University Chemistry — Year 1 · Bachelor Year 1

12Complexation

Pour a little ammonia solution into pale blue copper(II) sulfate: a pale blue precipitate forms at once. Pour more, and the precipitate dissolves into a solution of a deep, almost ink-like blue. Nothing in the acid–base or precipitation chapters explains the second step. The ammonia molecules have bound to the copper ion through their lone pairs and made a new species, a complex, more stable than the hydroxide and of a different colour. Complexes carry oxygen in the blood, hold the metal in chlorophyll and in vitamin B $_{12}$ , soften hard water and dissolve metals that nothing else will. This chapter treats them as one more exchange of particles in water, with its own constants and diagrams.

You already know

Lewis acids (electron-pair acceptors), Lewis bases (electron-pair donors) and the dative bond were defined in Chapter 3. The predominance diagrams and the predominant-reaction method come from Chapter 10; the solubility product and the condition of precipitation from Chapter 11.

Left: a suspension of pale blue copper(II) hydroxide (photograph Alvy16, CC BY 4.0). Right: with more ammonia the solid dissolves into the deep blue tetraamminecopper(II) ion; some undissolved hydroxide is still at the bottom of the tube (photograph Chemicalinterest, public domain). Both Wikimedia Commons.

12.1 Complexes and ligands

Definition 12.1 (Complex, central atom, ligand)

A complex is a species made of a central atom or ion, usually a metal cation acting as a Lewis acid, bound by dative bonds to molecules or ions acting as Lewis bases, the ligands. Its formula is written in square brackets, with its total charge: $[\ce{Cu(NH3)4}]^{2+}$ , $[\ce{Fe(CN)6}]^{4-}$ , $[\ce{Ag(NH3)2}]^{+}$ , $[\ce{FeSCN}]^{2+}$ .

The charge of the complex is the sum of the charges of the central ion and of the ligands: $\mathrm{Fe^{2+}}$ and six $\ce{CN-}$ make $[\ce{Fe(CN)6}]^{4-}$ . A ligand needs at least one lone pair: $\ce{NH3}$ , $\ce{H2O}$ , $\ce{OH-}$ , $\ce{Cl-}$ , $\ce{CN-}$ , $\ce{SCN-}$ (bound through S or N). Every metal ion in water is in fact already a complex with water molecules as ligands, $[\ce{Cu(H2O)6}]^{2+}$ or $[\ce{Fe(H2O)6}]^{3+}$ ; forming another complex means exchanging those water molecules for other ligands, and the water ligands are left out of the formulas, as $\ce{H3O+}$ is written for the hydrated proton.

Three complexes and their shapes. Each line is a dative bond from the lone pair of the ligand (the N of ammonia, the C of cyanide) to the metal ion; dashed lines outline the square of four ligands. The shapes of complexes are explained in the Year 2 volume.

Definition 12.2 (Polydentate ligand, chelate)

A ligand that binds the same central ion through several of its atoms at once is a polydentate ligand; the complex it forms, in which the ligand closes rings around the metal ion, is a chelate. Ethylenediamine $\ce{H2N-CH2-CH2-NH2}$ binds through two N atoms, the oxalate ion $\ce{C2O4^2-}$ through two O atoms, and the ethylenediaminetetraacetate ion (edta, written $\ce{Y^4-}$ ) through two N and four O atoms.

Left: the edta ion

\ce{Y^4-}

, with its two N atoms and four carboxylate groups. Right: the calcium–edta chelate

[\ce{CaY}]^{2-}

, schematic: the six donor atoms (blue bonds) sit at the corners of an octahedron around the ion, the two N atoms side by side, and the carbon chains of the ligand (orange) close five rings of five atoms each, the chelate rings.

The edta ion holds a calcium ion firmly enough to be the reagent of the titration of water hardness (Chapter 15).

12.2 Formation and dissociation constants

Definition 12.3 (Formation and dissociation constants)

For a central ion M and a ligand L (charges omitted), the overall formation constant of $\mathrm{ML}_n$ is the constant $\beta_n$ of $\mathrm{M} + n\,\mathrm{L} \rightleftharpoons \mathrm{ML}_n$ ,

\beta_n = \frac{[\mathrm{ML}_n]}{[\mathrm{M}][\mathrm{L}]^n} .

The successive formation constant $K_i$ is that of the addition of one ligand, $\mathrm{ML}_{i-1} + \mathrm{L} \rightleftharpoons \mathrm{ML}_i$ . The dissociation constant is the inverse, $K_d = 1/K_i$ , and $\mathrm{p}K_d = \log K_i$ , so that a complex/ligand pair is described exactly like an acid–base pair, with the ligand in place of the proton.

