How to Calculate the Minimum Ion Concentration Needed to Start Precipitation

A clear solution contains silver ions, \(\mathrm{Ag}^{+}\). You slowly add sodium chloride solution, supplying chloride ions, \(\mathrm{Cl}^{-}\). Eventually, a white solid called silver chloride forms.

But what is the minimum chloride ion concentration needed to start precipitation?

It is not necessarily equal to the silver ion concentration. The answer depends on the solubility product, \(K_{sp}\), and the concentration of the other ion already present.

For silver chloride, the threshold is:

\[ \boxed{[\mathrm{Cl}^{-}]_{\min}= \frac{K_{sp}}{[\mathrm{Ag}^{+}]}} \]

For example, if \(K_{sp}=1.80\times10^{-10}\) at \(25^\circ\mathrm{C}\), and the final solution contains \(2.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) silver ions:

\[ [\mathrm{Cl}^{-}]_{\min} =\frac{1.80\times10^{-10}}{2.00\times10^{-4}} \]

\[ \boxed{[\mathrm{Cl}^{-}]_{\min} =9.00\times10^{-7}\ \mathrm{mol\,L^{-1}}} \]

At this concentration, the solution is just saturated. A slightly higher chloride concentration makes precipitation thermodynamically favourable.

The important detail is that you must use the concentrations of the free ions in the final mixed solution, including any ions already present from other substances.

All numerical solubility product values and experimental data in this article are illustrative teaching values at \(25^\circ\mathrm{C}\). In an examination, use the \(K_{sp}\) value supplied in the question.

01The threshold: when does precipitation begin?

A sparingly soluble ionic compound can dissolve in water, but only to a limited extent.

Consider silver chloride:

\[ \ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)} \]

The double arrow represents a dynamic equilibrium. Silver chloride dissolves, releasing ions, while dissolved silver and chloride ions can also recombine to form solid silver chloride.

The solubility product, \(K_{sp}\), describes the equilibrium relationship between the concentrations of the dissolved ions when the solution is saturated.

For silver chloride:

\[ K_{sp}=[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \]

The value of \(K_{sp}\) is constant at a particular temperature.

However, the concentrations in a solution you’re currently examining do not necessarily satisfy this equilibrium relationship. To determine whether precipitation will occur, we calculate the ion product, \(Q_{sp}\).

The expression for \(Q_{sp}\) has exactly the same structure as the \(K_{sp}\) expression. The difference is that \(Q_{sp}\) uses the current ion concentrations, rather than necessarily using equilibrium concentrations.

For silver chloride:

\[ Q_{sp}=[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \]

Compare the two values:

This gives us the central calculation rule:

\[ \boxed{Q_{sp}=K_{sp}\text{ at the threshold}} \]

You calculate the ion concentration at this point to find the minimum concentration required to start precipitation.

Strictly, a solution at \(Q_{sp}=K_{sp}\) is just saturated. Precipitation becomes thermodynamically favoured when \(Q_{sp}\) rises above \(K_{sp}\). Visible precipitation can take slightly longer because solid particles must nucleate and grow.

02Use the dissolution equation to write the correct Ksp expression

Before calculating an unknown ion concentration, write the balanced dissolution equation.

This matters because the coefficients in the dissolution equation become the powers in the \(K_{sp}\) expression.

Consider three examples.

Notice that calcium fluoride produces two fluoride ions for every calcium ion.

Therefore, its solubility product is:

\[ K_{sp}=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]

The fluoride concentration is squared.

It is the stoichiometric coefficient, not the ionic charge, that determines the exponent.

The general threshold equation

Suppose a solid dissolves according to:

\[ \ce{A_pB_q(s) <=> pA(aq) + qB(aq)} \]

Its solubility product expression is:

\[ K_{sp}=[A]^p[B]^q \]

If the concentration of \(A\) is known and you need to calculate the minimum concentration of \(B\), rearrange:

\[ K_{sp}=[A]^p[B]^q \]

\[ [B]^q=\frac{K_{sp}}{[A]^p} \]

Therefore:

\[ \boxed{[B]_{\min}= \sqrt[q]{\frac{K_{sp}}{[A]^p}}} \]

Here, \(p\) and \(q\) are the coefficients in the balanced dissolution equation.

This is the general method. The exact powers and rearrangement depend on which solid is forming and which ion concentration you’re trying to find.

03Account for ions already present: the common-ion effect

This is where many precipitation calculations become slightly more complicated.

Imagine a solution that already contains calcium ions because calcium nitrate has dissolved in it.

