
A mixture of $HCl$ and ${H_3}P{O_4}$ is titrated with $1.0M\,NaOH$ .The first end point (Methyl red) occurs at $35.0mL$ and second endpoints (Bromothymolblue) occurs at $50.0\,mL$ .($15.0mL$ after the first end point). The number of millimoles of \[HCl\]present in the solution is___?
Answer
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Hint:Acid‐base titration is one of the most common operations in analytical chemistry. A solution containing an unknown amount of ionizable hydrogen can be titrated with a solution of standard base until all the hydrogen ions have been consumed .
Complete answer:Titration of $HCl\,\& \,{H_3}P{O_4}$ Mixture Using Indicators and a pH Meter.
The amount of \[NaOH\] used by phosphoric acid. = $15mL$
Therefore amount of phosphoric acid present is $15mL \times 1M$=$15\,{\text{millimoles}}$
[$\because molarity = 1M$]
During the first end point, out of $35mL$ , $15mL$ are used for neutralization of phosphoric acid.
$\therefore $ The amount of $NaOH$ used for neutralization of \[HCL\] ,
$
{\text{HCl = 20 mL}} \\
{\Rightarrow {HCl = 20mL \times 1M}} \\
{\Rightarrow \text{HCl = 20 millimoles}} \\
$
Hence the number of millimoles of \[HCl \] present in the solution is 20millimoles.
Additional Information:Double titration:
The method involves two indicator phenolphthalein and methyl orange. This is a titration of specific compounds. Let us consider a solid mixture of \[NaOH,{\text{ }}N{a_2}C{O_3}\] and inert impurities weighing. You are asked to find out the % composition of the mixture. You are also given a reagent that can react with the sample, say, \[HCL\] along with its concentration (\[{M_1}\]).
We first dissolve the mixture in water to make a solution and then we add two indicators in it, namely phenolphthalein and methyl orange. Now, we titrate this solution with \[HCL\] .
\[NaOH\] is a strong base while \[N{a_2}C{O_3}\] is a weak base. So it is safe to assume that \[NaOH\] reacts with \[HCL\] first, completely and only then does \[N{a_2}C{O_3}\] react.
Note:A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution.Typically,the titrant(the known solution) is added from a burette to a known quantity of analyte (the unknown solution) until the reaction is complete.
Complete answer:Titration of $HCl\,\& \,{H_3}P{O_4}$ Mixture Using Indicators and a pH Meter.
The amount of \[NaOH\] used by phosphoric acid. = $15mL$
Therefore amount of phosphoric acid present is $15mL \times 1M$=$15\,{\text{millimoles}}$
[$\because molarity = 1M$]
During the first end point, out of $35mL$ , $15mL$ are used for neutralization of phosphoric acid.
$\therefore $ The amount of $NaOH$ used for neutralization of \[HCL\] ,
$
{\text{HCl = 20 mL}} \\
{\Rightarrow {HCl = 20mL \times 1M}} \\
{\Rightarrow \text{HCl = 20 millimoles}} \\
$
Hence the number of millimoles of \[HCl \] present in the solution is 20millimoles.
Additional Information:Double titration:
The method involves two indicator phenolphthalein and methyl orange. This is a titration of specific compounds. Let us consider a solid mixture of \[NaOH,{\text{ }}N{a_2}C{O_3}\] and inert impurities weighing. You are asked to find out the % composition of the mixture. You are also given a reagent that can react with the sample, say, \[HCL\] along with its concentration (\[{M_1}\]).
We first dissolve the mixture in water to make a solution and then we add two indicators in it, namely phenolphthalein and methyl orange. Now, we titrate this solution with \[HCL\] .
\[NaOH\] is a strong base while \[N{a_2}C{O_3}\] is a weak base. So it is safe to assume that \[NaOH\] reacts with \[HCL\] first, completely and only then does \[N{a_2}C{O_3}\] react.
Note:A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution.Typically,the titrant(the known solution) is added from a burette to a known quantity of analyte (the unknown solution) until the reaction is complete.
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