Qualitative Analysis: Spot Tests and Microanalysis

• Qualitative analysis involves applying solutions and reagents to a substance sample.
• The reaction creates a colored spot, indicating the presence of a specific ion in the unknown sample.
• Spot tests may also be performed on special spot plates with indentations.

Ultramicro Analysis

• Detects less than 1 mg of a substance.
• Analytical operations are conducted under a microscope.

Microanalysis

• Uses approximately 0.01g or less of a substance for analysis.
• Sensitive reactions are employed to detect individual components, even in small quantities.
• Reactions performed with microcrystaloscopic or drop methods (spot tests).
• Microcrystaloscopic reactions occur on glass microscope slides.
• The determination of a given ion depends on the crystal form, which can be viewed under a microscope.
• Drop method reactions involve color changes or precipitate formation.
• These tests are commonly done on filter paper strips, using drops of unknown solution and reagents.

Salts

• Consist of basic/metallic and acidic/non-metallic radicals.
• Salts typically break apart in aqueous solutions.
• They form positively charged cations and negatively charged anions.
• Most quantitative analysis wet reactions involve ion reactions.

Qualitative Analysis Principles

• Solubility product and common ion effect.
• Solubility product and common ion effect are primarily used to modify ion concentrations and selectively precipitate intended ions.

Solubility Product

• Some ionic compounds have low water solubility, like AgCl, PbSO4, and BaSO4.
• These salts are sparingly soluble.
• A sparingly soluble salt AB, when dissolved in water, only dissolves a small amount.
• As strong electrolytes, the dissolved salt separates completely with an equilibrium between ionized and unionized forms: AB ⇔ A+ + B-.
• The formula for the law of mass action is K = [A+][B-]/[AB], where brackets denote active masses.
• Molar concentration generally measures active mass, but activity is a more accurate measure.

Solubility Product and Saturated Solutions

• Since the solution is saturated, the concentration of unionized salt remains constant at a constant temperature.
• The constant K can be expressed as Ksp = [A+][B-], where Ksp is the solubility product.
• At equilibrium in a saturated solution, the product of ionic concentrations remains constant at a given temperature.
• For a salt like MAqBrr, which dissociates into qM+ and rA-, Ksp = [Mq+]q[Ar-]r.
• The solubility product (Ksp) is the product of molar concentrations of constituent ions, each raised to the power of its number of ions in a saturated solution at a given temperature.

Determining Ion Dissolution

• Ksp value indicates a salt's extent of dissolution.
• Each salt has its unique solubility product, like AgCl with Ksp = 1.0 × 10-10.
• In AgCl saturated solution at 298K, the product of silver and chloride ion concentrations equals 1.0 × 10-10.
• AgCl precipitation occurs when the solution's [Ag+][Cl-] exceeds AgCl's solubility product (1.0 × 10-10).
• Solubility product important for precipitating ions from solution.
• Saturated solution results at the Ksp value.
• Precipitation occurs when the ionic product goes over the solubility product.
• A substance's solubility measures the amount dissolved in water at a given temperature.
• Solubility is expressed in mol dm-3 or g dm-3.
• A substance’s solubility product can be calculated accordingly.

Temperature and Solubility

• Increasing temperature increases the solubility and solubility product, generally.

Example

• Silver chloride's solubility in water measures at 1.435 x 10-3 g.dm-3 at 298K with a molar mass of 143.5, the silver chloride's solubility product can be calculated.
• First, molar concentration of saturated AgCl solution: CAgCl = (1.435×10-3) / 143.5 = 1.0 × 10-5 mol. dm-3.
• Each gram molecule of AgCl yields one gram ion of Ag+ and one of Cl-, so [Ag+] = [Cl-] = 1 × 10-5 mol. dm-3.
• AgCl Solubility product = (1.0 x 10-5) x (1.0 × 10-5) = 1.0 × 10-10.

Example

• Silver chromate's solubility product can also be calculated if one liter of saturated solution at 298K contains 4.3 × 10-2 g, with M.wt. of Ag2CrO4 at 331.8.
• Molar concentration of saturated Ag2CrO4 solution is 4.3×10-2 /331.8 = 1.3 × 10-4 mol dm-3.
• Each gram molecule of Ag2CrO4 gives 2 g ions of Ag+ and 1 g ion of CrO42-.
• The Ag2CrO4 equation is Ag2CrO4 ⇌ 2Ag+ + CrO42-.
• [Ag+]2 = 2(1.3 × 10-4) mol dm-3, and [CrO42-] = 1.3 × 10-4 mol dm-3
• Hence, Ksp = [Ag+]2 [CrO42-] = (2.6×10-4)2 (1.3 × 10-4) = 9.0 × 10-12
• Therefore, Ksp of Ag2CrO4 = 9.0 × 10-12.