You now introduce fluoride ions by adding sodium fluoride.

If calcium fluoride begins to precipitate, its dissolution equilibrium is:

\[ \ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)} \]

The solution already contains calcium ions before any calcium fluoride precipitates.

Calcium is therefore a common ion: an ion that is present in the solution from another source and also appears in the dissolution equilibrium of the sparingly soluble salt.

You must include this existing calcium concentration in your \(K_{sp}\) calculation.

For calcium fluoride:

\[ K_{sp}=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]

If the existing calcium concentration is known, substitute it directly.

You do not assume that the calcium concentration equals the amount of calcium fluoride that dissolves.

What if the unknown ion is already present?

Suppose the solution already contains some fluoride ions.

The threshold calculation gives you the total free fluoride concentration needed to reach saturation.

If the volume stays constant, the additional fluoride concentration required is:

\[ \boxed{ [\mathrm{F}^{-}]_{\mathrm{additional}} = [\mathrm{F}^{-}]_{\mathrm{threshold}} - [\mathrm{F}^{-}]_{\mathrm{initial}} } \]

This subtraction is important.

You are finding how much more fluoride must be supplied, not pretending the solution initially contained none.

If you are adding a stock solution, however, its addition also increases the total volume. You must account for that dilution rather than simply subtracting concentrations.

Do not confuse common-ion precipitation with pure-water solubility

There is an important difference between two types of questions.

Situation A: A solid dissolves in pure water.

For calcium fluoride, if \(s\) mol L\(^{-1}\) dissolves:

\[ [\mathrm{Ca}^{2+}]=s \]

\[ [\mathrm{F}^{-}]=2s \]

Therefore:

\[ K_{sp}=s(2s)^2=4s^3 \]

Situation B: Calcium ions are already present and fluoride is added.

Now the calcium concentration is not simply \(s\).

You use the existing calcium concentration:

\[ K_{sp}=[\mathrm{Ca}^{2+}]_{\mathrm{initial}} [\mathrm{F}^{-}]_{\mathrm{threshold}}^2 \]

These are different calculations answering different questions.

If you want to explore the first calculation in more detail, see How to Calculate Ksp from Calcium Fluoride Solubility.

For the reverse problem, where you calculate how much solid dissolves when a common ion is present, see How to Calculate Solubility with a Common Ion.

04Get the concentrations right before using Ksp

Even if your \(K_{sp}\) expression is perfect, an incorrect concentration will give you an incorrect precipitation threshold.

There are three things to check.

Check 1: Have you accounted for dilution?

When two solutions are mixed, their volumes combine, and their concentrations change.

The number of moles of an ion is:

\[ n=CV \]

where:

• \(n\) is the amount in moles,
• \(C\) is the concentration in mol L\(^{-1}\),
• \(V\) is the volume in litres.

After mixing:

\[ \boxed{C_{\mathrm{final}}= \frac{n_{\mathrm{total}}}{V_{\mathrm{total}}}} \]

For example, suppose you mix:

• 20.0 mL of \(0.0100\ \mathrm{mol\,L^{-1}}\) calcium nitrate,
• 80.0 mL of another solution.

Calcium nitrate dissociates according to:

\[ \ce{Ca(NO3)2(aq) -> Ca^2+(aq) + 2NO3-(aq)} \]

The initial calcium concentration in its stock solution is \(0.0100\ \mathrm{mol\,L^{-1}}\).

After mixing, the final calcium concentration is:

\[ [\mathrm{Ca}^{2+}] = \frac{0.0100\times0.0200}{0.1000} \]

\[ \boxed{[\mathrm{Ca}^{2+}]= 2.00\times10^{-3}\ \mathrm{mol\,L^{-1}}} \]

This is the concentration you use in the \(K_{sp}\) expression.

Do not use the original stock concentration unless the question specifically tells you that it is already the final concentration.

Check 2: Have you accounted for dissociation stoichiometry?

Some soluble salts produce more than one mole of a particular ion per mole of salt.

For example:

\[ \ce{CaCl2(aq) -> Ca^2+(aq) + 2Cl-(aq)} \]

A \(0.0200\ \mathrm{mol\,L^{-1}}\) calcium chloride solution supplies:

\[ [\mathrm{Ca}^{2+}]=0.0200\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{Cl}^{-}]=0.0400\ \mathrm{mol\,L^{-1}} \]

The chloride concentration is twice the salt concentration.

This is a different use of stoichiometry from the exponents in \(K_{sp}\), but both come from the balanced chemical equation.

