Complex Formation and Simultaneous Equilibria

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. There is a white precipitate of silver chloride at the bottom of a test tube. It hardly dissolves in water, right? Now you pour in ammonia solution. The white solid disappears! But ammonia brings no silver and no chloride. So where did the solid go? Today we find out: complex formation, and two equilibria working together. This idea does come up in JEE, so let's begin.

0:31Complex formation pulls a precipitate into solution

First, the equilibrium we know. Solid silver chloride, A g C l, with s in brackets for solid, gives A g plus, the silver ion, and C l minus, the chloride ion. The plus and minus up top are charges, and the double half arrows mean equilibrium. Its constant is K s p, the solubility product, with s p written small; it is really tiny. Okay? Now suppose something takes the silver ions away. Remember Q, the ionic product: like K s p, but with the actual concentrations. Is Q now above K s p, or below it? Below, yes, exactly! Q is less than K s p; each fat arrow means so. The solution is no longer saturated, so more solid dissolves. That is Le Chatelier's principle at work. With me? And who grabs the silver? Ammonia! One silver ion joins two ammonia molecules, N H three, with a small subscript three. These square brackets are not concentration: they hold the whole complex ion, with one plus charge. Its constant, K f, the formation constant, is large. Clear? Now add these two equations. The silver ion cancels. Solid silver chloride plus two ammonia gives the complex ion plus chloride. So what is the K of this overall reaction? Yes, exactly! Adding equilibria multiplies their constants, so K equals K s p times K f. Tiny K s p, times a large K f, gives a K far bigger than K s p. Puzzle solved: ammonia pulls the silver into the complex, and the precipitate dissolves. Got it?

2:10Hydroxides in excess alkali, and in ammonia

Now a second way. Take aluminium hydroxide, which hardly dissolves in water. By the common ion effect, extra hydroxide should lower its solubility. So in excess sodium hydroxide, it dissolves even less, right? No, no, careful! Aluminium hydroxide, A l O H three, is amphoteric: it reacts with acids, and with bases too. With excess O H minus, the hydroxide ion, it forms the aluminate ion: A l O H four in square brackets, with a minus charge, and that is soluble. The single arrow means the reaction goes forward. Okay? Zinc hydroxide, Z n O H two, does the same: with two hydroxide ions it gives Z n O H four, with a two minus charge. So both dissolve in excess alkali. Right? Now a trap. Aqueous ammonia is also a base. So aluminium hydroxide dissolves in excess ammonia too, yes? No, no, wait! Zinc hydroxide does dissolve in excess ammonia, because the zinc ion forms a stable complex with four ammonia molecules and a two plus charge. That lowers the zinc ion concentration, so Q falls below K s p. Same idea as silver chloride, okay? But aluminium hydroxide does not dissolve. The aluminium ion forms no ammine complex, that is, no complex with ammonia. And aqueous ammonia is too weakly basic to make aluminate. Remember this trap. With me?

3:36Two salts with a common ion

Last idea: two salts together. Take M X and M Y, sharing the cation M plus, both saturating the same water. K one is M plus times X minus; K two is M plus times Y minus. The small one and two just number the salts. Here the square brackets do mean concentration. Okay? Now the tempting shortcut: solve each salt as if it were alone, and add up. Will that work? No, careful! One solution means one M plus concentration, and it must satisfy both expressions. The M plus from one salt is a common ion for the other, so each dissolves less than it would alone. Right? So we need one more equation: the charge balance. Total positive charge equals total negative charge, so M plus equals X minus plus Y minus. Two K s p expressions, one charge balance. Got it?

4:27Example 1: solubility of AgCl in 1.0 M NH3

Now let's put numbers on our puzzle. Find the solubility of silver chloride in one point zero molar ammonia; capital M means molar. K s p is one point eight times ten to the minus ten, and K f of the complex is one point six times ten to the seven. Pause the video, and try it. Done? Okay, let's check. K equals K s p times K f. One point eight times one point six is about two point nine, and the powers of ten give ten to the minus three. The wavy equals sign means approximately. Okay? Now let s moles per litre dissolve. Each gives one complex ion and one chloride ion. So how much ammonia is used up? Two s, yes! Each silver ion takes two ammonia molecules. So the complex is s, chloride is s, and the ammonia left is one minus two s. Clear? Into K: complex times chloride, over ammonia squared. So s squared, over one minus two s whole squared, equals two point nine times ten to the minus three. Do we expand a quadratic? Or is there a faster way? Yes, exactly! The left side is a perfect square. Take the square root of both sides: s over one minus two s equals zero point zero five four. Topper's trick: no quadratic at all. With me? So the solubility is zero point zero five four molar, right? Oh no, careful! That is s over one minus two s, not s itself. Solve for s: about zero point zero four eight molar. Over three thousand times its solubility in water! That is why our precipitate disappeared. Okay?

6:09Example 2: water saturated with both AgCl and AgBr

Now a harder one. Water is saturated with both silver chloride, K s p one point eight times ten to the minus ten, and silver bromide, A g B r, K s p five point zero times ten to the minus thirteen. Find the silver, chloride and bromide ion concentrations. Pause and try. Hint: two K s p expressions, one charge balance. Done? Let's check. Call the two K s p values K one and K two. So chloride is K one over silver ion, and bromide is K two over silver ion. Silver is the only cation. Put both into the charge balance. What do you get? Very good! Silver ion equals chloride plus bromide: K one plus K two, all over silver ion. Silver ion on both sides, right? Multiply across: silver ion squared equals K one plus K two. One point eight times ten to the minus ten, plus five point zero times ten to the minus thirteen, gives one point eight zero five times ten to the minus ten. Clear? Take the square root: silver ion is one point three four times ten to the minus five molar. Chloride, K one over this, also comes to one point three four times ten to the minus five. So nearly all the silver comes from silver chloride. Okay? And bromide is five point zero times ten to the minus thirteen, over one point three four times ten to the minus five: three point seven times ten to the minus eight molar. Tiny! Silver from silver chloride is a common ion, and it pushes this solubility right down. That is why we never solve each salt alone. Got it?

7:44Quick revision: K = Ksp x Kf, amphoteric hydroxides

So, quick revision. One: a stable complex pulls a precipitate into solution, because K equals K s p times K f. Two: amphoteric aluminium and zinc hydroxides dissolve in excess alkali. Three: two salts with a common ion need both K s p expressions and one charge balance. Now try the two practice problems on this page. All the best!