Solution X contains Na_2CO_3 and NaHCO_3. 20 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For titration of $Na_2CO_3$ and $NaHCO_3$ with $HCl$ using phenolphthalein,the reaction is: $Na_2CO_3 + HCl \longrightarrow NaHCO_3 + NaCl$. At th

Vedclass · Accepted Answer

$0.1, 0.1$

Vedclass · Answer

(B) For titration of $Na_2CO_3$ and $NaHCO_3$ with $HCl$ using phenolphthalein,the reaction is: $Na_2CO_3 + HCl \longrightarrow NaHCO_3 + NaCl$. At the phenolphthalein end point,$20 \ mL$ of $0.1 \ M$ $HCl$ is consumed,which corresponds to half the $Na_2CO_3$. $Moles \ of \ Na_2CO_3 = Moles \ of \ HCl = 20 \ mL 	imes 0.1 \ M = 2 \ mmol$. Concentration of $Na_2CO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$. Using methyl orange,both $Na_2CO_3$ and $NaHCO_3$ are neutralized. Total $HCl$ consumed = $60 \ mL 	imes 0.1 \ M = 6 \ mmol$. $HCl$ consumed by $Na_2CO_3$ (complete) = $2 	imes (2 \ mmol) = 4 \ mmol$. $HCl$ consumed by $NaHCO_3$ = $6 \ mmol - 4 \ mmol = 2 \ mmol$. Concentration of $NaHCO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$.

Similar Questions

The position vectors of $P$ and $Q$ are respectively $\overrightarrow{a}$ and $\overrightarrow{b}$. If $R$ is a point such that $\overrightarrow{PR}=5 \overrightarrow{PQ}$,then the position vector of $R$ is

In the following circuit,find $I_1$ and $I_2$.

In the given branch $AB$ of a circuit, a current $I = (10t + 5) \, A$ is flowing, where $t$ is time in seconds. At $t = 0$, the potential difference between points $A$ and $B$ $(V_A - V_B)$ is......$V$.

If the distance $s$ travelled by a particle in time $t$ is given by $s=t^2-2t+5$,then its acceleration is

The $pH$ of $0.05 \ M$ acetic acid is $(K_a = 2 \times 10^{-5})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module