Check 3: Are your concentration units correct?

Solubility product calculations require ion concentrations in mol L\(^{-1}\).

If a question gives you a mass concentration, convert it before substituting.

For example:

\[ 40.0\ \mathrm{mg\,L^{-1}} = 0.0400\ \mathrm{g\,L^{-1}} \]

If the substance is calcium ions, divide by the molar mass of calcium:

\[ [\mathrm{Ca}^{2+}] = \frac{0.0400}{40.08} \]

\[ =9.98\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Do not substitute \(40.0\) directly into the \(K_{sp}\) expression.

A useful decision sequence

05A proposed experiment: predicting the first appearance of a precipitate

The calculation gives us a theoretical precipitation threshold. But how could we test whether that threshold is realistic?

Consider an investigation using silver nitrate and sodium chloride.

The aim is to estimate the chloride concentration at which silver chloride first forms.

Setting up the apparatus

Start with a beaker containing 100.0 mL of \(2.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) silver nitrate solution.

Place the beaker on a magnetic stirrer.

Above the beaker, position a burette containing \(1.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) sodium chloride solution.

A turbidity probe is placed in the beaker to detect the formation of suspended solid particles. Keep the solution at \(25^\circ\mathrm{C}\).

What should happen?

As sodium chloride enters the beaker, it supplies chloride ions.

These react with silver ions:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

Initially, the solution is unsaturated.

As more sodium chloride is added, the chloride concentration increases. Eventually, the ion product reaches the solubility product.

At that point:

\[ Q_{sp}=K_{sp} \]

The solution is just saturated.

A little more chloride makes precipitation thermodynamically favourable, producing suspended white silver chloride particles.

Predicting the volume at the threshold

Let’s calculate how much sodium chloride solution should be added.

The illustrative data are:

Let \(V\) be the volume of sodium chloride solution added, measured in litres.

The total volume becomes:

\[ V_{\mathrm{total}}=0.1000+V \]

The silver ion concentration decreases slightly because of dilution:

\[ [\mathrm{Ag}^{+}] = \frac{(2.00\times10^{-4})(0.1000)} {0.1000+V} \]

The chloride ion concentration is:

\[ [\mathrm{Cl}^{-}] = \frac{(1.00\times10^{-4})V} {0.1000+V} \]

At the threshold:

\[ [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] =1.80\times10^{-10} \]

Substituting:

\[ \frac{(2.00\times10^{-4})(0.1000)} {0.1000+V} \times \frac{(1.00\times10^{-4})V} {0.1000+V} =1.80\times10^{-10} \]

Solving for the smaller, physically relevant root gives:

\[ V\approx9.17\times10^{-4}\ \mathrm{L} \]

Therefore:

\[ \boxed{V\approx0.917\ \mathrm{mL}} \]

The model predicts that the solution should approach saturation after approximately 0.917 mL of sodium chloride solution has been added.

This is a predicted threshold, not a reported experimental measurement.

Why the experimental design matters

Suppose the sodium chloride is poured quickly into an unstirred beaker.

The region immediately surrounding the incoming drop can have a much higher chloride concentration than the rest of the solution. Silver chloride may form locally before the entire beaker reaches the predicted bulk concentration.

This makes the first visible cloudiness a poor measurement of the true bulk precipitation threshold.

A better design uses:

• A dilute sodium chloride solution.
• Slow addition from a burette.
• Continuous stirring.
• A turbidity probe to detect suspended particles.
• A controlled temperature.

Even then, the measured onset may differ slightly from the theoretical prediction because crystal nucleation and particle detection take time.

The important distinction is that \(K_{sp}\) predicts the equilibrium threshold, while the experiment detects the first observable solid.

06Worked examples

07Questions and solutions

The following questions move from straightforward threshold calculations towards more involved combinations of dilution, stoichiometry, competing precipitates, and acid-base equilibria.

All numerical equilibrium data are supplied for these questions. Assume \(25^\circ\mathrm{C}\), unless otherwise stated, and use concentrations as approximations to activities.

Question 1

At \(25^\circ\mathrm{C}\), the solubility product of silver chloride is:

\[ K_{sp}=1.80\times10^{-10} \]

A solution contains:

\[ [\mathrm{Ag}^{+}]=6.00\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{Cl}^{-}]=1.00\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

Calculate the additional chloride ion concentration required to just reach the precipitation threshold.

Assume the volume remains constant.

