# 0.7 Titration  (Page 4/4)

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## Oxalic acid sodium hydroxide

The reaction involved here is also completed in two stages. Oxalic acid has two furnishable hydrogen ions. Two hydrogen ions are replaced one after other in two stages :

${|}_{COOH}^{COOH}+NaOH\to {|}_{COOH}^{COONa}+{H}_{2}O$

${|}_{COOH}^{COONa}+NaOH\to {|}_{COONa}^{COONa}+{H}_{2}O$

Considering that sodium hydrooxide is titrant, we see that one of two hydrogens of oxalic acid is replaced by sodium in the first stage. It can be seen that methyl orange can detect the rise of pH value at equivalence at the end of first stage reaction. On the other hand, completion of second stage reaction is detected by phenolphthalein. Following earlier logic, we conclude :

$\text{geq of NaOH}=\text{geq of oxalic acid (indicator : phenolphthalein)}$

$\text{geq of NaOH}=\frac{1}{2}X\text{geq of oxalic acid (indicator : methyl orange)}$

## Titrating basic mixtures

We can titrate mixtures basic compounds with strong acid like HCl. Consider the combination of bases :

1: $NaOH\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

2: $NaHC{O}_{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

## Mixture of sodium hydro-oxide and sodium carbonate

The analysis depends on the particular indicator used. If we use methyl orange, then we know that it can detect completion of reaction of HCl with either of two basic compounds. Hence,

$\text{volume of HCl used}=\text{volume acid required for neutralization of NaOH and}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

This means that :

$\text{geq of HCl}=\text{geq of NaOH}+\text{geq of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}\text{(indicator : methyl orange)}$

If we use phenolphthalein, then we know that it can detect completion of reaction of HCl with NaOH. However, it can detect only half of the completion of reaction of HCl with sodium carbonate.

$\text{volume of HCl used}=\text{volume acid required for neutralization of NaOH}+\frac{1}{2}\text{volume acid required for neutralization of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

This means that :

$\text{geq of HCl}=\text{geq of NaOH}+\frac{1}{2}X\text{geq of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}\text{(indicator : phenolphthalein)}$

Q. A volume of 25 ml of 0.2 N HCl is titrated to completely neutralize 25 ml mixture of NaOH and $N{a}_{2}C{O}_{3}$ , using phenolphthalein as indicator. On the other hand, 60 ml of 0.1 N HCl is required to neutralize the equal volume of mixture, using methyl orange as indicator. Find the strengths of NaOH and $N{a}_{2}C{O}_{3}$ in the mixture.

Answer : Strength is expressed in terms of gm/litre. Thus, we need to know the mass of each component in the mixture. First titration uses phenolphthalein, which detects completion of reaction with NaOH and half of reaction with $N{a}_{2}C{O}_{3}$ . Hence,

$\text{meq of acid}=\text{meq of NaOH}+\frac{1}{2}X\text{meq of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

Clearly, it is helpful to assume unknowns in terms of milli-equivalents (meq) instead of mass. Once, meq are calculated, we convert the same finally in mass terms and strength of solution as required. Let x and y be the meq of NaOH and $N{a}_{2}C{O}_{3}$ in the mixture.

Putting values in the equation, we have :

$⇒{N}_{a}{V}_{a}=x+0.5y$

$⇒x+0.5y=0.2X25=5$

Second titration uses methyl orange, which detects completion of reaction with both NaOH and $N{a}_{2}C{O}_{3}$ . Hence,

$\text{meq of acid}=\text{meq of NaOH}+\text{meq of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

$⇒{N}_{a}{V}_{a}=x+y$

$⇒x+y=0.1X60=6$

Solving two equations (subtracting first from second equation), we have

$⇒0.5y=1$

$⇒y=2$

and

$⇒x=6-2=4$

Now, we need to convert meq into strength of solution. For that, we first convert meq to mass of solute (B) in gram.