Proposition 12.4 (Overall and successive constants)

$\beta_n = K_1 K_2 \cdots K_n$ , that is $\log\beta_n = \sum_{i=1}^n \log K_i$ and $\log K_i = \log\beta_i - \log\beta_{i-1}$ (with $\beta_0 = 1$ ).

Proof. The formation of $\mathrm{ML}_n$ from M is the sum of the $n$ successive additions; the constant of a sum of reactions is the product of their constants (Chapter 7). Directly: $K_1 K_2\cdots K_n = \frac{[\mathrm{ML}]}{[\mathrm{M}][\mathrm{L}]} \frac{[\mathrm{ML_2}]}{[\mathrm{ML}][\mathrm{L}]}\cdots \frac{[\mathrm{ML}_n]}{[\mathrm{ML}_{n-1}][\mathrm{L}]}$ , and the intermediate concentrations cancel. ∎

complex	$\log\beta_n$	$\log K_i$
$[\ce{Cu(NH3)_n}]^{2+}$ , $n = 1$ to 4	4.10, 7.51, 10.33, 12.36	4.10, 3.41, 2.82, 2.03
$[\ce{Ag(NH3)_n}]^{+}$ , $n = 1, 2$	3.32, 7.22	3.32, 3.90
$[\ce{Zn(OH)4}]^{2-}$	$\log\beta_4 = 14.45$
$[\ce{FeSCN}]^{2+}$ , $[\ce{FeF}]^{2+}$	2.96; 6.85
$[\ce{CaY}]^{2-}$ , $[\ce{MgY}]^{2-}$ , $[\ce{NiY}]^{2-}$	12.69, 10.90, 20.54

Formation constants at

25\,{}^{\circ}\mathrm{C}

. The ammine, hydroxo, thiocyanato and fluoro constants are computed from tabulated standard Gibbs energies of formation; the edta constants are critically selected values at zero ionic strength, as are the

\mathrm{p}K_a

of edta used below, 2.23, 3.15, 6.80 and 11.24.

The successive constants of the copper ammines decrease, $K_1 > K_2 > K_3 > K_4$ : each ammonia molecule added leaves fewer sites and makes the complex a little less eager for the next one. This is the usual order. Silver is an exception: $K_2 > K_1$ , with consequences drawn in the weekend problem.

12.3 Predominance in pL

Definition 12.5 (pL scale)

For a ligand L, $\mathrm{pL} = -\log[\mathrm{L}]$ , where $[\mathrm{L}]$ is the concentration of the free ligand. A pL scale is an axis of pL on which the domains of predominance of the central ion and of its complexes are drawn, as the domains of acids and bases are drawn on a pH axis.

Proposition 12.6 (Boundaries in pL)

For the pair $\mathrm{ML}_i/\mathrm{ML}_{i-1}$ ,

\mathrm{pL} = \log K_i + \log\frac{[\mathrm{ML}_{i-1}]}{[\mathrm{ML}_i]} ,

so that $\mathrm{ML}_{i-1}$ predominates for $\mathrm{pL} > \log K_i$ and $\mathrm{ML}_i$ for $\mathrm{pL} < \log K_i$ . When the $\log K_i$ decrease with $i$ , the diagram is a sequence of domains: $\mathrm{ML}_n$ at small pL (much ligand), then $\mathrm{ML}_{n-1}$ , down to M at large pL.

Proof. Take the logarithm of $K_i = [\mathrm{ML}_i]/([\mathrm{ML}_{i-1}][\mathrm{L}])$ : $\log K_i = \log\frac{[\mathrm{ML}_i]}{[\mathrm{ML}_{i-1}]} + \mathrm{pL}$ . The two forms are equal at $\mathrm{pL} = \log K_i$ , and the ratio moves by a factor ten per unit of pL. The domain of $\mathrm{ML}_i$ lies between $\log K_{i+1}$ and $\log K_i$ , which exists only if $\log K_{i+1} < \log K_i$ . ∎

Method 12.7 (Drawing a pL diagram)

From the $\log\beta_n$ , compute the successive $\log K_i$ .
If they decrease, place each at its boundary on the pL axis: the complex with the most ligands on the left (small pL), the free ion on the right.
If some $\log K_{i+1} > \log K_i$ , the complex $\mathrm{ML}_i$ has no domain: it reacts with itself into $\mathrm{ML}_{i-1}$ and $\mathrm{ML}_{i+1}$ . Remove it and draw a single boundary between $\mathrm{ML}_{i-1}$ and $\mathrm{ML}_{i+1}$ at $\frac12(\log K_i + \log K_{i+1})$ .