Solution 1

The additional chloride concentration required is:

\[ \boxed{2.00\times10^{-6}\ \mathrm{mol\,L^{-1}}} \]

We first need to find the total chloride concentration at saturation, then subtract the chloride already present.

Step 1: Write the dissolution equation and \(K_{sp}\) expression.

\[ \ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)} \]

\[ K_{sp}=[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \]

Step 2: Set the ion product equal to the solubility product.

\[ 1.80\times10^{-10} = (6.00\times10^{-5})[\mathrm{Cl}^{-}]_{\min} \]

Rearrange:

\[ [\mathrm{Cl}^{-}]_{\min} = \frac{1.80\times10^{-10}}{6.00\times10^{-5}} \]

\[ [\mathrm{Cl}^{-}]_{\min} =3.00\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

Step 3: Subtract the initial chloride concentration.

\[ [\mathrm{Cl}^{-}]_{\mathrm{additional}} = 3.00\times10^{-6} - 1.00\times10^{-6} \]

\[ \boxed{ [\mathrm{Cl}^{-}]_{\mathrm{additional}} =2.00\times10^{-6}\ \mathrm{mol\,L^{-1}} } \]

The solution initially contains some chloride, but not enough to cause precipitation.

A common mistake is to report \(3.00\times10^{-6}\ \mathrm{mol\,L^{-1}}\) as the additional concentration. That is the total threshold concentration, not the increase required.

Question 2

Calcium fluoride has a solubility product of:

\[ K_{sp}=3.90\times10^{-11} \]

A solution contains the following free-ion concentrations before additional fluoride is introduced:

\[ [\mathrm{Ca}^{2+}] =1.25\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{F}^{-}]_{\mathrm{initial}} =1.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Calculate the additional fluoride ion concentration needed to start calcium fluoride precipitation.

Assume the volume remains constant.

Solution 2

The additional fluoride concentration required is approximately:

\[ \boxed{7.66\times10^{-5}\ \mathrm{mol\,L^{-1}}} \]

The key is to account for the two fluoride ions in the dissolution equation.

Step 1: Write the balanced dissolution equation.

\[ \ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)} \]

Step 2: Write the solubility product expression.

\[ K_{sp}=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]

Step 3: Set the expression equal to \(K_{sp}\).

\[ 3.90\times10^{-11} = (1.25\times10^{-3})[\mathrm{F}^{-}]_{\min}^{2} \]

Rearrange:

\[ [\mathrm{F}^{-}]_{\min}^{2} = \frac{3.90\times10^{-11}} {1.25\times10^{-3}} \]

\[ [\mathrm{F}^{-}]_{\min}^{2} =3.12\times10^{-8} \]

Take the square root:

\[ [\mathrm{F}^{-}]_{\min} = \sqrt{3.12\times10^{-8}} \]

\[ [\mathrm{F}^{-}]_{\min} =1.77\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Step 4: Subtract the initial fluoride concentration.

\[ [\mathrm{F}^{-}]_{\mathrm{additional}} = 1.77\times10^{-4} - 1.00\times10^{-4} \]

\[ \boxed{ [\mathrm{F}^{-}]_{\mathrm{additional}} =7.66\times10^{-5}\ \mathrm{mol\,L^{-1}} } \]

The additional fluoride concentration must be approximately \(7.66\times10^{-5}\ \mathrm{mol\,L^{-1}}\).

The tempting mistake is to divide this result by two because calcium fluoride produces two fluoride ions.

That would be incorrect. The factor of two in the dissolution equation determines the exponent in \(K_{sp}\). We have already accounted for it by squaring the fluoride concentration.

Question 3

A student mixes two solutions at \(25^\circ\mathrm{C}\).

• 40.0 mL of \(5.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) silver nitrate.
• 60.0 mL of \(2.00\times10^{-5}\ \mathrm{mol\,L^{-1}}\) sodium chloride.

The solubility product of silver chloride is:

\[ K_{sp}=1.80\times10^{-10} \]

Assume the volumes are additive and there are no competing reactions.

(a) Determine whether silver chloride precipitates immediately after mixing.

(b) Calculate the equilibrium chloride ion concentration after precipitation has occurred.

(c) Calculate the mass of silver chloride formed. Use a molar mass of \(143.32\ \mathrm{g\,mol^{-1}}\).

Solution 3

Yes, silver chloride precipitates. The initial ion product is greater than \(K_{sp}\), and the equilibrium chloride concentration after precipitation is approximately:

\[ \boxed{9.53\times10^{-7}\ \mathrm{mol\,L^{-1}}} \]

The amount of silver chloride formed is approximately:

\[ \boxed{0.158\ \mathrm{mg}} \]

This question requires two different stages: first checking the threshold after dilution, then applying equilibrium and mole conservation after precipitation begins.