Precipitation Equilibria Factors

• Once a solid phase/solution equilibrium is achieved w/ a sparingly soluble salt, the solubility product is reached.
• If conditions prevent the ionic product from matching the solubility product, the system adjusts until they are equal, achieved through precipitation.
• The degree a weak electrolyte dissociates lowers with the addition of a strong electrolyte with a common ion.
• Consider the dissociation of NH4OH into NH4+ and OH-, where K = [NH4+][OH-]/[NH4OH].
• Adding NH4Cl, which is highly ionized, increases NH4+ concentration, shifting the equilibrium to the left, suppressing NH4OH dissociation.
• Common ion use systematically modifies concentration of ions in a soln, so intended ions are selectively precipitated from the solution.

Precipitation of Chlorides of Metals

• Such as Pb, Ag, using HCl for mixture of Group I metals (e.g., AgNO₃) with others (e.g., CuSO₄).
• Silver forms AgCl, copper forms CuCl₂
• AgCl has a low solubility product (1.5 × 10⁻¹⁰ at 298K), CuCl₂ has a high.
• High Cl⁻ concentration allows Ag⁺ to precipitates as AgCl.
• CuCl₂'s high solubility prevents it from precipitating, allowing Cu²⁺ to remain in the solution, if the maximum Cl⁻ concentration doesn't reach the product.

Precipitation of Hydroxides

• Fe3+, Al3+, and Cr3+ metal ions in the presence of metal ions like Ni2+, Co2+, Zn2+, Mn2+, Mg2+.
• Hydroxides of Fe3+, Cr3+, and Al3+ have lower solubility products than Ni2+, Zn2+, and Mg2+ hydroxides.
• When a mixture of salts with varying solubility products reacts with NH4Cl and NH4OH, a OH- concentration becomes low if its just enough to precipitate ions with the lower product.
• For example, if Fe3+ and Zn2+ are treated with NH4Cl and NH4OH, Fe(OH)3 is achieved as the solubility product alone, while Zn(OH)2 is not.
• Consequently the Fe(OH)3 precipitates and separated from the Zn2+. If NH4Cl is not added, there are high OH- concentrations for precipitate metals such as Ni2+, Co2+, Zn2+, Mn2+, Mg2+.

Precipitation of Sulphides

• Metal ions such as Cu2+, Cd2+, Pb2+ from H₂S gas in the presence of dilute HCl.
• H₂S acts as weak dibasic acid/ dissociates into H⁺ & S-2 ions.
• Presence of HCl reduces S-2 because of the common ion product.
• Second group exhibits sulphides with lower product.
• Result- reduced concentration of S-2 ions.
• Solubility- metal sulphides of group II exceeds, and precipitated.

Reagents

• NH4Cl, NH4OH, and H2S are used for metal ions such as Ni2+, Co2+, Zn2+.
• In the reaction, OH- ions comes from NH4OH and combine with H⁺ ions from H₂S to form undissociated H₂O and S²- ions is increased so the metal sulphides whose solubility products were exceeded are no more exceeded by passing H2S in HCl are precipitated.

pH and Solubility

• According to le Chatelier's, when the reaction starts, it continues with precipitating ions until the product equals its solubility product.
• All substances are more or less soluble, so the solubility product is never zero.
• The solubility is dependent on pH.
• The solubility of salt of a strong acid isn't affected with a change in concentration of H+ but it does affect precipitation of slightly soluble salts and metal hydroxides,
• Hydroxyl ion concentration plays an important part in precipitating the amount of hydroxides.
• Hydroxides increase with a need for a higher OH-concentration, so higher pH starts precipitation.

Calculations

• pH at when a certain hydroxide will start precipitation can be approximated from,the product.
• If Mg2+ in solution is 1 x 10-2 g/dm³, and the Magnesium hydroxide which is 5 x 10-12 is reached then.
• [Mg2+] [OH-]² = Kap Mg(OH)2 = 5 x 10-12
• The equation of product used to reach product for the start if product found using: [OH-] = (KspMg(OH)2/[Mg2+]) = 5×10-12/1×10-2 = 2.2 x 10-5 g ion/dm³.
• POH = -log 2.2 x 10-5 = -(0.3 – 5.0) = 4.7 - PH=14-POH = 14 – 4.7 = 9.3.
• pH = 9.3 -precipitation od Mg(OH)2 begins after in 0.01M and ends when it's past 11.3.
• The above concentration is needed to start the precipitation.
• H+ plays biggest influence in the salt being precipitated.
• You must have a pH that doesn't increase.pH- buffer mixture is resistant so it used.

Sparingly Soluble Salts

• They are of strong monobasic acids are precipitated with their ions.
• Their completeness isn't dependent of Ph so their change isn't affected by it.
• However theres a change to the salts of weak acids.
• Examples- oxalates.
• Anions combine with H, and anions, so in their concentration are dependent on it.

Effect of Complexation

• Ions can be bound by formation of stable or complex compounds. Complex has almost dissolved by how soluble the AgCl dissolves in NH4OH.
• Sugars in solutions are no longer precipitant when OH adds to analysis. You must convert it or find something to add to the interference and remain non-interfering.

Influence of Interionic Forces

• Interionic forces influence ions ability to take part in chemical action. Ability of ions influence chemical reaction.

Activity coefficient

• If activity in 0.1M HCl 0.0814 it ions act in chemical rxn and 0.1 is written as: v= 0.0814/0.1=0.814.
• There is a relationship: a=C... activity on of product