2.3 grams of sodium are placed into a beaker containing 500 ml HCl 0.1 M;
What is the pH of the final solution?
Note: Assume the volume of the solution remains constant.
(A)pH = 7
(B)pH = 1
(C)pH = 13
(D)The pH can not be determined using given data
(E)All answers (A-D)are wrong.
17 comments:
haven't learned this stuff yet .. but im gonna guess d or e
isn't it just b?
No right answer yet. I am still waiting.
it has to be A then cuz it has to be an acid if it has HCl
I would guess D, because there is no way to calculate the H+ concentration without more information
The answer is C
On the other hand, if your teacher caught you before the act, you would not be allowed to do it. So it remains as HCL, and so it is B.
See http://www.youtube.com/watch?v=Jw9p-5t8wWY&mode=related&search=
Ok guys.
this 2 reactions take place:
2Na+2HCl = 2NaCl + H2
2Na + 2H2O = 2NaOH + H2
We have 0.1 moles of Na and 0.05 moles of HCl.
Therefore we have an excess of Na which will react with the water from the solution; 0.05 moles of Na excess wil form 0.05 moles of NaOH.
[OH-]= 0.05/0.5= 0.1 moles.
pOH=1 and pH= 14-pOH = 13
"[OH-]= 0.05/0.5= 0.1 moles." .. where did the 0.5 come from ?
0.5 is the volume of the solution expressed in liters and it was assumed to remain constant despite the fact that hydrogen went out of the system.
ohh ok. and do .05 moles of NaOH form because there were .05 moles of Na left ?
but why would NaCl form? Na is soluble
For anonymous -yes
For Sylvia : Na is not soluble is a highly reactive metal which explosively reacts with both water and HCl.
but I thought the solubility rules said that Group 1A elements are ALWAYS soluble ..
but I thought the solubility rules said that Group 1A elements are ALWAYS soluble ..
3 reactions take place:
2Na+2HCl = 2NaCl + H2
2Na + 2H2O = 2NaOH + H2
NaOH + HCl|= NaCl +H2O
But the answer is pH= 13
The last reaction
NaOH + HCl|= NaCl +H2O
does not take place because there is no HCL left, all of it was already consumed.
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