Step 1: Calculate the total volume.

\[ V_{\mathrm{total}}=40.0+60.0 \]

\[ V_{\mathrm{total}}=100.0\ \mathrm{mL} =0.1000\ \mathrm{L} \]

Step 2: Calculate the silver ion concentration after mixing.

Silver nitrate dissociates in a 1:1 ratio:

\[ \ce{AgNO3(aq) -> Ag+(aq) + NO3-(aq)} \]

The concentration is:

\[ [\mathrm{Ag}^{+}] = \frac{(5.00\times10^{-4})(0.0400)} {0.1000} \]

\[ [\mathrm{Ag}^{+}] =2.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Step 3: Calculate the chloride ion concentration after mixing.

Sodium chloride also dissociates in a 1:1 ratio:

\[ \ce{NaCl(aq) -> Na+(aq) + Cl-(aq)} \]

Therefore:

\[ [\mathrm{Cl}^{-}] = \frac{(2.00\times10^{-5})(0.0600)} {0.1000} \]

\[ [\mathrm{Cl}^{-}] =1.20\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Step 4: Calculate the initial ion product.

\[ Q_{sp} = (2.00\times10^{-4})(1.20\times10^{-5}) \]

\[ Q_{sp}=2.40\times10^{-9} \]

Compare this with:

\[ K_{sp}=1.80\times10^{-10} \]

Since:

\[ 2.40\times10^{-9} > 1.80\times10^{-10} \]

the mixture is supersaturated.

Silver chloride precipitation is thermodynamically favoured.

Step 5: Calculate the equilibrium concentrations.

Once precipitation begins, silver and chloride ions are removed from the solution in a 1:1 mole ratio.

Let \(x\) represent the equilibrium chloride concentration.

The initial concentrations were:

\[ [\mathrm{Ag}^{+}]_{\mathrm{initial}} =2.00\times10^{-4} \]

\[ [\mathrm{Cl}^{-}]_{\mathrm{initial}} =1.20\times10^{-5} \]

Because the reaction consumes equal amounts of silver and chloride ions:

\[ [\mathrm{Ag}^{+}]_{\mathrm{eq}} = 2.00\times10^{-4} - (1.20\times10^{-5}-x) \]

Simplifying:

\[ [\mathrm{Ag}^{+}]_{\mathrm{eq}} = 1.88\times10^{-4}+x \]

At equilibrium, the solution is saturated:

\[ K_{sp}=[\mathrm{Ag}^{+}]_{\mathrm{eq}}[\mathrm{Cl}^{-}]_{\mathrm{eq}} \]

Substitute:

\[ 1.80\times10^{-10} = (1.88\times10^{-4}+x)x \]

Expand:

\[ x^2+1.88\times10^{-4}x -1.80\times10^{-10}=0 \]

Using the quadratic formula and taking the positive root:

\[ x=9.53\times10^{-7}\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ \boxed{ [\mathrm{Cl}^{-}]_{\mathrm{eq}} =9.53\times10^{-7}\ \mathrm{mol\,L^{-1}} } \]

Step 6: Calculate the amount of silver chloride formed.

The initial amount of chloride ions was:

\[ n(\mathrm{Cl}^{-}) = (2.00\times10^{-5})(0.0600) \]

\[ n(\mathrm{Cl}^{-})=1.20\times10^{-6}\ \mathrm{mol} \]

The amount of chloride remaining at equilibrium is:

\[ n(\mathrm{Cl}^{-})_{\mathrm{eq}} = (9.53\times10^{-7})(0.1000) \]

\[ n(\mathrm{Cl}^{-})_{\mathrm{eq}} =9.53\times10^{-8}\ \mathrm{mol} \]

The difference has formed silver chloride:

\[ n(\ce{AgCl}) = 1.20\times10^{-6} - 9.53\times10^{-8} \]

\[ n(\ce{AgCl}) =1.105\times10^{-6}\ \mathrm{mol} \]

Convert to mass:

\[ m=nM \]

\[ m=(1.105\times10^{-6})(143.32) \]

\[ m=1.58\times10^{-4}\ \mathrm{g} \]

Therefore:

\[ \boxed{m(\ce{AgCl})=0.158\ \mathrm{mg}} \]

The important distinction is that the initial ion product determines whether precipitation starts. Once it does, you need a mole balance and the equilibrium expression to determine how much solid forms.