Top: fractions of copper(II) present as Cu2+ and as [ Cu(NH3)_n]2+ (n = 1 to 4) against pNH_3. Middle: the predominance diagram of the copper ammines, boundaries at the K_i. Bottom: silver, where K_2 = 3.90 > K_1 = 3.32 (dotted): [ AgNH3]+ has no domain and a single boundary lies at 1/2 _2 = 3.61. — Top: fractions of copper(II) present as $\ce{Cu^2+}$ and as $[\ce{Cu(NH3)_n}]^{2+}$ ( $n = 1$ to 4) against $\mathrm{pNH_3}$ . Middle: the predominance diagram of the copper ammines, boundaries at the $\log K_i$ . Bottom: silver, where $\log K_2 = 3.90 > \log K_1 = 3.32$ (dotted): $[\ce{AgNH3}]^{+}$ has no domain and a single boundary lies at $\frac12\log\beta_2 = 3.61$ .

Example 12.8 (Copper in $0.010\,\mathrm{mol}/\mathrm{L}$ ammonia)

At free $[\ce{NH3}] = 0.010\,\mathrm{mol}/\mathrm{L}$ , $\mathrm{pNH_3} = 2.0$ lies in the domain of $[\ce{Cu(NH3)3}]^{2+}$ but close to the boundary 2.03 with $[\ce{Cu(NH3)4}]^{2+}$ : the two are nearly equal. The fractions are proportional to $\beta_n[\ce{NH3}]^n = 10^{\log\beta_n - 2n}$ , that is $1 : 10^{2.10} : 10^{3.51} : 10^{4.33} : 10^{4.36}$ , so $48\,\%$ of $[\ce{Cu(NH3)4}]^{2+}$ , $45\,\%$ of $[\ce{Cu(NH3)3}]^{2+}$ , $7\,\%$ of $[\ce{Cu(NH3)2}]^{2+}$ and almost no free $\ce{Cu^2+}$ .

12.4 Competitions

A ligand can be claimed by two metal ions, a metal ion by two ligands, and both are also engaged in precipitation and acid–base equilibria. Each competition is one more reaction whose constant combines the constants already known; the predominant-reaction method of Chapter 10 then applies unchanged.

Example 12.9 (Two ligands for iron(III))

A drop of thiocyanate in an iron(III) solution gives the blood-red $[\ce{FeSCN}]^{2+}$ ; fluoride ions decolourise it:

[\ce{FeSCN}]^{2+} + \ce{F-} \rightleftharpoons [\ce{FeF}]^{2+} + \ce{SCN-}, \qquad K = \frac{\beta(\ce{FeF})}{\beta(\ce{FeSCN})} = 10^{6.85 - 2.96} = 10^{3.89} .

The stronger complex (larger $\beta$ ) takes the metal ion, exactly as the stronger acid gives its proton to the stronger base.

Proposition 12.10 (Complexation against precipitation)

A solid MX dissolves in a ligand L by $\mathrm{MX(s)} + n\,\mathrm{L} \rightleftharpoons \mathrm{ML}_n + \mathrm{X}$ , of constant $K = \beta_n K_s$ . If the complex is the only form of M in solution and the free ligand concentration is $[\mathrm{L}]$ , the solubility is $s = \sqrt{K}\,[\mathrm{L}]^{n/2}$ .

Proof. The reaction is the sum of the dissolution (constant $K_s$ ) and of the formation of $\mathrm{ML}_n$ ( $\beta_n$ ). At saturation $[\mathrm{ML}_n] = [\mathrm{X}] = s$ and $K = s^2/[\mathrm{L}]^n$ . ∎

Method 12.11 (Dissolving a precipitate by complexation)

Write the dissolution reaction into the complex and compute $K = \beta_n K_s$ .
If $K$ is large, the dissolution is quantitative as long as enough ligand is present; if it is small, compute the free ligand concentration needed for the amount of solid to dissolve: $[\mathrm{L}]^n = s^2/K$ .
Add the ligand bound in the complex, $n s$ , to obtain the total ligand to introduce.
Check that the free metal ion is negligible: $[\mathrm{M}] = K_s/s \ll s$ .

With ammonia, silver chloride ( $K = 10^{7.22 - 9.75} = 10^{-2.53}$ ) dissolves in moderately concentrated solution while silver iodide ( $K = 10^{-8.85}$ ) does not: the weekend problem uses this to tell the halides apart.

Proposition 12.12 (Complexation against acidity)

If the ligand is a base $\mathrm{L}$ of an acid $\mathrm{HL}$ (and possibly more protonated forms), only the fraction $[\mathrm{L}]/c_L = 1/\alpha$ of the ligand not bound to the metal is available, with

\alpha = 1 + \frac{[\ce{H3O+}]}{K_{an}} + \frac{[\ce{H3O+}]^2}{K_{an}K_{a(n-1)}} + \cdots

( $K_{an}$ the last acidity constant). At a fixed pH the complexation then behaves as if its constant were the conditional constant $\beta' = \beta/\alpha$ , smaller the lower the pH.

