Hercules Chem questions :D

[HERculina]

Another fun fun chem question : which of the following species are amphiprotic in aqueous solutions: HS-, HNO3 and HC2O4-           Plz show working out too so i can get the gist of it

[Milkshake]

Another fun fun chem question : which of the following species are amphiprotic in aqueous solutions: HS-, HNO3 and HC2O4-           Plz show working out too so i can get the gist of it

Amphiprotic means that they can either gain or lose a H+ atom, so that will be HS- and HC2O4-, since you can both add one H+ or remove one H+ from them

[HERculina]

What about HNO3 ? Why isn't it one? It could turn into NO3- or H2NO3+ ?

[REBORN]

Why would it want to turn into H2NO3+. It wouldn't be 'happy' (stable). It has a very very very strong tendency to donate the H+ only (hence strong acid)

[HERculina]

Oh ok i think i get you What about this question: which of the followong acids is both diprptic and amphiprotic? A)H2SO4 b)H2PO4- c)H3PO4 d)HSO4-   Is it b?

[|ll|lll|]

Oh ok i think i get you What about this question: which of the followong acids is both diprptic and amphiprotic? A)H2SO4 b)H2PO4- c)H3PO4 d)HSO4-   Is it b?

Yeah you're right. 2 H atoms and can accept/donate protons

[HERculina]

A sample of contaminated hydrated copper sulfate CuSO4.5H20 was tested for purity. A 15.0 gram sample was dissolved in water and filtered to remove insoluble impurities. The sulfate ions were precipitated by the addition of excess barium chloride and the resulting precipitated was collected, dried and weighed. If the final mass is 4.95 grams find the percentage of purity of the sample.

[BlueSky_3]

Is the answer 22.6 % ?

[HERculina]

nah i don't have the answer to this question, but i got 35.3%. How did you work it out?

[HERculina]

Another question if anyone is free : 0.6238 g of copper (II) sulphate cyrstals with formula CuSO4.xH2O was dissolved in water, and black copper (II) oxide was precipiated by treatment with boiling NaOH solution. The precipitate was collected by filtration, washed, dried and weighed. if the precipitate weighs 0.1988g, calculate the value of x in the formula CuSO4.xH2O. (i don't quite get the boiling part - would hydrogen gas form when you boil a salt)

[Mao]

^^See if that helps

[BlueSky_3]

Mao can please check my answer to the previous question?

[HERculina]

Ok so is this what i do next: n(CuO)= 0.1988/(63.6+16)=0.002497 mol and then n(CuSO4.xH2O)=n(CuO)=0.002497 mol and then use n=m/M ---> 0.002497=0.6238/(63.6+32.1+4(16)+x(18)) then solve for x and x= 5.000 . but what did the (x+1) before the H2O mean? And where did the Na from NaOH go?

[BlueSky_3]

Ok Hercules this is what I did for the previous question:

n(CuSO4)=n(BaSO4) = 0.0212....mol
m(CuSO4)=3.38513....g
the purity= m(CuSO4)/15 grams
               = 22.57%

You have to write the precipitation equation....but I cbs writing it soz 

[Mao]

but what did the (x+1) before the H2O mean? And where did the Na from NaOH go?

You determined x=5, and thus on the RHS there are 6 waters.

The Na+ is a spectator ion, for all intents and purposes we can ignore them. (They're present as Na+(aq) on both sides of the equation. A chemical equation is just like a math equation, you can cancel things out which are on both sides)