Quantitative Analysis of Ions Revision Notes for HSC SSCE Chemistry

Quantitative Analysis of Ions

Introduction

In previous sections, we learned how to identify specific ions in solution. However, chemists often need to determine not just what ions are present, but how much of each ion exists. This section explores quantitative analysis techniques that measure the concentration of particular ions in solution.

Two main approaches are used for quantitative ion analysis:

Precipitation titrations: Use volume measurements and precipitation reactions to calculate ion concentrations
Gravimetric analysis: Use mass measurements of precipitates to determine ion quantities

Both methods rely on precipitation reactions where ions form insoluble compounds that can be separated and measured.

Precipitation Titrations

Precipitation titrations determine ion concentrations by using precipitation reactions in a titration setup. The technique and equipment are similar to acid-base and redox titrations you've studied before. Most precipitation titrations for anions use silver nitrate solution, which provides $\text{Ag}^+$ ions that precipitate with the anion being analyzed.

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A key challenge in precipitation titrations is identifying the end point. Unlike acid-base titrations where colour changes are clear, it's difficult to see when precipitation stops occurring. Three different methods have been developed to solve this problem.

Mohr's method

Mohr's method is a direct titration used to determine chloride, bromide, and cyanide ion concentrations. This method uses chromate ions ( $\text{CrO}_4^{2-}$ ) as the indicator.

How it works:

The sample solution containing the halide ion is titrated with a standardized silver nitrate solution. The silver ions first precipitate with the halide ions. Once all the halide ions have reacted, the next drop of silver nitrate reacts with the chromate indicator to form a red-brown precipitate of silver chromate ( $\text{Ag}_2\text{CrO}_4$ ). This permanent colour change signals the end point.

Chemical reactions:

First reaction (main titration): $\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$

End point reaction: $2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq) \rightarrow \text{Ag}_2\text{CrO}_4(s)$

Blank titration requirement:

The end point is detected slightly past the equivalence point because extra titrant is needed to produce the visible colour change. To correct for this error, a blank titration is performed. In the blank titration, the silver nitrate is titrated against a solution containing only the indicator (no halide), often with calcium carbonate added to mimic the white precipitate. The volume from the blank titration is subtracted from the sample titration volume before calculations.

chatImportant

Critical Conditions for Mohr's Method:

pH must be between 6 and 9: At higher pH, silver hydroxide precipitates. At lower pH, chromate ions become protonated forming chromic acid, making the concentration too low for precipitation at the equivalence point.
Constant indicator amount: The same amount of chromate indicator must be used each time. Too little indicator requires more $\text{Ag}^+$ to produce colour; too much makes the colour change hard to see.
Not suitable for iodide: This method doesn't work well for iodide ions.

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Worked Example: Mohr's Method Calculation

A solution of unknown chloride concentration was analyzed. A $25 \text{ mL}$ sample required $33.70 \text{ mL}$ of $1.00 \text{ mol L}^{-1}$ silver nitrate to reach the end point. A blank titration required $0.50 \text{ mL}$ to achieve colour change.

Step 1: Correct for blank titration

$\text{Volume to end point} = 33.70 - 0.50 = 33.20 \text{ mL}$

Step 2: Calculate moles of silver ions used

$n = cV = 1.00 \times 0.03320 = 0.0332 \text{ mol}$

Step 3: Use stoichiometry

$\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$

Since the mole ratio is 1:1:

$n(\text{Cl}^-) = 0.0332 \text{ mol}$

Step 4: Calculate concentration

$c = \frac{n}{V} = \frac{0.0332}{0.025} = 1.33 \text{ mol L}^{-1}$

The chloride concentration is 1.33 mol L⁻¹.

Volhard's method

Volhard's method uses a back titration approach in acidic solution. It can determine many different anions including halides ( $\text{Cl}^-$ , $\text{Br}^-$ , $\text{I}^-$ ), phosphate, chromate, sulfide, carbonate, and cyanide. It's also effective for direct titration of $\text{Ag}^+$ ions.

How it works:

A known excess amount of silver nitrate is added to the sample, precipitating all the ions being investigated. The excess (unreacted) silver nitrate is then titrated with a standard potassium thiocyanate solution. Iron(III) ions ( $\text{Fe}^{3+}$ ) serve as the indicator. When all excess silver ions have reacted with thiocyanate, the next drop of thiocyanate reacts with the iron(III) ions to produce a distinctive blood-red complex ( $\text{Fe(SCN)}^{2+}$ ), indicating the end point.