Proof. The ligand not bound to the metal is shared among L, HL, $\ce{H2L}$ and so on, with $[\mathrm{HL}] = [\mathrm{L}][\ce{H3O+}]/K_{an}$ and similar relations for the next forms (Chapter 10). Summing gives $c_L = \alpha[\mathrm{L}]$ , and $\beta = [\mathrm{ML}]/([\mathrm{M}][\mathrm{L}]) = \alpha[\mathrm{ML}]/([\mathrm{M}]c_L)$ , so $[\mathrm{ML}]/([\mathrm{M}]c_L) = \beta/\alpha$ . ∎

For edta at pH 10, $\alpha = 1 + 10^{11.24 - 10} + 10^{11.24 + 6.80 - 20} = 18.4$ , so the calcium complex keeps $\log\beta' = 12.69 - 1.26 = 11.43$ ; at pH 7 it falls to 8.24 (Exercise 12.9). This is why a titration of calcium with edta is run in an ammonia buffer at pH 10. Ammonia, too, is a base: in acid it becomes $\ce{NH4+}$ , which has no lone pair, and the ammine complexes fall apart. Adding nitric acid to $[\ce{Ag(NH3)2}]^{+}$ in the presence of chloride brings back the white precipitate of silver chloride.

Photographic fixers do the same with another ligand of the silver ion, the thiosulfate ion $\ce{S2O3^2-}$ : they dissolve the silver halide that light has not touched, so that the picture no longer darkens. Its constants are not among the data of this volume; ammonia shows the same chemistry with silver chloride.

12.5 Exercises

Exercise 12.1 ★

For each complex give the central ion, its charge, the ligands and the atom through which each ligand binds: $[\ce{Cu(NH3)4}]^{2+}$ , $[\ce{Fe(CN)6}]^{4-}$ , $[\ce{Fe(CN)6}]^{3-}$ , $[\ce{Ag(NH3)2}]^{+}$ , $[\ce{FeSCN}]^{2+}$ , $[\ce{Zn(OH)4}]^{2-}$ , $[\ce{CaY}]^{2-}$ .

Solution

Solution of Exercise 12.1.

$[\ce{Cu(NH3)4}]^{2+}$ : $\ce{Cu^2+}$ , four $\ce{NH3}$ bound through N. $[\ce{Fe(CN)6}]^{4-}$ : $\ce{Fe^2+}$ , six $\ce{CN-}$ bound through C; $[\ce{Fe(CN)6}]^{3-}$ : the same with $\ce{Fe^3+}$ . $[\ce{Ag(NH3)2}]^{+}$ : $\ce{Ag+}$ , two $\ce{NH3}$ through N. $[\ce{FeSCN}]^{2+}$ : $\ce{Fe^3+}$ , one $\ce{SCN-}$ , bound through N in spite of the way the formula is written. $[\ce{Zn(OH)4}]^{2-}$ : $\ce{Zn^2+}$ , four $\ce{OH-}$ through O. $[\ce{CaY}]^{2-}$ : $\ce{Ca^2+}$ , one edta ion through two N and four O.

Exercise 12.2 ★

From the $\log\beta_n$ of the copper ammines, compute the four $\log K_i$ and the four $\mathrm{p}K_d$ . Write the dissociation reaction to which the last $\mathrm{p}K_d$ refers.

Solution

Solution of Exercise 12.2.

$\log K_i = \log\beta_i - \log\beta_{i-1}$ : 4.10, 3.41, 2.82, 2.03, and $\mathrm{p}K_{d,i} = \log K_i$ takes the same values. The last refers to $[\ce{Cu(NH3)4}]^{2+} \rightleftharpoons [\ce{Cu(NH3)3}]^{2+} + \ce{NH3}$ .

Exercise 12.3 ★

Draw the pL diagram of the copper ammines. Which copper species predominates at $[\ce{NH3}] = 1.0$ , $3.2 \times 10^{-3}$ , $3.2 \times 10^{-4}$ and $1.0 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}$ ?

Solution

Solution of Exercise 12.3.

Domains: $[\ce{Cu(NH3)4}]^{2+}$ below pNH3 2.03, then $n = 3$ up to 2.82, $n = 2$ up to 3.41, $n = 1$ up to 4.10, $\ce{Cu^2+}$ above. pNH3 = 0: $[\ce{Cu(NH3)4}]^{2+}$ ; 2.5: $[\ce{Cu(NH3)3}]^{2+}$ ; 3.5: $[\ce{CuNH3}]^{2+}$ ; 5.0: $\ce{Cu^2+}$ .

