Common-Ion Solubility Numericals

Introduction

Adding an ion already produced by a dissolving salt usually lowers that salt's molar solubility. The arithmetic is easy to mishandle because the concentration of the common ion is not always identical to either the amount added or the amount released by the solid. An equilibrium table keeps those contributions separate and provides a clear test for a useful approximation.

Core explanation

Use a generic sparingly soluble fluoride, MF₂(s) ⇌ M²⁺(aq) + 2F⁻(aq), with Ksp = [M²⁺][F⁻]² in a dilute concentration model. In pure water, if s mol L⁻¹ dissolves, [M²⁺] = s and [F⁻] = 2s. Therefore Ksp = s(2s)² = 4s³. With the illustrative value Ksp = 4.0 × 10⁻¹¹, s = (Ksp/4)^(1/3) = (1.0 × 10⁻¹¹)^(1/3) ≈ 2.15 × 10⁻⁴ mol L⁻¹. The factor of two on fluoride must be included before squaring.

Now suppose the solid is placed in 0.0100 mol L⁻¹ NaF solution. Sodium fluoride supplies an initial free-fluoride concentration of approximately 0.0100 mol L⁻¹ under the simple full-dissociation model. If s mol L⁻¹ of MF₂ dissolves, it adds s mol L⁻¹ M²⁺ and 2s mol L⁻¹ F⁻. The actual equilibrium concentrations are [M²⁺] = s and [F⁻] = 0.0100 + 2s. The exact concentration equation is Ksp = s(0.0100 + 2s)².

When 2s is much smaller than 0.0100, estimate s ≈ Ksp/(0.0100)² = 4.0 × 10⁻⁷ mol L⁻¹. Then 2s = 8.0 × 10⁻⁷ mol L⁻¹, only 0.008% of the fluoride initially supplied. The assumption is self-consistent to the reported precision. Compared with 2.15 × 10⁻⁴ mol L⁻¹ in pure water, the common-ion solution holds roughly five hundred times less dissolved MF₂ in this idealised comparison. The large reduction arises because the added fluoride appears squared in Ksp.

The added-ion concentration must be the concentration after any dilution or mixing. If 50 mL of 0.0200 M NaF is combined with 50 mL of water before contact with the solid, the initial common-ion concentration is 0.0100 M, not 0.0200 M. Likewise, an initial concentration of the metal ion from a soluble metal salt would enter [M²⁺] as a starting term, while the fluoride would still gain 2s from dissolution. Write the actual starting inventory rather than copying a memorised formula.

There are practical limits. If the calculated 2s is not small compared with the added fluoride, retain it in the equation and solve numerically or algebraically. If another reaction binds fluoride or the metal ion, free-ion concentrations differ from analytical totals. At appreciable ionic strength, activity coefficients modify the thermodynamic solubility product; the simple concentration comparison is an educational approximation. The model predicts equilibrium dissolved amount, not how quickly the solid dissolves or whether a visible particle remains.

Step-by-step reasoning

1. Write the dissolution equation and Ksp using the correct powers. 2. Convert any added soluble-salt amount to its concentration in the final volume. 3. Define s as moles of solid dissolved per litre, and add its stoichiometric ion contributions. 4. Substitute the full equilibrium concentrations into Ksp. 5. Try neglecting the dissolved-solid contribution to the abundant common ion, then calculate s. 6. Compare the neglected term with the added-ion term; solve the full equation if the ratio is too large.

Visual explanation

Draw a tall bar labelled 0.0100 M added F⁻ and a much shorter bar labelled 2s from MF₂. Together they make equilibrium [F⁻]. A separate bar labelled s represents M²⁺. The short bar may be omitted in a first estimate only after its height is calculated and shown to be negligible next to the tall bar.

Real-world analogy

Suppose a reservoir already contains ten thousand litres of water, and a small inlet adds less than one litre. Treating the total as ten thousand litres is a sound approximation after checking the scale. If the inlet adds thousands of litres, that shortcut fails. In the common-ion calculation, added fluoride is the reservoir and 2s is the inlet.

