# 0.8 Back titration  (Page 2/2)

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## Back titration and dilution

While dealing dilution in titration calculation, we need only to remember that dilution does not change geq or meq of a solution.

Problem : 2.75 gram of a sample of dolomite containing $CaC{O}_{3}$ and $MgC{O}_{3}$ is dissolved in 80 ml 1N HCl solution. The solution is then diluted to 250 ml. 25 ml of this solution requires 20 ml of 0.1 N NaOH solution for complete neutralization. Calculate percentage composition of the sample.

Solution : We shall first determine the amount of HCl required to react with the sample by using back titration data. Here, we make use the fact that dilution does not change meq of a solution.

$⇒\text{meq of 25 ml of diluted excess HCl solution}=\text{meq of NaOH solution}=0.1X20=2$

$⇒\text{meq of 250 ml of diluted excess HCl solution}=20$

$⇒\text{total meq of HCl}=NV=80X1=80$

$\text{meq of HCl used to react with dolomite}=80-20=60$

Mass of HCl used in reaction with dolomite is calculated using expression of meq :

$meq=\frac{g}{E}X1000=\frac{xg}{{M}_{O}}X1000$

$⇒g=\frac{meqX{M}_{O}}{1000x}=\frac{60X36.5}{1000X1}=2.19\phantom{\rule{1em}{0ex}}gm$

In order to find the composition, we apply mole concept. The reactions involved are :

$CaC{O}_{3}+2HCl\to CaC{l}_{2}+C{O}_{2}+{H}_{2}O$

$MgC{O}_{3}+2HCl\to MgC{l}_{2}+C{O}_{2}+{H}_{2}O$

Let mass of $CaC{O}_{3}$ in the sample is x gm. Then mass of $MgC{O}_{3}$ is 2.75-x gm.

$\frac{x}{{M}_{CaC{O}_{3}}}\phantom{\rule{1em}{0ex}}\text{moles of}\phantom{\rule{1em}{0ex}}CaC{O}_{3}=\frac{x}{100}\phantom{\rule{1em}{0ex}}\text{moles of}\phantom{\rule{1em}{0ex}}CaC{O}_{3}=\frac{2x}{100}\phantom{\rule{1em}{0ex}}\text{moles of HCl}$

$⇒\text{mass of HCl required for}\phantom{\rule{1em}{0ex}}CaC{O}_{3}=\frac{2xX{M}_{HCl}}{100}=\frac{2xX36.5}{100}=\frac{73x}{100}=0.73x\phantom{\rule{1em}{0ex}}gm$

Similarly for $MgC{O}_{3}$ ,

$⇒\text{mass of HCl required for}\phantom{\rule{1em}{0ex}}MgC{O}_{3}=\frac{2\left(2.75-x\right){M}_{HCl}}{84}=\frac{2X\left(2.75-x\right)36.5}{84}=\frac{73\left(2.75-x\right)}{84}=2.39-0.87x\phantom{\rule{1em}{0ex}}gm$

$⇒\text{total HCl required}=0.73x+2.39-0.87x=2.19$

$⇒0.14x=0.2$

$x=1.43\phantom{\rule{1em}{0ex}}gm$

$⇒\text{% mass of}\phantom{\rule{1em}{0ex}}CaC{O}_{3}=\frac{1.43}{2.75}X100=52$

$⇒\text{% mass of}\phantom{\rule{1em}{0ex}}MgC{O}_{3}=100-52=48$

## Back titration and decomposition

Back titration is analyzed using geq or meq concept. On the other hand, decomposition is analyzed using mole concept. We combine two concepts by converting final geq or meq data of back titration into either mass or moles.

Problem : 4.08 gm of a mixture of BaO and an unknown carbonate $MC{O}_{3}$ is heated strongly. The residue weighs 3.64 gram. The residue is then dissolved in 100 ml of 1 N HCl. The excess acid requires 16 ml of 2.5 NaOH for complete neutralization. Identify the metal M, if atomic weight of Ba is 138.

Solution : We need to determine atomic weight of M to identify it. From the question, we see that first part involves decomposition, which can be analyzed using mole concept. Note that difference of mass of mixture and residue is mass of $C{O}_{2}$ . On the other hand, analysis of back titration gives us meq of HCl required for neutralization of residue. To combine meq data with decomposition data, we shall convert required meq in milli-moles.