Exercise 12.4 ★

A solution contains iron(III) at $1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ and thiocyanate at $0.10\,\mathrm{mol}/\mathrm{L}$ . Neglecting the thiocyanate bound, what fraction of the iron is present as $[\ce{FeSCN}]^{2+}$ ?

Solution

Solution of Exercise 12.4.

$[\ce{FeSCN^2+}]/[\ce{Fe^3+}] = \beta_1[\ce{SCN-}] = 10^{2.96} \times 0.10 = 91$ : the fraction complexed is $91/92 = 98.9\,\%$ .

Exercise 12.5 ★★

Prove that in a solution of a metal ion M and its complexes $\mathrm{ML}_n$ , the fraction of $\mathrm{ML}_n$ is $\beta_n[\mathrm{L}]^n / \sum_{k} \beta_k[\mathrm{L}]^k$ . Show that, if only $\mathrm{ML}_{i-1}$ , $\mathrm{ML}_i$ and $\mathrm{ML}_{i+1}$ are present in significant amounts, the fraction of $\mathrm{ML}_i$ is largest where $[\mathrm{ML}_{i-1}] = [\mathrm{ML}_{i+1}]$ , at $\mathrm{pL} = \frac12(\log K_i + \log K_{i+1})$ , and compute this pNH3 for $[\ce{Cu(NH3)2}]^{2+}$ .

Solution

Solution of Exercise 12.5.

Each $[\mathrm{ML}_k] = \beta_k[\mathrm{M}][\mathrm{L}]^k$ ; dividing by the sum over $k$ (the total metal) eliminates $[\mathrm{M}]$ . With three species, $f_i = 1/(1 + [\mathrm{ML}_{i-1}]/[\mathrm{ML}_i] + [\mathrm{ML}_{i+1}]/[\mathrm{ML}_i]) = 1/(1 + 1/(K_i x) + K_{i+1}x)$ with $x = [\mathrm{L}]$ . The denominator is smallest when its derivative $-1/(K_i x^2) + K_{i+1}$ vanishes, $x^2 = 1/(K_iK_{i+1})$ , where the two neighbours are equal, and $\mathrm{pL} = \frac12(\log K_i + \log K_{i+1})$ . For $[\ce{Cu(NH3)2}]^{2+}$ : $\frac12(3.41 + 2.82) = 3.12$ .

Exercise 12.6 ★★

To the red solution of Exercise 12.4, sodium fluoride is added until $[\ce{F-}] = 0.010\,\mathrm{mol}/\mathrm{L}$ (free). Compute the constant of the exchange of ligands and the ratio $[\ce{FeF^2+}]/[\ce{FeSCN^2+}]$ . What does one see?

Solution

Solution of Exercise 12.6.

$[\ce{FeSCN}]^{2+} + \ce{F-} \rightleftharpoons [\ce{FeF}]^{2+} + \ce{SCN-}$ , $K = 10^{6.85 - 2.96} = 10^{3.89}$ . Ratio $= K[\ce{F-}]/[\ce{SCN-}] = 10^{3.89} \times 0.010/0.10 = 780$ : the red complex is almost entirely replaced and the colour fades.

Exercise 12.7 ★★

Calcium ions are added to a solution of $[\ce{MgY}]^{2-}$ , in equal amount. Write the exchange reaction, compute its constant and the fraction of magnesium set free.

Solution

Solution of Exercise 12.7.

$[\ce{MgY}]^{2-} + \ce{Ca^2+} \rightleftharpoons [\ce{CaY}]^{2-} + \ce{Mg^2+}$ , $K = 10^{12.69 - 10.90} = 10^{1.79} = 62$ . With equal amounts, $x^2/(1 - x)^2 = 62$ , $x/(1 - x) = 7.9$ , $x = 0.89$ : $89\,\%$ of the magnesium is set free.

Exercise 12.8 ★★

Silver oxide $\ce{Ag2O}$ is a brown solid: $\ce{Ag2O(s) + H2O <=> 2Ag+ + 2OH-}$ has $\mathrm{p}K = 15.43$ . Write its dissolution in ammonia into $[\ce{Ag(NH3)2}]^{+}$ , compute the constant, and explain why the brown precipitate formed by a little ammonia in silver nitrate dissolves in more ammonia.

Solution

Solution of Exercise 12.8.

$\ce{Ag2O(s) + H2O + 4NH3 <=> 2[Ag(NH3)2]+ + 2OH-}$ , $K = \beta_2^2 \times 10^{-15.43} = 10^{14.44 - 15.43} = 10^{-0.99}$ . The constant is not small: as soon as the free ammonia reaches a few tenths of a mole per litre, the brown oxide formed by the first drops (the ammonia acting as a base) dissolves into the silver ammine.