Question 4

A solution at \(25^\circ\mathrm{C}\) contains:

\[ [\mathrm{Cl}^{-}] =2.00\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{CrO_{4}}^{2-}] =1.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Silver nitrate is slowly added, increasing the silver ion concentration.

Two possible precipitates are silver chloride and silver chromate.

Their solubility products are:

\[ K_{sp}(\ce{AgCl})=1.80\times10^{-10} \]

\[ K_{sp}(\ce{Ag2CrO4})=1.10\times10^{-12} \]

(a) Calculate the minimum silver ion concentration required to start forming each precipitate.

(b) Determine which precipitate forms first.

(c) Explain why comparing the numerical \(K_{sp}\) values alone would give an unreliable prediction.

Assume the initial anion concentrations remain effectively constant until the first precipitation event.

Solution 4

Silver chloride forms first, when the silver ion concentration reaches:

\[ \boxed{ [\mathrm{Ag}^{+}] =9.00\times10^{-6}\ \mathrm{mol\,L^{-1}} } \]

Silver chromate requires a higher silver ion concentration:

\[ \boxed{ [\mathrm{Ag}^{+}] =3.32\times10^{-5}\ \mathrm{mol\,L^{-1}} } \]

The key is that the two solids have different dissolution stoichiometries and therefore different \(K_{sp}\) expressions.

Step 1: Calculate the silver ion threshold for silver chloride.

The dissolution equation is:

\[ \ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)} \]

Therefore:

\[ K_{sp}=[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \]

Rearrange:

\[ [\mathrm{Ag}^{+}]_{\min} = \frac{K_{sp}}{[\mathrm{Cl}^{-}]} \]

Substitute:

\[ [\mathrm{Ag}^{+}]_{\min} = \frac{1.80\times10^{-10}} {2.00\times10^{-5}} \]

\[ \boxed{ [\mathrm{Ag}^{+}]_{\min} =9.00\times10^{-6}\ \mathrm{mol\,L^{-1}} } \]

Step 2: Calculate the silver ion threshold for silver chromate.

The dissolution equation is:

\[ \ce{Ag2CrO4(s) <=> 2Ag+(aq) + CrO4^2-(aq)} \]

Therefore:

\[ K_{sp} = [\mathrm{Ag}^{+}]^2[\mathrm{CrO_{4}}^{2-}] \]

Rearrange:

\[ [\mathrm{Ag}^{+}]^2 = \frac{K_{sp}}{[\mathrm{CrO_{4}}^{2-}]} \]

Substitute:

\[ [\mathrm{Ag}^{+}]^2 = \frac{1.10\times10^{-12}} {1.00\times10^{-3}} \]

\[ [\mathrm{Ag}^{+}]^2 =1.10\times10^{-9} \]

Take the square root:

\[ [\mathrm{Ag}^{+}]_{\min} = \sqrt{1.10\times10^{-9}} \]

\[ \boxed{ [\mathrm{Ag}^{+}]_{\min} =3.32\times10^{-5}\ \mathrm{mol\,L^{-1}} } \]

Step 3: Compare the two thresholds.

Silver chloride requires:

\[ 9.00\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

Silver chromate requires:

\[ 3.32\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Silver chloride therefore begins precipitating first.

Step 4: Check the other possible precipitate at the first threshold.

At the point where silver chloride starts forming, the silver ion concentration is \(9.00\times10^{-6}\ \mathrm{mol\,L^{-1}}\).

Calculate the silver chromate ion product:

\[ Q_{sp}(\ce{Ag2CrO4}) = (9.00\times10^{-6})^2 (1.00\times10^{-3}) \]

\[ Q_{sp}=8.10\times10^{-14} \]

Compare:

\[ 8.10\times10^{-14} < 1.10\times10^{-12} \]

Silver chromate has not reached its precipitation threshold.

This confirms that silver chloride forms first.

Step 5: Explain the misleading \(K_{sp}\) comparison.

Silver chromate has a smaller numerical \(K_{sp}\) than silver chloride.

A student might conclude that silver chromate must therefore precipitate first.

However, this ignores two important factors:

• The initial concentrations of the anions are different.
• The silver ion has a squared exponent in the silver chromate expression, but only a first-power exponent in the silver chloride expression.

The order of precipitation depends on the complete ion product expressions and the existing ion concentrations, not simply on which solid has the smaller \(K_{sp}\).