$⇒\text{total meq of HCl}=NV=1X100=100$

$⇒\text{meq of excess HCl}=NV=2.5X16=40$

$⇒\text{meq of HCl used for neutralization of residue}=100-40=60$

$⇒\text{milli-moles of HCl for neutralization of residue}=xXmeq=1X60=60$

Now, we analyze the data which is given for decomposition of mixture. On heating, BaO is not decoposed. Only $MC{O}_{3}$ is decomposed as :

$MC{O}_{3}=MO+C{O}_{2}$

Carbon dioxide is released and is not part of residue. The amount of $C{O}_{2}$ is equal to difference of mass of mixture and residue. Hence, moles of $C{O}_{2}$ released are :

$⇒{n}_{C{O}_{2}}=\frac{4.08-3.64}{\left(12+2X16\right)}=\frac{0.44}{44}=0.01\phantom{\rule{1em}{0ex}}mole$

Applying mole concept, moles of MO in the residue is 0.01 mole i.e. 10 milli-moles. Now, MO reacts with HCl as :

$MO+2HCl\to MC{l}_{2}+{H}_{2}O$

Moles of HCl required for MO is 2X0.01 moles = 20 milli-moles of HCl. But total milli-moles that reacted with residue is 60 milli-moles. Hence, milli-moles of HCl that reacted with BaO is 60-20 = 40 milli-moles of HCl. Applying mole concept,

$BaO+2HCl\to BaC{l}_{2}+{H}_{2}O$

The milli-moles of BaO is half of that of HCl. Hence, millimoles of BaO in the mixture is 20 milli-moles.

$⇒\text{amount of BaO in the mixture}=\frac{\text{milli-moles}X{M}_{BaO}}{1000}=\frac{20X\left(138+16\right)}{1000}=3.08\phantom{\rule{1em}{0ex}}gm$

$⇒\text{amount of MO in the residue}=3.64-3.08=0.56\phantom{\rule{1em}{0ex}}gm$

Using formula :

$⇒g=\frac{\text{milli-moles}X{M}_{MO}}{1000}=\frac{10X{M}_{MO}}{1000}$

$⇒{M}_{MO}=100g=100X0.56=56$

Clearly, atomic weight of M is 56-16 = 40. The element, therefore, is calcium.

## Back titration and multiple neutralizations

The concentration of titrant used to determine the excess reactant is not directly given. We make use of subsequent neutralization data to ultimately determine the concentration of titrant.

Problem : A sample of $MnS{O}_{4,}4{H}_{2}O$ (molecular wt : 223) is heated in air. The residue is ${\mathrm{Mn}}_{3}{O}_{4}$ , whose valency and molecular weight are 2 and 229 respectively. The residue is dissolved in 100 ml of 0.1 N $FeS{O}_{4}$ solution containing dilute sulphuric acid. The resulting solution reacts completely with 50 ml of $KMn{O}_{4}$ solution. 25 ml of the $KMn{O}_{4}$ solution requires 30 ml of 0.1N $FeS{O}_{4}$ solution for complete oxidation. Calculate the weight of $MnS{O}_{4,}4{H}_{2}O$ in the given sample.

Solution : The excess of $FeS{O}_{4}$ completely reacts with 50 ml of $KMn{O}_{4}$ solution of unknown concentration. We need to know this concentration. From the data of complete oxidation, however, we can know concentration. Let 1 and 2 subscripts denote $KMn{O}_{4}$ and $FeS{O}_{4}$ , then

${N}_{1}{V}_{1}={N}_{2}{V}_{2}$

$⇒{N}_{1}X25=0.1X30$

$⇒{N}_{1}=\frac{3}{25}=0.12N$

Now, considering back titration,

$⇒\text{total meq of}\phantom{\rule{1em}{0ex}}FeS{O}_{4}\phantom{\rule{1em}{0ex}}\text{solution}=0.1X100=10$

$⇒\text{meq of excess}\phantom{\rule{1em}{0ex}}FeS{O}_{4}\phantom{\rule{1em}{0ex}}\text{solution}=0.12X50=6$

$⇒\text{meq of}\phantom{\rule{1em}{0ex}}FeS{O}_{4}\phantom{\rule{1em}{0ex}}\text{solution used for residue}=10-6=4$

$⇒\text{meq of}\phantom{\rule{1em}{0ex}}M{n}_{2}{O}_{3}\phantom{\rule{1em}{0ex}}\text{in the residue}=4$

Mass of ${\mathrm{Mn}}_{3}{O}_{4}$ is determined using meq expression :

$meq=\frac{g}{E}X1000=\frac{xg}{{M}_{O}}X1000$

$⇒g=\frac{meqX{M}_{O}}{1000x}=\frac{4X229}{1000X2}=0.458\phantom{\rule{1em}{0ex}}gm$

The decomposition of $MnS{O}_{4,}4{H}_{2}O$ takes place as :

$3MnS{O}_{4,}10{H}_{2}O\to M{n}_{3}{O}_{4}+3S{O}_{2}+10{H}_{2}O$

Applying mole concept, the amount of $MnS{O}_{4,}4{H}_{2}O$ is :

$⇒\text{Mass of}\phantom{\rule{1em}{0ex}}MnS{O}_{4,}4{H}_{2}O=\frac{3X223}{229}X0.458=1.338\phantom{\rule{1em}{0ex}}gm$

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