$\text{meq}=\text{valence factor}X\text{milli-moles}$

$⇒\text{meq}=\frac{\text{valence factor}X{g}_{B}X1000}{{M}_{B}}$

$⇒{g}_{B}=\frac{{M}_{B}X\text{meq}}{\text{valence factor}X1000}$

Thus, strength of component in 25 ml solution is :

$⇒S=\frac{{g}_{B}X1000}{25}=\frac{{M}_{B}X\text{meq}X1000}{25X\text{valence factor}X1000}=\frac{{M}_{B}X\text{meq}}{25X\text{valence factor}}$

Putting values, strength of NaOH is :

$⇒S=\frac{{M}_{B}X\text{meq}}{25X\text{valence factor}}=\frac{40X4}{25X1}=6.4\phantom{\rule{1em}{0ex}}gm/l$

Similarly, strength of $N{a}_{2}C{O}_{3}$ is :

$⇒S=\frac{{M}_{B}X\text{meq}}{25X\text{valence factor}}=\frac{106X4}{25X2}=4.24\phantom{\rule{1em}{0ex}}gm/l$

## Mixture of sodium bicarbonate and sodium carbonate

The analysis again depends on the particular indicator used. If we use methyl orange, then we know that it can detect completion of reaction of HCl with either of two basic compounds. Hence,

$\text{volume of HCl used}=\text{volume acid required for neutralization of}\phantom{\rule{1em}{0ex}}NaHC{O}_{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

This means that :

$\text{geq of HCl}=\text{geq of}\phantom{\rule{1em}{0ex}}NaHC{O}_{3}+\text{geq of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}\text{(indicator : methylorange)}$

If we use phenolphthalein, then we know that it can detect one half of completion of reaction of HCl with $N{a}_{2}C{O}_{3}$ . However, it can not detect completion of reaction of HCl with $NaHC{O}_{3}$ .

$\text{volume of HCl used}=\frac{1}{2}X\text{volume of acid required for neutralization of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

This means that :

$\text{geq of HCl}=\frac{1}{2}X\text{geq of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}\text{(indicator : phenolphthalein)}$

Q. A volume of 40 ml of 0.1 N HCl is titrated to completely neutralize 25 ml of basic solution containing $N{a}_{2}C{O}_{3}$ and $NaHC{O}_{3}$ , using methyl orange as indicator. On the other hand, 15 ml of 0.1 N HCl is required to neutralize equal volume of basic solution, using phenolphthalein as indicator. Find the strength of $N{a}_{2}C{O}_{3}$ and $NaHC{O}_{3}$ in the solution.

Answer : First titration uses methyl orange, which detects completion of reaction with both $N{a}_{2}C{O}_{3}$ and $NaHC{O}_{3}$ . Hence,

$\text{meq of acid}=\text{meq of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}+\text{meq of}\phantom{\rule{1em}{0ex}}NaHC{O}_{3}$

Let x and y be the meq of $N{a}_{2}C{O}_{3}$ and $NaHC{O}_{3}$ in the mixture. Putting values in the equation, we have :

$⇒{N}_{a}{V}_{a}=x+y$

$⇒x+y=0.1X40=4$

Second titration uses phenolphthalein, which detects completion of half reaction with $N{a}_{2}C{O}_{3}$ . It does not detect completion of reaction with respect to NaHCO3. Hence,

$\text{meq of acid}=\frac{1}{2}X\text{meq of}\phantom{\rule{1em}{0ex}}N{a}_{2}C{O}_{3}$

$⇒{N}_{a}{V}_{a}=\frac{1}{2}x$

$⇒\frac{1}{2}x=0.1X15=1.5$

$⇒x=3$

Putting in the first equation, we have

$⇒y=4-3=1$

Now, we need to convert meq into strength of solution. Using formula as derived earlier,

$⇒S=\frac{{M}_{B}X\text{meq}}{{V}_{CC}X\text{valence factor}}$

Putting values, strength of $N{a}_{2}C{O}_{3}$ is :

$⇒S=\frac{{M}_{B}X\text{meq}}{{V}_{CC}X\text{valence factor}}=\frac{106X3}{25X2}=6.36\phantom{\rule{1em}{0ex}}gm/l$

Similarly, strength of $NaHC{O}_{3}$ is :

$⇒S=\frac{{M}_{B}X\text{meq}}{{V}_{CC}X\text{valence factor}}=\frac{84X1}{25{X}_{1}}=3.36\phantom{\rule{1em}{0ex}}gm/l$

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