[GeLe5000]

Hello.

It's a mathematical problem probably as old as Chemistry itself.

You mix an unknown amount of NaCl (s) with an unknown amount of KCl (s).

You weigh the mixture : 1,25 g.

You dissolve it in water and add AgNO₃.

All the chlorine ions are consumed in the formation of an AgCl precipitate. It weighs 2,5 g.

The question is : what are the amounts of NaCl and KCl in the initial mixture ?

When I read the question I told to myself that it was impossible to know.

Beginning with x = amount of NaCl, 1,25 – x = amount of KCl, and using the NaCl, KCl and AgCl molecular weights, you get

since amount of NaCl + amount of KCl = amount of AgCl (in moles)

x/58,44 + (1,25-x)/74,55 = 2,5/143,3

NaCl = 14 %, KCl = 86 %

(In fact, it could as well be NaCl=86 %, KCl=14 %, if x=amount of KCl)

Don't you find it astonishing that a simple equation leads to the amounts of Na and K whilst these ions deliver the same chlorine ions reacting indistinctively with the Ag+ ions to give the final AgCl precipitate ?

It looks like science-fiction.

With other NaCl/KCl ratios, we will have different quantities of AgCl. Why ?

[sjb]

With other NaCl/KCl ratios, we will have different quantities of AgCl. Why ?

Consider for instance a mixture of elephants and mice. If you have a total mass of 10 tons, then the number of legs you have will vary depending on whether you have mostly elephants, or mostly mice.

Sorry, a little convoluted but I hope you see the analogy

[GeLe5000]

I suppose you mean 10 tons of elephants  and ducks.

Anyway, the problem is that a chlorine ion is a chlorine ion. The counterion (Na⁺ or K⁺) has no importance.

[Corribus]

Forget mixtures for a second. Consider pure substances.

If you have 1 gram of NaCl in one beaker and 1 gram of KCl in another, both dissolved in the equivalent amount of water, and you add the same amount of silver nitrate to each beaker, which one will form more precipitate?  (Hint, which contains more chloride, 1 gram of NaCl or 1 gram of KCl)?)

[Dan]

The counterion (Na⁺ or K⁺) has no importance.

The cation doesn't participate in the reaction, but the amount (moles) of chloride present depends on the amount of (moles) of cations present. Let's forget the legs. Say you have elephants and mice and each animal has been implanted with an identical identification chip. Na and K have different relative masses (like mice and elephants), and there is more chloride in 1 g NaCl than in 1 g KCl (similarly there are more identification chips in a 10 ton crate of mice than in a 10 ton crate of elephants). No magic.

[GeLe5000]

With your help, I think that I begin to see some light far away.

According to the values : mass % NaCl = 14,4, mass % KCl = 85,6, it's only 0,18 g NaCl (0,00308 mole chloride),  and 1,07 g KCl (0,01435 mole chloride), that will give 0.01744 mole chloride in AgCl.

No magic, but still wonderful. Happy those who can say "I only believe in mathematics".

There just remains the problem that the author begins the calculation saying "let's say that x is the amount of NaCl in grams". I suppose that he couldn't have said the opposite (x is the amount of KCl in grams).

Thank you very much.

[Dan]

There just remains the problem that the author begins the calculation saying "let's say that x is the amount of NaCl in grams". I suppose that he couldn't have said the opposite (x is the amount of KCl in grams).

Try it, what do you get? You should get the same result (14% NaCl, 86% KCl) - post your working if not.