Precipitation and Qualitative Analysis

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. I take a solution with chloride ions and chromate ions, both zero point zero one molar, and add silver nitrate, drop by drop. Silver chloride is a white solid; silver chromate is red. Silver chromate has the smaller solubility product, so the red solid comes out first, right? No! The white silver chloride comes out first, and the red appears only when almost all the chloride is gone. How? One simple comparison answers it, and it also runs the salt analysis lab. Questions on this come up often in JEE, so let's begin.

0:39The ionic product Q

So, first, a new quantity: the ionic product, written as capital Q on screen. It has the form of K s p, the solubility product; that is capital K with a small s p below it. But Q uses the actual concentrations in the solution right now. Okay? For silver chloride, Q is silver ion, A g plus, times chloride ion, C l minus. Square brackets mean concentration in moles per litre; the small plus and minus up top are the charges. Right? And one very important point. If two solutions are mixed, use the concentrations after mixing, with the new, diluted volume. Many students forget this. Got it?

1:25Compare Q with Ksp

Now, the main rule. Suppose Q is bigger than K s p. What will happen in the solution? Yes, exactly! The solid precipitates, and keeps coming out until Q falls back to K s p. The sign on screen, open towards Q, means greater than. With me? If Q is equal to K s p, the solution is just saturated. Clear? And if Q is less than K s p, the sign turned the other way, there is no precipitate at all. So this whole topic is one comparison: Q against K s p. Okay?

1:58Selective precipitation

Now, what if two ions in the solution can both precipitate with the same reagent? We add the reagent slowly, drop by drop. Right? So tell me, which salt will come out first? The one whose K s p is reached first, correct! As the reagent builds up, Q rises for both salts. Whichever touches its own K s p first, precipitates first. Got it? So, for each salt, find how much reagent it needs to reach its K s p. The smaller need comes out first. Not simply the smaller K s p; Example two shows why.

2:34Group II and group IV: H2S

Okay? Now the same idea runs the salt analysis lab. Group two, written with the Roman numeral two: copper, lead, cadmium, bismuth, mercury, arsenic, antimony and tin; their symbols are on screen. With me? They are precipitated by H two S gas in dilute hydrochloric acid, H C l; the small two counts the hydrogen atoms. The acid keeps sulphide ion concentration, S two minus, very low; two minus is its charge. Right? Now think. With so little sulphide, Q stays small. Which sulphides can still precipitate? Correct! Only the sulphides with a very small K s p. For them, even this tiny Q is bigger than K s p. Got it? Next, group four, the Roman numeral four: zinc, manganese, cobalt and nickel. In the acid, with so little sulphide, these stay dissolved. Clear? So we pass H two S into an ammoniacal solution, made basic with ammonia. Now the sulphide ion concentration is much larger, Q crosses their K s p, and they precipitate. Same gas, different conditions! Okay?

3:45Group III: NH4OH with NH4Cl

Now group three: iron three plus, aluminium three plus and chromium three plus. They are precipitated by ammonium hydroxide, N H four O H, in the presence of ammonium chloride, N H four C l. Right? But why do we add ammonium chloride first? The common ion, yes! Ammonium chloride gives N H four plus, the same ion that ammonium hydroxide gives. So the ionisation of ammonium hydroxide is held back, and hydroxide ion, O H minus, stays low. Got it? So hydroxides of higher K s p, like magnesium hydroxide and zinc hydroxide, stay dissolved; the small two after the bracket means two O H groups. Only iron, aluminium and chromium come down. Clear?

4:31Example 1: will AgCl precipitate on mixing?

Okay, Example one. Equal volumes of silver nitrate and sodium chloride, each two times ten to the minus four molar, are mixed; the cross means times, and capital M means molar. Will silver chloride precipitate? Its K s p is one point eight times ten to the minus ten. Pause the video, and try it first. Done? Okay, let's check. Silver ion is two times ten to the minus four molar, chloride is the same, and Q is their product, right? No, no, careful! That is the trap. Equal volumes are mixed, so the volume doubles, and each concentration is halved. Both ions become ten to the minus four molar. The wrong Q still says yes here, but in a closer case it misleads. Okay? Now Q is ten to the minus four, times ten to the minus four: ten to the minus eight. Right? Compare with K s p. Ten to the minus eight is far bigger than one point eight times ten to the minus ten. Q is greater than K s p, so yes, silver chloride precipitates. Very good!

5:40Example 2: AgCl or Ag2CrO4 first, Mohr's method

Now let us solve our puzzle as a real problem. Silver nitrate is added slowly to a solution that is zero point zero one molar in both chloride and chromate, C r O four two minus. K s p of silver chloride is one point eight times ten to the minus ten; of silver chromate, one point one times ten to the minus twelve. Which precipitates first, and what is the chloride concentration when the second begins? Pause and try. Hint: find the silver ion each salt needs. Done? Silver chromate has the smaller K s p, so it precipitates first, right? No, no, wait! The formulas are different, so we do not compare K s p values directly. We find the silver ion each salt needs. For silver chloride: one point eight times ten to the minus ten, over zero point zero one, which is one point eight times ten to the minus eight molar. Okay? For silver chromate, A g two C r O four, there are two silver ions, so silver ion comes in squared. So we take the square root, that tick-shaped sign, of one point one times ten to the minus twelve over zero point zero one. That is one point zero five times ten to the minus five molar. Right? Now compare: ten to the minus eight against ten to the minus five. Silver chloride needs hundreds of times less silver ion, so the white silver chloride precipitates first. Mystery solved! With me? Second part. Red silver chromate appears when silver ion reaches one point zero five times ten to the minus five molar. Then chloride is one point eight times ten to the minus ten, over that: one point seven times ten to the minus five molar. Clear? Look at that! From zero point zero one molar chloride, only one point seven times ten to the minus five is left. So over ninety nine point eight percent of the chloride has precipitated before the red colour appears. This is the basis of Mohr's method for chloride. Very good!

7:42Quick revision: Q vs Ksp, selective precipitation, groups

So, quick revision. Q uses the actual concentrations, after mixing: if Q is greater than K s p, a precipitate forms; if less, none. Added slowly, the salt that needs the least reagent to reach its K s p comes out first. Group two: H two S in acid; group four: ammoniacal H two S; group three: ammonium hydroxide with ammonium chloride. Now go and try the practice problems on this page. You can do them. All the best!