Chemical reactions:

First reaction (precipitation of analyte): $\text{Ag}^+(aq) + \text{X}^-(aq) \rightarrow \text{AgX}(s)$

Second reaction (back titration of excess $\text{Ag}^+$ ): $\text{Ag}^+(aq) + \text{SCN}^-(aq) \rightarrow \text{AgSCN}(s)$

End point indicator reaction: $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightarrow \text{Fe(SCN)}^{2+}(aq)$ (blood-red colour)

chatImportant

Solubility Considerations - Critical Source of Error:

An important source of error occurs when the initial precipitate (e.g., $\text{AgCl}$ ) is more soluble than the end point precipitate ( $\text{AgSCN}$ ). In this case, the initial precipitate can dissolve as thiocyanate is added:

$\text{AgCl}(s) + \text{SCN}^-(aq) \rightarrow \text{AgSCN}(s) + \text{Cl}^-(aq)$

This gives falsely high measurements of excess silver, leading to an incorrectly low calculated concentration of the analyte ion.

To determine if this is a problem, compare the solubility product constants ( $K_{sp}$ ) of the precipitates:

If $K_{sp}$ of initial precipitate > $K_{sp}$ of AgSCN: The precipitate will dissolve and must be filtered out before titration
If $K_{sp}$ of initial precipitate < $K_{sp}$ of AgSCN: The precipitate won't dissolve and doesn't need removal

Ksp values for common silver precipitates:

Precipitate	$K_{sp}$
AgSCN (end point)	$1.03 \times 10^{-12}$
AgBr	$5.35 \times 10^{-13}$
$\text{Ag}_2\text{CO}_3$	$8.46 \times 10^{-12}$
AgCl	$1.77 \times 10^{-10}$
$\text{Ag}_2\text{CrO}_4$	$1.12 \times 10^{-12}$
AgCN	$5.97 \times 10^{-17}$
AgI	$8.52 \times 10^{-17}$
$\text{Ag}_3\text{PO}_4$	$8.89 \times 10^{-17}$
$\text{Ag}_2\text{SO}_4$	$1.20 \times 10^{-5}$
$\text{Ag}_2\text{S}$	$8 \times 10^{-51}$

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From this table, we can see that $\text{AgCl}$ is more soluble than $\text{AgSCN}$ , so it must be filtered before titration. However, $\text{AgBr}$ , $\text{AgI}$ , and most other silver salts are less soluble than $\text{AgSCN}$ and don't need to be removed.

Important conditions:

Low pH required: Prevents precipitation of $\text{Fe(OH)}_3$ from the iron(III) indicator
Possible transfer errors: If precipitate must be filtered before titration, losses can occur during transfer

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Worked Example: Volhard's Method Calculation

A student analyzed an iodide solution by adding $50.0 \text{ mL}$ of $0.890 \text{ mol L}^{-1}$ silver nitrate to $25 \text{ mL}$ of the sample. The excess silver nitrate was titrated with $42.3 \text{ mL}$ of $0.881 \text{ mol L}^{-1}$ sodium thiocyanate solution, using $\text{Fe}^{3+}$ as indicator.

Step 1: Write balanced equations

$\text{Ag}^+(aq) + \text{I}^-(aq) \rightarrow \text{AgI}(s)$

$\text{Ag}^+(aq) + \text{SCN}^-(aq) \rightarrow \text{AgSCN}(s)$

Step 2: Calculate original moles of silver added

$n(\text{AgNO}_3) = 0.890 \times 0.0500 = 0.0445 \text{ mol}$

Step 3: Calculate moles of excess silver (that reacted with thiocyanate)

$n(\text{SCN}^-) = 0.881 \times 0.0423 = 0.0373 \text{ mol}$

$n(\text{Ag}^+ \text{ excess}) = 0.0373 \text{ mol}$

Step 4: Calculate moles of silver that reacted with iodide

$n(\text{Ag}^+ \text{ reacted with I}^-) = 0.0445 - 0.0373 = 0.00720 \text{ mol}$

Step 5: Calculate iodide concentration

$n(\text{I}^-) = 0.00720 \text{ mol}$

$c = \frac{n}{V} = \frac{0.00720}{0.025} = 0.288 \text{ mol L}^{-1}$

The iodide ion concentration is 0.288 mol L⁻¹.

Fajan's method

Fajan's method involves a direct titration where the end point is detected by a colour change using an absorption indicator. Unlike the chemical indicators in Mohr's and Volhard's methods, absorption indicators work by physically adsorbing onto the surface of the precipitate at the end point and changing colour.

How it works:

The indicator molecules absorb onto the precipitate surface at the equivalence point, causing a visible colour change. The specific indicator used depends on which halide ion is being analyzed.

