chemistry about pH?

Question

I would like to ask the following 3 questions:

what is the answer if you are asked to arrange the pH of NH4Cl, NaCl, HCl, H2SO4 and NaCH3COO in descending order (all are aqueous solution and 1M)?

which of the the four 1M solution of the following salts has the highest pH , NaNO3, Na2CO3, NH4Cl, NaHCO3?

How do you determine which of them have the highest pH, especially those NH4Cl, NaNO3, NaCl (not typical acid or alkali like HCl or NaOH)?

thanks a lot

i am not abusing this place to search for answer to my HW. those questions are from my exmination and i got them wrong, answer to question one is NH4Cl>NaCl>NaCH3COO>HCl>H2SO4
answer to question two is Na2CO3.

i just want to learn how to tackle these kind of quesitons and please help and don't tell me to ask my teacher for it's summer holiday in my place now.

answer to question one should be NaCH3COO>NaCl>NH4Cl>HCl>H2SO4

Answer 1

pH is a convinient way to represent how much acidity something has. As you know, this is directly proportional to the number of H+ and since pH=-log[H+], Lower the number, more acidic it is.

Having that said, out of NH4Cl, NaCl, HCl, H2SO4 and NaCH3COO lets see which should have the highest pH. (Basic).

You have to consider that all the mentioned chemicals above in water (aqueous soln) breaks up into its ions.

Sodium Acetate breaks up into --> Na+ CH3COO-.
The CH3COO- (Acetate ion) is a conjugate base of a weak acid (Acetic acid) and is a fairly strong base. This will give it a high pH. (Above 7)

Next is NaCl. This in water does not create any H+ or OH-, therefore its neutral. (About 7)

Next is NH4Cl. This will break up into NH4+ and Cl-. Notice that ammonium ion is created. As you should know, ammonium ion is a relatively weak acid. (Lower than 7)

The next two HCl and H2SO4 are quite tricky. These two are commonly known as strong acids, and the only way to know is either have a Ka Chart of deeply ponder on how a polyprotic acid will behave in water. If my memory serves me correctly, I think HCl should actually be stronger than H2SO4. Are you sure you provided the correct answer? I think the Ka for HCl should be about 1000000 vs Ka for H2SO4 ~1000. Are you sure the answer key you have is correct? Logically thinking, the monoprotic acid HCl should be stronger than diprotic H2SO4. The reason for it is more in depth than I care to explain right now so I'll provide a link with along with Ka for both strong acids.

To determine the pH of the second set of chemicals you listed, you use the same method and break those chemicals up into ions and see what you have.

NaNO3, Na2CO3, NH4Cl, NaHCO3

NaNO3 will produce NO3-, which is a conjuage base of a strong acid therfore it is a weak base.

Na2CO3 will produce CO3--, which is a conjugate base of a weak acid (carbonic acid will TWO H+ removed), therefore it is a VERY Strong base.

NH4CL will produce NH4+, and as I mentioned earlier, it is a weak acid.

NaHCO3 is initial conjugate base of (weak) carbonic acid with only one proton removed so it is a relatively strong base.

To summerize this you would get:
Na2CO3 > NaHCO3 > NaNO3 > NH4Cl
in order of decreasing pH (Basic to acidic)

Good luck in your Chemistry studying and if you need more help, don't hesitate to ask. Cheers.

Answer 2

sure ,you may join for Ph.D by way of distance gaining understanding of , yet he has to justify the supply of required laboratory centers and different appropriate centers at his place of place of abode . alongside-with this he has to spend a fastened time together with his handbook each and every 3 hundred and sixty 5 days. The handbook chosen could desire to be an authorized handbook by the college in contact.

Answer 3

when you have such Q's look for the conj. acid or base , then sort this conj. based in there strength . write the upset order that you fond it well be the answer ( JUST ALL THEIR CONJ. MUST BE ALL ACIDS OR ALL BASES )

ex NH4----> HH3
NAOH ---- > HOH
AND SO ON

Answer 4

it would suck if this answer was part of your job but i suggest doing your homework rather than hope some one else will do it for you because next thing u know its part of a more complicated question. Good luck