Exercise 12.9 ★★

Compute $\alpha$ and the conditional constant $\log\beta'$ of $[\ce{CaY}]^{2-}$ at pH 10 and at pH 7 (Proposition 12.12). Why is a titration of calcium by edta impossible in neutral solution if a conditional constant of at least $10^{10}$ is required?

Solution

Solution of Exercise 12.9.

pH 10: $\alpha = 1 + 10^{1.24} + 10^{-1.96} = 18.4$ , $\log\beta' = 12.69 - 1.26 = 11.43$ . pH 7: $\alpha = 1 + 10^{4.24} + 10^{4.04} + 10^{0.19} + \cdots = 10^{4.45}$ , $\log\beta' = 8.24$ . At pH 7 the conditional constant is below $10^{10}$ : the reaction of calcium with edta is not complete enough for a titration, because most of the edta is protonated.

Exercise 12.10 ★★★

Copper(II) oxide $\ce{CuO}$ is the solid that copper(II) hydroxide turns into on standing; $\ce{CuO(s) + H2O <=> Cu^2+ + 2OH-}$ has $\mathrm{p}K = 20.64$ .

Write the dissolution of $\ce{CuO}$ in ammonia into $[\ce{Cu(NH3)4}]^{2+}$ and compute its constant.
In $1.0\,\mathrm{mol}/\mathrm{L}$ ammonia alone, the pH is fixed by ammonia ( $\mathrm{p}K_a(\ce{NH4+}) = 9.25$ ). Compute the pH and the concentration of copper dissolved.
Same question in a buffer of ammonia and ammonium chloride, both at $1.0\,\mathrm{mol}/\mathrm{L}$ . Conclude.

Solution

Solution of Exercise 12.10.

1. $\ce{CuO(s) + H2O + 4NH3 <=> [Cu(NH3)4]^2+ + 2OH-}$ , $K = \beta_4 \times 10^{-20.64} = 10^{12.36 - 20.64} = 10^{-8.28}$ . 2. $\mathrm{pH} = 7 + \frac12(9.25 + \log 1.0) = 11.63$ , $[\ce{OH-}] = 10^{-2.37}$ , and $[\ce{Cu(NH3)4^2+}] = 10^{-8.28}/10^{-4.75} = 3.0 \times 10^{-4}\,\mathrm{mol}/\mathrm{L}$ : very little dissolves. 3. $\mathrm{pH} = 9.25$ , $[\ce{OH-}] = 10^{-4.75}$ , and the formula gives $10^{-8.28 + 9.50} = 10^{1.22}$ , far more than the ammonia can carry: the oxide dissolves until the ammonia, not the equilibrium, runs out. The ammonium ions consume the hydroxide that the dissolution releases; this is why ammonium salts help ammonia dissolve copper compounds.

Exercise 12.11 ★★★

Silver and ammonia.

Show that $[\ce{AgNH3}]^{+}$ reacts with itself, $\ce{2[AgNH3]+ <=> Ag+ + [Ag(NH3)2]+}$ , and compute the constant.
Compute the largest fraction of silver ever present as $[\ce{AgNH3}]^{+}$ , and the pNH3 where it occurs.

Solution

Solution of Exercise 12.11.

1. The reaction is the second formation minus the first: $K = K_2/K_1 = 10^{3.90 - 3.32} = 10^{0.58} = 3.8 > 1$ , so $[\ce{AgNH3}]^{+}$ reacts with itself when it is the main species. 2. By Exercise 12.5, the largest fraction is at $\mathrm{pNH_3} = \frac12(3.32 + 3.90) = 3.61$ , where $\beta_1 x = 10^{-0.29} = 0.51$ and $\beta_2x^2 = 1.0$ : $f = 0.51/(1 + 0.51 + 1.0) = 0.20$ . Never more than $20\,\%$ of the silver is $[\ce{AgNH3}]^{+}$ .

Exercise 12.12 ★★★

A solution of edta, $0.10\,\mathrm{mol}/\mathrm{L}$ of $\ce{Y^4-}$ at high pH, is used to dissolve calcium carbonate scale.

Write the reaction and compute its constant.
What mass of calcium carbonate can one litre dissolve?
Why must the solution be basic? (The carbonate and the edta ions are both bases.)

Solution

Solution of Exercise 12.12.