Question 5

This final problem combines solubility equilibrium with acid-base equilibrium.

A buffered solution at \(25^\circ\mathrm{C}\) contains:

\[ [\mathrm{Ca}^{2+}] =1.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Fluoride ions can react with hydrogen ions to form hydrofluoric acid:

\[ \ce{HF(aq) <=> H+(aq) + F-(aq)} \]

The acid dissociation constant of HF is:

\[ K_a=6.80\times10^{-4} \]

The solubility product of calcium fluoride is:

\[ K_{sp}=3.90\times10^{-11} \]

The total dissolved fluoride concentration is:

\[ C_F=[\mathrm{HF}]+[\mathrm{F}^{-}] \]

Assume the buffer maintains the specified pH, calcium remains at its stated concentration until precipitation begins, and no other complexes or side reactions occur.

(a) Calculate the minimum total fluoride concentration needed to start calcium fluoride precipitation at pH 2.00.

(b) Calculate the minimum total fluoride concentration needed to start precipitation at pH 5.00.

(c) A solution contains \(3.00\times10^{-3}\ \mathrm{mol\,L^{-1}}\) total dissolved fluoride. Determine whether calcium fluoride precipitation is favoured at pH 2.00.

(d) The pH is then raised to 5.00 without changing the total fluoride concentration. Determine whether precipitation is now favoured.

Explain why the result changes even though the total fluoride concentration stays constant.

Solution 5

The minimum total fluoride concentrations are approximately:

At pH 2.00:

\[ \boxed{ C_{F,\min}=3.10\times10^{-3}\ \mathrm{mol\,L^{-1}} } \]

At pH 5.00:

\[ \boxed{ C_{F,\min}=2.00\times10^{-4}\ \mathrm{mol\,L^{-1}} } \]

The solution containing \(3.00\times10^{-3}\ \mathrm{mol\,L^{-1}}\) total fluoride does not precipitate calcium fluoride at pH 2.00, but precipitation becomes thermodynamically favoured when the pH rises to 5.00.

The crucial idea is that \(K_{sp}\) uses the concentration of free fluoride ions, not the total concentration of fluoride-containing species.

Step 1: Calculate the free fluoride concentration required for precipitation.

Calcium fluoride dissolves according to:

\[ \ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)} \]

Its solubility product is:

\[ K_{sp}=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]

At the precipitation threshold:

\[ 3.90\times10^{-11} = (1.00\times10^{-3})[\mathrm{F}^{-}]_{\min}^{2} \]

Rearrange:

\[ [\mathrm{F}^{-}]_{\min}^{2} = \frac{3.90\times10^{-11}}{1.00\times10^{-3}} \]

\[ [\mathrm{F}^{-}]_{\min}^{2} =3.90\times10^{-8} \]

Take the square root:

\[ \boxed{ [\mathrm{F}^{-}]_{\min} =1.97\times10^{-4}\ \mathrm{mol\,L^{-1}} } \]

This free fluoride concentration is the same threshold for both pH conditions because the calcium concentration and \(K_{sp}\) remain unchanged.

However, the total fluoride concentration needed to produce this amount of free fluoride depends on the pH.

Step 2: Calculate the HF-to-fluoride ratio at pH 2.00.

The acid dissociation expression is:

\[ K_a=\frac{[\mathrm{H}^{+}][\mathrm{F}^{-}]}{[\mathrm{HF}]} \]

Rearrange:

\[ \frac{[\mathrm{HF}]}{[\mathrm{F}^{-}]} = \frac{[\mathrm{H}^{+}]}{K_a} \]

At pH 2.00:

\[ [\mathrm{H}^{+}]=10^{-2.00} \]

\[ [\mathrm{H}^{+}]=1.00\times10^{-2}\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ \frac{[\mathrm{HF}]}{[\mathrm{F}^{-}]} = \frac{1.00\times10^{-2}} {6.80\times10^{-4}} \]

\[ \frac{[\mathrm{HF}]}{[\mathrm{F}^{-}]} =14.7 \]

This means there are approximately 14.7 moles of HF for every mole of free fluoride ions.

Step 3: Calculate the total fluoride concentration required at pH 2.00.

We know:

\[ C_F=[\mathrm{HF}]+[\mathrm{F}^{-}] \]

Since:

\[ [\mathrm{HF}]=14.7[\mathrm{F}^{-}] \]

we can write:

\[ C_F=14.7[\mathrm{F}^{-}]+[\mathrm{F}^{-}] \]

\[ C_F=15.7[\mathrm{F}^{-}] \]

At the precipitation threshold:

\[ C_{F,\min} = 15.7(1.97\times10^{-4}) \]

\[ \boxed{ C_{F,\min} =3.10\times10^{-3}\ \mathrm{mol\,L^{-1}} } \]

This is the minimum total dissolved fluoride concentration at pH 2.00.

