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Class 10 Chemistry (India)
Given salt, find acid and base
Given a salt, let's see how to find the acid and base that could have formed it. Created by Ram Prakash.
Want to join the conversation?
- For the second one, why is it Cu(OH)2 instead of CuO?(1 vote)
- copper hydroxide is formed when copper oxide reacts with water
CuO+H2O->Cu(OH)2
unlike copper oxide which is formed during the combustion of copper in air
(2Cu+O2->2CuO)(3 votes)
- For those who are wondering why the opposite reaction does not occur, water does not split into OH and H that easily. 2:36(2 votes)
- unless electrolysis?(1 vote)
- So basically, the chemical name of water can be written as Hydrogen Hydroxide?(2 votes)
- There is no chemical or specific name for water.(1 vote)
- I would really like to know why the ion of CuSO4 is +2 and -2 instead of just + and -, like in the other examples. Why does this happen?(1 vote)
- the + and - depends on the valency of the compound or element!(1 vote)
- In the 2nd example, I don't understand how can Cu replace H from H2SO4? Because I beleive that Cu is below in reactivity series than H. So, generally it should now displace H? Please resolve my query.(1 vote)
Video transcript
- If I give you a salt can you guess what acid and based would've reacted to give us this salt? See we have seen in earlier videos that an acid and a base, they
react to form salt and water. And I'm pretty sure if I give you an example of an acid and a base reacting you will be able to guess what
salt is going to be formed. But is this video we are
going to do the opposite. Given a salt, we will try to find out what acid and base would have reacted to give us this salt. So let's begin with an example. First of all let's recall how acids and bases react
to give us salt and water. So here I've taken H-Cl
acid and Na-OH base. Now when I mix them in water, H-Cl is going to dissociate and give me H plus ion
and chlorine minus ion. Similarly Na-OH, when
we put this in water, this is going to dissociate and give us Na plus ion and O-H minus ion. Now the ions are going
to exchange position. A double displacement
reaction is going to happen. The H plus ion will now get
attracted to the O-H minus ion. And this will form water, H-O-H or H2O. Let me write down like this. And sodium ion will in turn now get attracted to chlorine minus ion and form this salt, Na-Cl. Na-Cl, salt. So this is how we are getting
salt and water as the product. The ions are exchanging position. So if the ions of acid and base, they exchange position,
we get a salt and water. So that means if we exchange
the ions of salt and water we can get back our acid and base. Let's try that out. So here the ions and salt in Na-Cl will be Na plus and chlorine minus. And similarly I can write H-O-H, water as H plus and O-H minus ion. Now let's exchange the ions. Now sodium will get
attracted towards O-H minus, and we will get Na-O-H. And we know that Na-O-H is a base because it has a O-H in it. It can increase the concentration of O-H ions in an alcohol solution. And similarly now H plus will get attracted to chlorine minus. And this will form H-Cl which is an acid. This has hydrogen in it and it can increase the H plus
ion concentration in water. So yes, we have got our acid and base that can react and form
the give salt Na-Cl. By the way I would like to clarify here that this is just a technique to get the parent acid
and base of a given salt. It's not that if you put the salt in water it will give you acid and
base and you can use it. That is not going to happen. Now let's look at one more example. So this time I have the
salt copper sulfate, and we need to find
out which acid and base could have reacted to give us this salt. So we have just seen acid and bases, they undergo double displacement, or the ions between them, they exchange position to
give us salt and water. So we will exchange the
ions of salt and water to get back the acid and base. So lets do that. So water formalize H2O or I'm
going to write it as H-O-H. Now let's write down the ions
that are present over here. So copper sulfate will
have copper plus two ion, and sulfate minus two ion. And water I can write as H
plus ion and O-H minus ion. Now lets exchange the ions. So copper will now get
attracted to O-H minus ion, and we will get the ions of a base. Copper plus two and O-H minus one. And the reason why these
are the ions of a base, because a base is they increase the O-H
minus sign concentration. And this has O-H minus sign in it. Okay now Cu-O-H this is not
going to be the final formula. Because see copper has
a charge of plus two, and O-H has a charge of minus one. This molecule is not going
to be electrically neutral. For that, copper plus two will have to attract one more O-H minus. Because see then we'll only
have plus two and minus two, electrically neutral. So our final formula of the base is going to be Cu-O-H, whole twice. You can also get this by
exchanging of the charges, like two comes over here,
one comes over here. Cu-O-H, whole twice. Now let's go and exchange the other ions. So now H plus will get attracted
to sulfate ion over here. And we will get the ions of an acid. H plus and sulfate ions,
S-O four minus two. And why these are the ions of an acid? Because acid, they dissociate to increase the concentration of H plus, and this has H plus in it. So this can only be the ions of an acid. Okay now this is not going
to be the final formula, H-S-O four is not the final formula. Because this is not electrically neutral. See this other charge of plus one and this other charge of minus two. So for this to be electrically neutral, there has to be two H plus. Then only we will have
plus two and minus two. So that means the final
formula of the acid will be H two, S-O four. H two S-O four is our acid. Now see we can also get this
by exchanging of the charges. Two comes over here, one comes over here. H two, S-O four. Now this is the acid and this is the base, and when they both react we will get copper sulfate salt and
water will be formed. So yes we have solved our problem. Now lets do one last problem. So this time I'm given with the salt, K-N-O three, potassium nitrate. And we have to find
out which acid and base could react to give us this salt. So I want you to pause the video and try this by yourself first. Give it a try. Now if you have tried it, lets see. So we know that the
ions of acid and bases, they exchange position to
give us a salt and water. So if we exchange the
ions of salt and water we can get back the acid and base. So lets do that. So let me write down water. Water is H two O, or H-O-H. Let me use same color. H-O-H. Now lets break the salt
and water into their ions. So K-N-O three will have potassium ion, K plus and nitrate ion, N-O three minus. From water I'll get H plus and O-H minus. Now lets rearrange,
lets exchange the ions. So potassium will get
attracted to O-H minus, and this will give us the ions of a base. K plus and O-H minus. And why these are the ions of a base? Because it has O-H minus right? Bases increase the O-H
minus sign concentration. And similarly H plus will now get attracted to
N-O three, nitrate ion. And we'll get the ions of an acid. H plus and N-O three minus. And these are the ions of an
acid because they have H plus. Acids they increase the H
plus ion concentration, right? Now with this let's try to find out the formula of the acid and the base. See over here if we exchange the charges H-N-O three we will get
the acid as H-N-O three. Similarly we exchange the
ions, the charges of K-O-H. So one comes here, one
comes here, we'll get K-O-H. So we have found out the acid and the base that will react to get give
us the salt K-N-O three. So if you are given a salt and you have to find out the
acid and base that can make it, then break the salt into its ions and exchange these ions
with the ions of water. You will get the ions
of the acid and base. And from there you can
build the chemical formula of the acid and the base
by exchanging the charges.