1. $\ce{CaCO3(s) + Y^4- <=> [CaY]^2- + CO3^2-}$ , $K = \beta K_s = 10^{12.69 - 8.30} = 10^{4.39}$ . 2. $x^2/(0.10 - x) = 10^{4.39}$ gives $x = 0.10\,\mathrm{mol}/\mathrm{L}$ to within $4 \times 10^{-7}$ : all the edta is used, $10.0\,\mathrm{g}$ of calcium carbonate per litre. 3. In acid, edta is protonated ( $\mathrm{p}K_{a4} = 11.24$ ) and its conditional constant drops; the carbonate is also protonated, which helps, but the complex is what holds the calcium. Keeping the pH high keeps the edta as $\ce{Y^4-}$ .

12.6 Problem: Ammonia and the Three Silver Halides

Problem 12.1

Weekend problem — the silver ammines and their inverted constants, dissolving silver chloride, bromide and iodide in ammonia, the confirmatory test, and the least ammonia that dissolves one gram of silver chloride

An analyst has three white-to-yellow precipitates and must say which is the chloride, the bromide and the iodide of silver; then she must dissolve $1.0\,\mathrm{g}$ of silver chloride in $1.00\,\mathrm{L}$ with as little ammonia as possible. Data at $25\,{}^{\circ}\mathrm{C}$ : $\log\beta_1 = 3.32$ and $\log\beta_2 = 7.22$ for the silver ammines; $\mathrm{p}K_s$ of $\ce{AgCl}$ 9.75, $\ce{AgBr}$ 12.27, $\ce{AgI}$ 16.07; $\mathrm{p}K_a(\ce{NH4+}) = 9.25$ ; molar masses $\ce{AgCl}$ $143.4\,\mathrm{g}/\mathrm{mol}$ , $\ce{AgBr}$ $187.8\,\mathrm{g}/\mathrm{mol}$ , $\ce{AgI}$ $234.8\,\mathrm{g}/\mathrm{mol}$ .

Part I — The silver ammines.

Write the formation reactions of $[\ce{AgNH3}]^{+}$ and $[\ce{Ag(NH3)2}]^{+}$ and the expressions of $\beta_1$ and $\beta_2$ .
Compute $\log K_1$ and $\log K_2$ .
Draw the pL diagram naively with these two boundaries. What is wrong with it?
Write the reaction of $[\ce{AgNH3}]^{+}$ with itself and compute its constant.
Draw the correct diagram; where is its single boundary?
At $[\ce{NH3}] = 0.10\,\mathrm{mol}/\mathrm{L}$ , compute $[\ce{Ag(NH3)2+}]/[\ce{Ag+}]$ .
Give the overall dissociation constant $\mathrm{p}K_d$ of $[\ce{Ag(NH3)2}]^{+}$ .

Part II — Silver chloride in ammonia.

Write the dissolution of $\ce{AgCl}$ in ammonia and compute its constant.
Express the solubility $s$ as a function of the free ammonia concentration and compute it for $[\ce{NH3}] = 1.0\,\mathrm{mol}/\mathrm{L}$ .
By how much has the solubility increased compared with pure water?
What mass of silver chloride does one litre then dissolve?
Check that free $\ce{Ag+}$ is negligible.
Now $1.0\,\mathrm{mol}/\mathrm{L}$ is the total ammonia introduced. Taking the ammonia bound into account, compute $s$ again.
Ammonia is a base; explain why its protonation can be neglected in this solution.

Part III — Bromide and iodide.

Compute the constant of dissolution of $\ce{AgBr}$ in ammonia.
Compute the solubility of $\ce{AgBr}$ in $1.0\,\mathrm{mol}/\mathrm{L}$ free ammonia, in $\mathrm{mol}/\mathrm{L}$ and $\mathrm{g}/\mathrm{L}$ .
Same for $\ce{AgI}$ , in $\mathrm{mg}/\mathrm{L}$ .
Deduce a way of identifying the three precipitates.
What free ammonia concentration would dissolve $1.0\,\mathrm{g}$ of $\ce{AgBr}$ in $1.00\,\mathrm{L}$ ?
Same for $1.0\,\mathrm{g}$ of $\ce{AgI}$ . Comment.

Part IV — One gram of silver chloride.

To the ammoniacal solution of silver chloride, nitric acid is added. Write the overall reaction and compute its constant. What does the analyst see?
What free ammonia concentration is needed to dissolve $1.0\,\mathrm{g}$ of $\ce{AgCl}$ in $1.00\,\mathrm{L}$ ?
How much ammonia is bound in the complex?
Check that free $\ce{Ag+}$ is negligible in this solution.
What is the least total concentration of ammonia that dissolves $1.0\,\mathrm{g}$ of silver chloride in $1.00\,\mathrm{L}$ ?

Solution

Solution of Problem 12.1.