Common indicators for Fajan's method:

Indicator	Colour change	Use
Fluorescein	Yellow-green to pink	All halides, pH 7-10
Eosin	Pink to red-violet	Sample must not contain $\text{Cl}^-$ , pH > 1
Dichlorofluorescein	Orange to blue	$\text{I}^-$ only, pH 4-7
Diiododimethylfluorescein	Yellow-green to red	$\text{Cl}^-$ and $\text{Br}^-$ , pH 4-7

chatImportant

Important Conditions for Fajan's Method:

pH must be strictly controlled: Each indicator works only at specific pH ranges
Not suitable for low concentrations: Insufficient precipitate means not enough surface area for the indicator to produce a visible colour change
Cannot have high levels of other ions: Non-reacting ions may cause the indicator to coagulate with them instead of adsorbing onto the target precipitate

Comparison of the three methods

Method	Titration type	Indicator and change	Species analyzed	Limitations
Mohr	Direct titration	Yellow chromate to red-brown $\text{Ag}_2\text{CrO}_4(s)$	$\text{Cl}^-$ , $\text{Br}^-$ , $\text{CN}^-$	Needs pH 6-9 Not suitable for $\text{I}^-$ Blank titration needed
Volhard	Direct for $\text{Ag}^+$ Back titration for anions	Colourless $\text{SCN}^-$ to blood red $\text{Fe(SCN)}^{2+}$	$\text{Cl}^-$ , $\text{Br}^-$ , $\text{I}^-$ , $\text{CN}^-$ , $\text{PO}_4^{3-}$ , $\text{Cr}_2\text{O}_7^{2-}$ , $\text{S}^{2-}$ , $\text{CO}_3^{2-}$	Need low pH Precipitate may need removal Possible transfer errors
Fajan	Direct titration	Specific indicator for each species	$\text{Cl}^-$ , $\text{Br}^-$ , $\text{I}^-$	Needs specific pH for indicator Cannot use for low concentrations Cannot use if high levels of other ions present

Investigation 14.4: Determination of iodide ion concentration

This investigation uses Volhard's method to determine iodide ion concentration in a solution. Since silver iodide has a much lower $K_{sp}$ than silver thiocyanate, the precipitate doesn't need to be filtered before the back titration.

Materials:

$20 \text{ mL}$ of $1.0 \text{ mol L}^{-1}$ nitric acid
$50 \text{ mL}$ of $0.1 \text{ mol L}^{-1}$ silver nitrate
$5 \text{ mL}$ saturated ferric ammonium sulfate solution
$50 \text{ mL}$ of $0.1 \text{ mol L}^{-1}$ sodium iodide sample
$100 \text{ mL}$ of $0.1 \text{ mol L}^{-1}$ sodium thiocyanate solution
Distilled water
Dropper, measuring cylinders
3 × $250 \text{ mL}$ conical flasks
$50 \text{ mL}$ burette
Burette clamp and retort stand
Filter funnel
2 × $10 \text{ mL}$ pipette
Pipette filler
Safety glasses

Risk assessment:

Risk	Safety precautions
Nitric acid is corrosive	Wear safety glasses and wash hands at the end of the experiment
Silver salts are poisonous; silver nitrate causes brown stains	Avoid contact with skin and clothes; wash hands thoroughly after use; dispose as directed
Glassware could break	Keep glassware away from bench edge; leave pipette filler on when not in use; inform teacher if breakage occurs

Method:

Pipette $10 \text{ mL}$ of sample solution into a conical flask
Pipette $10 \text{ mL}$ of silver nitrate solution into the flask. Add $5 \text{ mL}$ of nitric acid and $1 \text{ mL}$ of ferric ammonium sulfate solution
Repeat steps 1-2 until there are three flasks
Fill the burette with $0.1 \text{ mol L}^{-1}$ sodium thiocyanate solution
Record the initial burette reading
Titrate the solution with thiocyanate until a blood-red colour appears. Record the final volume
Repeat steps 4-6 until three consistent readings are obtained

Analysis:

Calculate the average volume of sodium thiocyanate used
Calculate the average moles of sodium thiocyanate
Calculate the moles of silver nitrate that reacted with thiocyanate (excess silver)
Calculate the moles of silver nitrate originally added
Calculate the moles of silver that reacted with iodide (difference between total and excess)
Calculate the moles of iodide ions (1:1 ratio with silver)
Calculate the concentration of sodium iodide in the sample

Gravimetric Analysis

Gravimetric analysis is a quantitative analytical technique that uses mass measurements rather than volume. The species being analyzed must be separated from the mixture, typically by precipitation, then filtered, dried, and weighed.

