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  3. Chemistry
  4. 40 mL of 0.05 M solution of sodium sesquicarbonate ( Na 2 CO 3 NaHCO 3 ; 2 H 2 O ) is titrated against 0.05 M HCl. x mL of HCl is used when phenolphthalein is the indicator and y mL of HCl is used when methyl orange is the indicator in two separate titrations. Hence e (y-x) is

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Q. $40 \,mL$ of $0.05 \,M$ solution of sodium sesquicarbonate $\left( Na _2 CO _3 NaHCO _3 ; 2 H _2 O \right)$ is titrated against $0.05 \,M \, HCl. x \,mL$ of $HCl$ is used when phenolphthalein is the indicator and $y \,mL$ of $HCl$ is used when methyl orange is the indicator in two separate titrations. Hence $e (y-x)$ is

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Solution:

$0.05\, M \,Na _2 CO _3=1 \,N\, Na _2 CO _3$
$0.05 \,M \,NaHCO _3=0.05 \,N \,NaHCO$
$40 L$ of $0.1 Na _2 CO _3 \equiv 40 \,mL$ of $0.1 \,N \,HCl$
(For complete reaction) $=80 \,mL$ of $0.05 \,M \,HCl$
For $50 \%$ reaction $=40 \,mL$ of $0.05 \,M \,HCl$
With phenolphthalein $=x=40\, mL$
$40\, mL$ of $0.05 \,NaHCO _3=40\, mL$ of $0.05\, M \,HCl$
With methyl orange, $y=80+40=120 \,mL$
$\therefore(y-x)=80\, mL$