Real-world example

Precipitation-based separation can use a common ion to make a sparingly soluble salt less soluble. For instance, changing chloride concentration affects the dissolution equilibrium of silver chloride. In a real separation the effect can be complicated by complex ions at very high chloride concentration, so a simple common-ion prediction should be checked against full solution chemistry before designing an analytical procedure.

Why?

Why does adding fluoride suppress MF₂ dissolution? Dissolution produces fluoride. When fluoride is already abundant, a smaller concentration of dissolved M²⁺ is enough for the product [M²⁺][F⁻]² to reach Ksp. The solid can remain while the saturated solution contains fewer metal ions than it would in pure water.

Common misconception

“The common-ion concentration stays exactly at its added value.” Dissolving solid always contributes its own ions; the added value is an approximation to the total when that contribution is small. Another error is to set [F⁻] = s for MF₂, overlooking that every dissolved formula unit releases two fluoride ions.

Worked example

For MF₂ with Ksp = 4.0 × 10⁻¹¹, find its molar solubility in 0.0100 M NaF. Let s be the dissolved MF₂ concentration. Then Ksp = s(0.0100 + 2s)². First set 2s aside: s ≈ 4.0 × 10⁻¹¹/(0.0100)² = 4.0 × 10⁻⁷ M. Check 2s/0.0100 = 8.0 × 10⁻⁵, or 0.008%. This validates the approximation for the given significant figures. Pure-water solubility from 4s³ = Ksp is 2.15 × 10⁻⁴ M, so the common ion sharply lowers solubility in this model.

Quick check

1. If 0.0010 M fluoride was added to an MF₂ solution and s = 2.0 × 10⁻⁵ M, is 2s negligible at a five-percent guideline? Answer: Yes; 2s/0.0010 = 0.040, or 4%, so the approximation is within that rough guideline.

Exam focus

State what s represents before writing concentrations. An examiner can distinguish a correct setup, s(F₀ + 2s)², from an unjustified answer that simply divides Ksp by an initial concentration. Show the approximation check and use the concentration after mixing, not the stock-bottle label.

Advanced insight

The thermodynamic solubility product uses ion activities, and adding a salt changes ionic strength as well as the common-ion concentration. These effects can oppose or modify the simple concentration-based prediction. Complexation can even increase total dissolved metal at high ligand concentration while free metal ion remains low. For this reason, “common ion lowers solubility” is a strong statement only within a specified equilibrium model and range of conditions.

Summary

For MF₂ in a solution initially containing F₀ fluoride, Ksp = s(F₀ + 2s)². The approximate expression s ≈ Ksp/F₀² is useful only after checking that 2s is small compared with F₀. In pure water the relation is instead Ksp = 4s³. Keep added-ion concentration, dissolved-solid contribution and final-volume dilution distinct throughout the calculation.

Practice questions

1. Write the exact Ksp equation for MX(s) in a solution initially containing 0.020 M X⁻, using s as molar solubility. Answer: MX releases one M⁺ and one X⁻, so Ksp = s(0.020 + s). 2. A salt MX has Ksp = 2.0 × 10⁻⁸. Estimate s in 0.020 M X⁻ and test the approximation. Answer: s ≈ 2.0 × 10⁻⁸/0.020 = 1.0 × 10⁻⁶ M; s/0.020 = 0.005%, so neglecting s in the X⁻ term is sound. 3. Why does pure-water MF₂ give Ksp = 4s³ rather than s³? Answer: Each formula unit produces two fluoride ions, so [F⁻] = 2s and [M²⁺][F⁻]² = s(2s)² = 4s³. 4. A common-ion mixture was made by diluting equal volumes of a 0.020 M NaF stock and water. Which fluoride concentration enters the initial equilibrium table? Answer: The post-mixing concentration is 0.010 M, assuming additive volumes and full dissolution of NaF.