General procedure:

Dissolve the sample in water
Add a chemical that precipitates the species being analyzed
Filter the solid precipitate from the solution
Dry the precipitate thoroughly
Weigh the dried precipitate
Calculate the percentage composition or concentration

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Common Applications

Gravimetric analysis is commonly used to analyze fertilizer composition. For example, sulfate content in lawn fertilizers can be determined because sulfur is an important plant nutrient. In alkaline soils, ammonium sulfate helps lower pH and promote better plant growth.

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Worked Example: Gravimetric Analysis of Sulfate

A $10.0 \text{ g}$ fertilizer sample containing ammonium sulfate is dissolved in water. Barium chloride is added, and the barium ions react with sulfate ions to form solid barium sulfate. The precipitate is filtered, dried, and weighed, giving a mass of $8.25 \text{ g}$ .

Step 1: Write the balanced equation $\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$

Step 2: Calculate the mass of sulfate ion $\text{Mass of SO}_4^{2-} = \frac{\text{formula mass of SO}_4^{2-}}{\text{formula mass of BaSO}_4} \times \text{mass BaSO}_4$

$\text{Mass of SO}_4^{2-} = \frac{96.1}{233.4} \times 8.25 = 3.4 \text{ g}$

Step 3: Calculate the percentage of sulfate in fertilizer $\% \text{ mass sulfate} = \frac{\text{mass of sulfate}}{\text{mass of fertilizer}} \times 100$

$\% \text{ mass sulfate} = \frac{3.4}{10.0} \times 100 = 34\%$

The fertilizer contains 34% sulfate by mass.

Errors in gravimetric analysis

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Major Sources of Error in Gravimetric Analysis:

Several sources of error can affect the accuracy of gravimetric analysis:

Contamination with other species: Impurities in the precipitate give falsely high mass readings
Loss of sample: Inefficient filtering, spillage, or incomplete rinsing of containers causes loss of precipitate
Incomplete precipitation: Not all the desired species precipitates from solution, giving low results
Water remaining in sample: Insufficient drying leaves water in the precipitate, increasing the measured mass

To minimize errors:

Use proper filtering techniques
Rinse all glassware thoroughly
Ensure complete precipitation by adding excess reagent
Dry the precipitate to constant mass (repeated drying until mass no longer changes)
Handle samples carefully to avoid losses during transfers

Investigation 14.5: Measuring sulfate content of lawn fertiliser

This investigation uses gravimetric analysis to determine the percentage of sulfate in a lawn fertilizer sample.

Overview:

You will design an experiment to precipitate sulfate ions from a fertilizer sample using barium ions in acidic solution, producing a white precipitate of barium sulfate. The precipitate will be filtered, dried, and weighed to calculate the percentage of sulfate.

Key hints:

Materials: You'll need chemicals to produce $\text{Ba}^{2+}$ ions (e.g., barium chloride solution) and an acid to keep the solution acidic. Use approximately $1 \text{ g}$ of fertilizer sample.
Safety: Note that acids are corrosive and barium compounds are toxic. Dispose of all waste as directed by your teacher.
Method considerations:
- Add barium solution gradually (drop by drop) until no more precipitate forms
- Gently heat (don't boil) the precipitate solution for about one hour to facilitate filtration
- The precipitate must be thoroughly dried before weighing
Analysis steps:
- Write the balanced equation: $\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$
- Calculate the mass of $\text{BaSO}_4$ formed
- Calculate the mass of sulfate ion using the formula mass ratio
- Calculate the percentage: $\frac{\text{mass of sulfate}}{\text{original mass of fertilizer}} \times 100$

Discussion points:

Compare your result with the stated amount on the fertilizer packet. Consider how these factors affect accuracy:

Solubility of barium sulfate (small amount remains in solution)
Contamination of the precipitate with impurities
Fine particles passing through the filter paper
Loss of precipitate during transfer from beaker to filter paper
Difficulty achieving complete drying of the filter paper

bookmarkSummary

Key Points to Remember:

Precipitation titrations use precipitation reactions to quantitatively determine ion concentrations, mainly using silver nitrate to analyze anions
Mohr's method is a direct titration using chromate indicator that changes to red-brown silver chromate at the end point; requires pH 6-9 and a blank titration correction
Volhard's method is a back titration in acidic solution using excess silver, then titrating with thiocyanate using ferric ion indicator; may require precipitate removal depending on $K_{sp}$ values
Fajan's method uses absorption indicators that change colour when adsorbed onto the precipitate surface; different indicators are needed for different halides
Gravimetric analysis determines ion quantities by mass measurements rather than volume; the species is precipitated, filtered, dried, and weighed
Sources of error in gravimetric analysis include contamination, sample loss, incomplete precipitation, and retained water in the precipitate

Quantitative Analysis of Ions (HSC SSCE Chemistry): Revision Notes