Back Titration

Back titration is an analytical chemistry technique used to determine the concentration of an analyte (a substance to be analyzed) by reacting it with a known excess amount of a standard reagent (titrant), and then titrating this excess with another standard solution. It is particularly useful when the analyte is either insoluble or reacts very slowly, making direct titration impractical.

Understanding Back Titration

The process of back titration involves several steps:

A known excess amount of a standard solution (titrant A) is added to the analyte.
The reaction between the analyte and titrant A goes to completion.
The remaining excess of titrant A is then titrated with another standard solution (titrant B).
The point at which the reaction between titrant A and titrant B is complete is known as the endpoint, which can be detected by a color change with an indicator or by some instrumental method.
The amount of titrant B used to reach the endpoint is measured.
The original concentration of the analyte can be calculated by considering the amount of titrant A that reacted with the analyte, which is the difference between the initial amount of titrant A and the excess amount titrated by titrant B.

The Process in Equations

Let's denote the following:

$C_A$: concentration of titrant A
$V_A$: volume of titrant A added to the analyte
$C_B$: concentration of titrant B
$V_B$: volume of titrant B used to titrate the excess of titrant A

The number of moles of titrant A added to the analyte is:

$$ n_A = C_A \cdot V_A $$

The number of moles of titrant B used to titrate the excess of titrant A is:

$$ n_B = C_B \cdot V_B $$

If the stoichiometry of the reaction between titrant A and titrant B is 1:1, then the moles of titrant A that reacted with the analyte is:

$$ n_{A,reacted} = n_A - n_B $$

From this, the amount of analyte can be calculated if the stoichiometry of the reaction between the analyte and titrant A is known.

Advantages and Disadvantages of Back Titration

Advantages	Disadvantages
Useful for substances that are insoluble or react slowly	More complex than direct titration
Can be faster than waiting for a slow reaction to complete	Requires accurate knowledge of the stoichiometry
Allows for the use of indicators that are not suitable for the analyte-titrant reaction	Potential for greater error due to multiple steps

Example of Back Titration

Let's consider an example where we want to determine the purity of a sample of calcium carbonate (CaCO₃), which is insoluble in water. We can use back titration to analyze it.

A known mass of CaCO₃ is treated with an excess of hydrochloric acid (HCl), which reacts to form calcium chloride (CaCl₂), water (H₂O), and carbon dioxide (CO₂):

$$ \text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) $$
The excess HCl is then titrated with a standard solution of sodium hydroxide (NaOH). The reaction is:

$$ \text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l) $$
If we added 50.00 mL of 0.1000 M HCl to the CaCO₃ and then used 10.00 mL of 0.1000 M NaOH to titrate the excess HCl, we can calculate the amount of HCl that reacted with CaCO₃:

$$ n_{\text{HCl, added}} = 0.1000 \, \text{M} \times 50.00 \, \text{mL} = 5.000 \, \text{mmol} $$ $$ n_{\text{NaOH}} = 0.1000 \, \text{M} \times 10.00 \, \text{mL} = 1.000 \, \text{mmol} $$

Since the stoichiometry of the reaction between HCl and NaOH is 1:1, the moles of HCl that reacted with CaCO₃ is:

$$ n_{\text{HCl, reacted}} = n_{\text{HCl, added}} - n_{\text{NaOH}} = 5.000 \, \text{mmol} - 1.000 \, \text{mmol} = 4.000 \, \text{mmol} $$
Now, we can calculate the amount of CaCO₃ in the sample:

Since 1 mole of CaCO₃ reacts with 2 moles of HCl, the moles of CaCO₃ is half the moles of HCl that reacted:

$$ n_{\text{CaCO}3} = \frac{n{\text{HCl, reacted}}}{2} = \frac{4.000 \, \text{mmol}}{2} = 2.000 \, \text{mmol} $$

If the mass of the CaCO₃ sample was 0.200 g, the purity of the sample can be calculated as:

$$ \text{Purity} = \frac{n_{\text{CaCO}_3} \times \text{molar mass of CaCO}_3}{\text{mass of sample}} \times 100\% $$

$$ \text{Purity} = \frac{2.000 \, \text{mmol} \times 100.09 \, \text{g/mol}}{0.200 \, \text{g}} \times 100\% \approx 100.09\% $$

(Note: The purity can be over 100% due to experimental errors or impurities that also react with HCl.)

Back titration is a versatile technique that can be adapted to various scenarios in analytical chemistry, especially when direct titration is not feasible. Understanding the principles and calculations involved is essential for accurate determination of analyte concentrations.