Most of the fluoride is present as HF, so a relatively large total concentration is needed to produce the required free fluoride concentration.

Step 4: Repeat the calculation at pH 5.00.

At pH 5.00:

\[ [\mathrm{H}^{+}]=10^{-5} \]

Using the same acid dissociation expression:

\[ \frac{[\mathrm{HF}]}{[\mathrm{F}^{-}]} = \frac{1.00\times10^{-5}} {6.80\times10^{-4}} \]

\[ \frac{[\mathrm{HF}]}{[\mathrm{F}^{-}]} =0.0147 \]

Therefore:

\[ C_F=0.0147[\mathrm{F}^{-}]+[\mathrm{F}^{-}] \]

\[ C_F=1.0147[\mathrm{F}^{-}] \]

At the precipitation threshold:

\[ C_{F,\min} = 1.0147(1.97\times10^{-4}) \]

\[ \boxed{ C_{F,\min} =2.00\times10^{-4}\ \mathrm{mol\,L^{-1}} } \]

The required total fluoride concentration is now much lower.

At the higher pH, much less fluoride is tied up as HF, so a greater proportion exists as free \(\mathrm{F}^{-}\).

Step 5: Determine whether precipitation occurs at pH 2.00 when total fluoride is \(3.00\times10^{-3}\ \mathrm{mol\,L^{-1}}\).

At pH 2.00:

\[ C_F=15.7[\mathrm{F}^{-}] \]

Therefore:

\[ [\mathrm{F}^{-}] = \frac{3.00\times10^{-3}}{15.7} \]

\[ [\mathrm{F}^{-}] =1.91\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Calculate the ion product:

\[ Q_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]

\[ Q_{sp} = (1.00\times10^{-3}) (1.91\times10^{-4})^2 \]

\[ Q_{sp}=3.65\times10^{-11} \]

Compare:

\[ 3.65\times10^{-11} < 3.90\times10^{-11} \]

Therefore, calcium fluoride precipitation is not thermodynamically favoured at pH 2.00.

The total fluoride concentration is slightly below the threshold calculated in part (a).

Step 6: Determine what happens when the pH increases to 5.00.

The total fluoride concentration remains:

\[ C_F=3.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

However, the HF-to-fluoride ratio has changed.

At pH 5.00:

\[ C_F=1.0147[\mathrm{F}^{-}] \]

Therefore:

\[ [\mathrm{F}^{-}] = \frac{3.00\times10^{-3}}{1.0147} \]

\[ [\mathrm{F}^{-}] =2.96\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Now calculate the ion product:

\[ Q_{sp} = (1.00\times10^{-3}) (2.96\times10^{-3})^2 \]

\[ Q_{sp}=8.74\times10^{-9} \]

Compare:

\[ 8.74\times10^{-9} > 3.90\times10^{-11} \]

Therefore, calcium fluoride precipitation is thermodynamically favoured at pH 5.00.

Why did the result change?

The tempting prediction is that the precipitation behaviour should remain unchanged because the total fluoride concentration has not changed.

But \(K_{sp}\) does not use total fluoride concentration. It uses free fluoride ion concentration.

Increasing the pH reduces the hydrogen ion concentration. The HF equilibrium consequently shifts towards producing more free fluoride ions:

\[ \ce{HF(aq) <=> H+(aq) + F-(aq)} \]

The increased free fluoride concentration raises \(Q_{sp}\) above \(K_{sp}\), making calcium fluoride precipitation favourable.

This is the important distinction: a change in pH can change the concentration of an ion participating in a precipitation equilibrium without changing the total concentration of that element in solution.

08Where to go next

Once you can calculate the minimum concentration needed to start precipitation, the next step is working out how much solid forms after that threshold is crossed.

That requires combining the \(K_{sp}\) expression with mole ratios and mass conservation, as in Question 3.

Keep the two tasks separate:

• Precipitation threshold: Set \(Q_{sp}=K_{sp}\) and solve for the required free-ion concentration.
• Amount of precipitate formed: Use equilibrium concentrations and stoichiometric mole balances to determine how much solid is produced.

Being able to distinguish these two calculations makes unfamiliar precipitation problems much easier to approach.