1. $\ce{Ag+ + NH3 <=> [AgNH3]+}$ , $\beta_1 = [\ce{AgNH3+}]/([\ce{Ag+}][\ce{NH3}])$ ; $\ce{Ag+ + 2NH3 <=> [Ag(NH3)2]+}$ , $\beta_2 = [\ce{Ag(NH3)2+}]/([\ce{Ag+}][\ce{NH3}]^2)$ . 2. $\log K_1 = 3.32$ , $\log K_2 = 7.22 - 3.32 = 3.90$ . 3. $[\ce{AgNH3}]^{+}$ would predominate between pNH3 3.90 and 3.32, an interval that runs backwards: it has no domain. 4. $\ce{2[AgNH3]+ <=> Ag+ + [Ag(NH3)2]+}$ , $K = K_2/K_1 = 10^{0.58} = 3.8$ . 5. $[\ce{Ag(NH3)2}]^{+}$ below and $\ce{Ag+}$ above $\frac12\log\beta_2 = 3.61$ . 6. $\beta_2[\ce{NH3}]^2 = 10^{7.22 - 2.00} = 10^{5.22} = 1.7 \times 10^{5}$ . 7. $\mathrm{p}K_d = \log\beta_2 = 7.22$ . 8. $\ce{AgCl(s) + 2NH3 <=> [Ag(NH3)2]+ + Cl-}$ , $K = \beta_2K_s = 10^{7.22 - 9.75} = 10^{-2.53}$ . 9. $s = \sqrt{K}\,[\ce{NH3}] = 10^{-1.265} \times 1.0 = 0.054\,\mathrm{mol}/\mathrm{L}$ . 10. In water $s = 10^{-4.875} = 1.3 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}$ : about 4000 times more. 11. $0.0543 \times 143.4 = 7.8\,\mathrm{g}$ . 12. $[\ce{Ag+}] = K_s/[\ce{Cl-}] = 10^{-9.75}/0.054 = 3.3 \times 10^{-9}\,\mathrm{mol}/\mathrm{L} \ll s$ . 13. $s = \sqrt{K}(1.0 - 2s)$ , $s = 0.0543/(1 + 2 \times 0.0543) = 0.049\,\mathrm{mol}/\mathrm{L}$ . 14. $1.0\,\mathrm{mol}/\mathrm{L}$ ammonia has pH 11.6, more than $9.25 + 1$ : only $0.4\,\%$ of it is $\ce{NH4+}$ . 15. $K = 10^{7.22 - 12.27} = 10^{-5.05}$ . 16. $s = 10^{-2.525} = 3.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ , $0.56\,\mathrm{g}/\mathrm{L}$ . 17. $K = 10^{-8.85}$ , $s = 10^{-4.425} = 3.8 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}$ , $8.8\,\mathrm{mg}/\mathrm{L}$ . 18. Treat each precipitate with dilute ammonia: only the chloride dissolves. Concentrated ammonia dissolves the bromide, more slowly and partly; the iodide does not dissolve. 19. $s = 1.0/187.8 = 5.3 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ ; $[\ce{NH3}] = s/\sqrt{K} = 5.3 \times 10^{-3}/10^{-2.525} = 1.8\,\mathrm{mol}/\mathrm{L}$ . 20. $s = 4.3 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ and $[\ce{NH3}] = 4.3 \times 10^{-3}/10^{-4.425} = 113\,\mathrm{mol}/\mathrm{L}$ , twice the concentration of water in pure water ( $55\,\mathrm{mol}/\mathrm{L}$ ): impossible. Ammonia cannot dissolve silver iodide. 21. $\ce{[Ag(NH3)2]+ + Cl- + 2H3O+ -> AgCl(s) + 2NH4+ + 2H2O}$ , $K = 10^{2.53} \times (10^{9.25})^2 = 10^{21.03}$ : the white precipitate reappears, which confirms silver chloride. 22. $s = 1.0/143.4 = 6.97 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ ; $[\ce{NH3}] = s/\sqrt{K} = 6.97 \times 10^{-3}/10^{-1.265} = 0.128\,\mathrm{mol}/\mathrm{L}$ . 23. $2s = 0.0139\,\mathrm{mol}/\mathrm{L}$ . 24. $[\ce{Ag+}] = 10^{-9.75}/6.97 \times 10^{-3} = 2.6 \times 10^{-8}\,\mathrm{mol}/\mathrm{L}$ , negligible beside $s$ . 25. $0.128 + 0.014 =$ $0.142\,\mathrm{mol}/\mathrm{L}$ of ammonia: $0.142\,\mathrm{mol}$ in the litre dissolves exactly $1.0\,\mathrm{g}$ of silver chloride.

Terms defined in this chapter

See all 852 terms in the glossary