The purpose of double titration is to determine the percentage composition of an alkali mixture or an acid mixture. In the present case, we will find the percentage composition of an alkali mixture. Let us consider a solid mixture of `NaOH, Na_2CO_3` and some inert impurities, weighing `'w' g`. We are required to find the `%` composition of this alkali mixture. We are also given an acid reagent (`HCl`) of known concentration `M_1` that can react with the alkali sample.
We first dissolve this mixture in water to make an alkaline solution and then we add two indicators, (Indicators are substances that indicate colour change of solution when a reaction gets completed), namely phenolphthalein and methyl orange to the solution. Now, we titrate this alkaline solution with standard `HCl`.
`NaOH` is a strong base while `Na_2CO_3` is a weak base. So it is obvious that `NaOH` reacts first with `HCl` completely and `Na_2CO_3` reacts only after complete `NaOH` is neutralized.
`NaOH+HCl -> NaCl + H_2O`............(i)
Once `NaOH` has reacted completely, then `Na_2CO_3` starts reacting with `HCl` in two steps, shown as
`Na_2CO_3+HCl -> NaHCO_3+NaCl`.........(ii)
`NaHCO_3+HCl -> NaCl + CO_2 + H_2O`............(iii)
It is clear that when we add `HCl` to the alkaline solution, alkali is neutralized and the `pH` of the solution decreases. Initially the `pH` decrease would be rapid as strong base `(NaOH)` is neutralized completely. When `Na_2CO_3` is converted to `NaHCO_3` completely, the solution is still weakly basic due to the presence of `NaHCO_3` (which is weaker as compared to `Na_2CO_3`). At this point, phenolphthalein changes colour since it requires this weakly basic solution to show its colour change. When `HCl` is further added, the `pH` again decreases and when all the `NaHCO_3` reacts to form `NaCl, CO_2` and `H_2O` the solution becomes weakly acidic due to the presence of the weak acid `(H_2CO_3)`. At this point, methyl orange changes colour as it requires this weakly acidic solution to show its colour change.
Thus in general, phenolphthalein shows colour change when the solution contains weakly basic `NaHCO_3` along with other neutral substances while methyl orange shows colour change when solution contains weakly acidic `H_2CO_3` along with other neutral substances.
Let the volume of `HCl` used up for the first and the second reaction be `V _1` litre {this is the volume of `HCl` used from the beginning of the titration up to the point when phenolphthalein shows colour change) and the volume of `HCl` required for the third reaction be `V_2` litre (this is the volume of `HCl` used from the point where phenolphthalein had changed colour upto the point when methyl orange shows colour change). Then,
Moles of `HCl` consumed by `NaHCO_3` = Moles of `NaHCO_3` reacted= `M_1V_2`
Moles of `NaHCO_3` formed from `Na_2CO_3 = M_1 V_2`
Moles of `Na_2CO_3` in the mixture `= M_1 V_2`
Mass of `Na_2CO_3` in the mixture `= M_1 V_2 xx 106`
`%` of ` Na_2CO_3` in the mixture `=(M_1V_2 xx 106)/w xx 100`
Moles of `HCl` used in the reaction (i) and (ii) `= M _1 V_1`
Moles of `HCl` used in reaction (ii) `= M_1 V_2`
Moles of `HCl` used in reaction (i) `= (M_1V_1 - M_1V_2)`
Moles of `NaOH = (M_1 V_1 -M_1 V_2)`
Mass of `NaOH = (M_1V_1 - M_1V_2)xx40`
`%` of `NaOH` in the mixture `= ((M_1V_1 -M_1V_2)xx40)/w xx 100`
The purpose of double titration is to determine the percentage composition of an alkali mixture or an acid mixture. In the present case, we will find the percentage composition of an alkali mixture. Let us consider a solid mixture of `NaOH, Na_2CO_3` and some inert impurities, weighing `'w' g`. We are required to find the `%` composition of this alkali mixture. We are also given an acid reagent (`HCl`) of known concentration `M_1` that can react with the alkali sample.
We first dissolve this mixture in water to make an alkaline solution and then we add two indicators, (Indicators are substances that indicate colour change of solution when a reaction gets completed), namely phenolphthalein and methyl orange to the solution. Now, we titrate this alkaline solution with standard `HCl`.
`NaOH` is a strong base while `Na_2CO_3` is a weak base. So it is obvious that `NaOH` reacts first with `HCl` completely and `Na_2CO_3` reacts only after complete `NaOH` is neutralized.
`NaOH+HCl -> NaCl + H_2O`............(i)
Once `NaOH` has reacted completely, then `Na_2CO_3` starts reacting with `HCl` in two steps, shown as
`Na_2CO_3+HCl -> NaHCO_3+NaCl`.........(ii)
`NaHCO_3+HCl -> NaCl + CO_2 + H_2O`............(iii)
It is clear that when we add `HCl` to the alkaline solution, alkali is neutralized and the `pH` of the solution decreases. Initially the `pH` decrease would be rapid as strong base `(NaOH)` is neutralized completely. When `Na_2CO_3` is converted to `NaHCO_3` completely, the solution is still weakly basic due to the presence of `NaHCO_3` (which is weaker as compared to `Na_2CO_3`). At this point, phenolphthalein changes colour since it requires this weakly basic solution to show its colour change. When `HCl` is further added, the `pH` again decreases and when all the `NaHCO_3` reacts to form `NaCl, CO_2` and `H_2O` the solution becomes weakly acidic due to the presence of the weak acid `(H_2CO_3)`. At this point, methyl orange changes colour as it requires this weakly acidic solution to show its colour change.
Thus in general, phenolphthalein shows colour change when the solution contains weakly basic `NaHCO_3` along with other neutral substances while methyl orange shows colour change when solution contains weakly acidic `H_2CO_3` along with other neutral substances.
Let the volume of `HCl` used up for the first and the second reaction be `V _1` litre {this is the volume of `HCl` used from the beginning of the titration up to the point when phenolphthalein shows colour change) and the volume of `HCl` required for the third reaction be `V_2` litre (this is the volume of `HCl` used from the point where phenolphthalein had changed colour upto the point when methyl orange shows colour change). Then,
Moles of `HCl` consumed by `NaHCO_3` = Moles of `NaHCO_3` reacted= `M_1V_2`
Moles of `NaHCO_3` formed from `Na_2CO_3 = M_1 V_2`
Moles of `Na_2CO_3` in the mixture `= M_1 V_2`
Mass of `Na_2CO_3` in the mixture `= M_1 V_2 xx 106`
`%` of ` Na_2CO_3` in the mixture `=(M_1V_2 xx 106)/w xx 100`
Moles of `HCl` used in the reaction (i) and (ii) `= M _1 V_1`
Moles of `HCl` used in reaction (ii) `= M_1 V_2`
Moles of `HCl` used in reaction (i) `= (M_1V_1 - M_1V_2)`
Moles of `NaOH = (M_1 V_1 -M_1 V_2)`
Mass of `NaOH = (M_1V_1 - M_1V_2)xx40`
`%` of `NaOH` in the mixture `= ((M_1V_1 -M_1V_2)xx40)/w xx 100`