)$Mass of $Na_{2}CO_{3} = \text{milliequivalents} \times \text{equivalent weight} \times 10^{-3}$$= 0.15 \times 53 \times 10^{-3} = 0.00795 \ g$Weight percentage of $Na_{2}CO_{3} = \frac{0.00795}{0.4} \times 100 = 1.9875 \%$Rounding off to the nearest integer,we get $2 \%$.

Class 12 Chemistry Principles Related to Practical Chemistry Titrations

Medium

$0.4 \ g$ mixture of $NaOH, Na_{2}CO_{3}$ and some inert impurities was first titrated with $\frac{N}{10} \ HCl$ 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 $Na_{2}CO_{3}$ in the mixture is .......... (Rounded-off to the nearest integer)

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$2$
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$4$
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$8 \ mL$ of $\frac{N}{10} \ HCl$ are required to neutralize $20 \ mL$ of $Na_2CO_3$ solution in water. The normality of the $Na_2CO_3$ solution is: (in $N$)

Medium

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Two acids $A$ and $B$ are titrated separately. $25 \ mL$ of $0.5 \ M \ Na_2CO_3$ solution requires $10 \ mL$ of $A$ and $40 \ mL$ of $B$ for complete neutralization. The volumes (in $L$) of $A$ and $B$ required to produce $1 \ L$ of $1 \ N$ acid solution respectively are:

MediumAP EAMCET 2025

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$A$ $100 \ mL$ solution of $0.1 \ N \ HCl$ was titrated with $0.2 \ N \ NaOH$ solution. The titration was discontinued after adding $30 \ mL$ of $NaOH$ solution. The remaining titration was completed by adding $0.25 \ N \ KOH$ solution. The volume of $KOH$ required for completing the titration is $..... \ mL$

Medium

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$10 \, mL$ of concentrated $HCl$ is diluted to $1 \, L$. If $20 \, mL$ of this diluted solution is completely neutralized by $25 \, mL$ of $0.1 \, N$ sodium hydroxide solution,then the normality of the original concentrated hydrochloric acid is:

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$A$ 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 $0.2 \ N$ $HCl$ is required. The amount of $NaOH$ present in the solution is ................. $g$ $(NaOH = 40, Na_2CO_3 = 106)$.

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$8 \ mL$ of $\frac{N}{10} \ HCl$ are required to neutralize $20 \ mL$ of $Na_2CO_3$ solution in water. The normality of the $Na_2CO_3$ solution is: (in $N$)

Two acids $A$ and $B$ are titrated separately. $25 \ mL$ of $0.5 \ M \ Na_2CO_3$ solution requires $10 \ mL$ of $A$ and $40 \ mL$ of $B$ for complete neutralization. The volumes (in $L$) of $A$ and $B$ required to produce $1 \ L$ of $1 \ N$ acid solution respectively are:

$10 \, mL$ of concentrated $HCl$ is diluted to $1 \, L$. If $20 \, mL$ of this diluted solution is completely neutralized by $25 \, mL$ of $0.1 \, N$ sodium hydroxide solution,then the normality of the original concentrated hydrochloric acid is:

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