Sulphate and the Sodium Carbonate Extract

Transcript

0:00Introduction

Hello everyone! Let us start with a small lab puzzle. Take two test tubes with two different salts. You add barium chloride to each, and both give a white precipitate. One salt is a sulphate, and the other is a carbonate. How will you tell which is which, with just one more reagent? And here is a second puzzle. What if your salt does not dissolve in water at all, or it is blue? How do you test its anion then? These tests are part of JEE practical chemistry, so let's begin.

0:33Sulphate: the barium chloride test

So, let us start with sulphate. Sulphate gives no gas with dilute acid, and no gas with concentrated acid either. So we must test it in solution. Okay? Here is the test. Take the salt solution, acidify it with dilute hydrochloric acid, H C l, and then add barium chloride solution, B a C l two. The small two below is a subscript, which means two chlorine atoms. Right? Now let me ask you something. Sulphuric acid is also a strong acid. So why do we acidify with hydrochloric acid, and never with sulphuric acid? Yes, exactly, sulphuric acid, H two S O four, would itself supply sulphate ions. On screen, the sulphate ion is S O four with two minus up top: the small four counts oxygen atoms, and two minus is the charge. So we would get a precipitate even with no sulphate in the salt. Clear? Now look at the reaction. B a two plus, the barium ion with a charge of two plus, meets the sulphate ion, and the arrow means they give B a S O four, barium sulphate. It is a white precipitate, and it does not dissolve in hydrochloric acid, or in nitric acid, HNO three. Got it? So, a white precipitate with barium chloride means sulphate, right? But wait, be careful here, because carbonate also gives a white precipitate with barium chloride. So how do we tell the two apart? We add dilute hydrochloric acid to the precipitate. Barium carbonate, B a C O three, dissolves with effervescence, which means fizzing. Barium sulphate is unaffected. One reagent, and the first puzzle is solved. With me?

2:19A second test: lead acetate

Now, there is a second test for sulphate. Add lead acetate in acetic acid, and we get white lead sulphate, P b S O four. Okay? And here is something interesting: this white lead sulphate dissolves in hot ammonium acetate solution. But lead sulphate is insoluble in water, so why should it dissolve here? The reason is that the acetate ions tie up the lead as lead acetate, which is soluble but very weakly ionised. So the free lead ions, P b two plus, are removed from the solution. That carries the solubility equilibrium of lead sulphate forward, and the precipitate dissolves. Clear?

3:00The sodium carbonate extract

Alright, now our second puzzle. What if the salt does not dissolve in water? Then we prepare the sodium carbonate extract: we boil the salt with sodium carbonate solution, N a two C O three, and then filter it. With me? During the boiling, a double decomposition takes place. Write the salt as M A, where M is the metal cation and A is the anion, okay? M A plus N a two C O three gives MCO three, with a down arrow, which means it comes out as a precipitate, plus N a two A, the sodium salt of the anion. Got it? So the cation is removed as an insoluble carbonate, and it stays behind as the residue. And the anion passes into the filtrate as its sodium salt, ready for testing. Right? Now, suppose a blue salt is to be tested for chloride. Why use the sodium carbonate extract, and not the original solution? The blue colour comes from the copper ion, C u two plus. In the extract, it is removed as insoluble basic copper carbonate. So the colours of the precipitates are not masked. Okay? So here is the rule on screen. We use the extract when the salt is insoluble in water, or when it has coloured or interfering cations. Clear?

4:18Neutralise the extract before each test

Now one important detail. The extract contains excess carbonate, C O three two minus, so it must be neutralised before each test. With me? Here is a question. Before adding silver nitrate, why do we neutralise the extract with nitric acid? Otherwise, the excess carbonate precipitates white silver carbonate, A g two C O three, as soon as we add silver nitrate, A g N O three. Nitric acid removes the carbonate, and it adds no interfering ion. Got it? In the same way, for the barium chloride test we use dilute hydrochloric acid, and for the lead acetate test we use acetic acid. Learn these pairs. Okay? Now let me show you a trap. Suppose a student adds dilute acid to the extract, sees effervescence, and writes that carbonate is present. Is that correct? That would be wrong, because of course it fizzes: that carbonate came from our own sodium carbonate. So the extract cannot be used to test for carbonate. Right?

5:18Example 1: PbCl2 boiled with Na2CO3, residue and filtrate

Now, example one. Read it with me. Write the equation for the preparation of the extract from lead chloride, and say what the residue and the filtrate contain. Pause the video and try it first. Done? Okay, let's check. Lead chloride is P b C l two, lead with two chlorides. Follow our pattern: P b C l two plus N a two C O three gives P b C O three, plus two N a C l. Two sodiums and two chlorines on each side, so it is balanced. With me? The lead is removed as insoluble lead carbonate, so the residue is lead carbonate, P b C O three. Okay? And the filtrate contains sodium chloride, together with the excess sodium carbonate. To test this chloride with silver nitrate, we first neutralise it with dilute nitric acid, just as we learnt.

6:11Example 2: mass of BaSO4 from 1.42 g of Na2SO4

Clear? Now a numerical, a little harder. Find the mass of barium sulphate precipitated from one point four two grams of anhydrous sodium sulphate. Anhydrous means it has no water of crystallisation, okay? Pause the video and try it first. Done? Okay, let's check. First, the molar mass, capital M, of sodium sulphate, N a two S O four. Two sodiums at twenty three each, one sulphur at thirty two, and four oxygens at sixteen each; the brackets mean times. That gives one hundred and forty two grams per mole, which is what the g mol to the power minus one on screen means. Right? Next, the number of moles, small n, is the mass over the molar mass. One point four two over one hundred and forty two gives zero point zero one zero mole. Got it? Now, each sulphate ion gives one B a S O four, so the ratio is one to one. That means we get the same amount, zero point zero one zero mole of barium sulphate. Okay? Now the molar mass of barium sulphate: one hundred and thirty seven for barium, thirty two for sulphur, and four times sixteen for the oxygens. That gives two hundred and thirty three grams per mole. With me? So the mass is zero point zero one zero times two hundred and thirty three, which gives two point three three grams of barium sulphate. That is our answer, and well done if you got it!

7:38Quick revision: BaSO4, PbSO4 and the Na2CO3 extract

So, quick revision. Sulphate gives white barium sulphate, which does not dissolve in hydrochloric acid or in nitric acid. Lead acetate gives white lead sulphate, which dissolves in hot ammonium acetate. And for the extract: boil with sodium carbonate, filter, and neutralise it before each test, but never test it for carbonate. Now go and try the practice problems on this page. All the best!