i have a test tomorrow. and i have absolutely no idea how to do these problems. pleaseee help me!! PLEASE!! These are a variety of problems and you dont have to answer all of them if you dont want to, any help would be appreciated!
1) write a balanced chemical reaction by writing out the reactants and then determining the products of each reaction. if no reaction will occur, write no reaction.
solid sodium reacts with sulfuric acid.
2) Given the following [H30+] values, determine the pH and pOH of each solution. then determine if the solution is an acid, base, or neutral.
1.0 x 10^-12
3) calculate the molarity of H3O+ in a solution that has a pH of 8.9. Calculate the pOH in the solution.
4) Given a hydronium ion concentration of a 5.0 x 10^-9 M. What is the pOH of the solution?
once again, u dont have to answer all of these...if u can help me at all that would be great!!
1) write a balanced chemical reaction by writing out the reactants and then determining the products of each reaction. if no reaction will occur, write no reaction.
solid sodium reacts with sulfuric acid.
2) Given the following [H30+] values, determine the pH and pOH of each solution. then determine if the solution is an acid, base, or neutral.
1.0 x 10^-12
3) calculate the molarity of H3O+ in a solution that has a pH of 8.9. Calculate the pOH in the solution.
4) Given a hydronium ion concentration of a 5.0 x 10^-9 M. What is the pOH of the solution?
once again, u dont have to answer all of these...if u can help me at all that would be great!!
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Question Three
===========
pOH + pH = 14
pOH + 8.9 = 14
pOH = 14 - 8.9
pOH = 5.1
[H3O+] = 10^-8.9
[H3O+] = 1.26 * 10^-9
Question Four
===========
pH = - log[5*10^-9]
pH = 8.30
pH + pOH = 14
8.3 + pOH = 14
pOH = 4.7
Question 2
========
The solution is basic (very)
[H+] = 1* 10^-12
[OH-] = x
[1*10^-12] * [OH-] = 1* 10^-14
[OH-] = 1*10^-2
pH = - log[1*10^-12]
pH = 12
pOH = 2
See if you can get these last two yourself.
Problem One
==========
This is so rective, that I have no idea how it would be done really.
2Na + H2SO4 ===> Na2SO4 + H2
Edit
=====
I do agree with the poster above me. I thought of putting the same thing. This section is not that hard. You need only 3 equations.
pH + pOH = 14
pH = - log[H+ concentration]
[H+ concetration] = - 10^-pH
The resent is what are you asked for and what are you given? If you guess, you have a 33.33 chance of getting it right.
[OH-] = 10^-pOH
pOH = - log(OH- concentration) are not included because they are almost the same as the three equations above.
===========
pOH + pH = 14
pOH + 8.9 = 14
pOH = 14 - 8.9
pOH = 5.1
[H3O+] = 10^-8.9
[H3O+] = 1.26 * 10^-9
Question Four
===========
pH = - log[5*10^-9]
pH = 8.30
pH + pOH = 14
8.3 + pOH = 14
pOH = 4.7
Question 2
========
The solution is basic (very)
[H+] = 1* 10^-12
[OH-] = x
[1*10^-12] * [OH-] = 1* 10^-14
[OH-] = 1*10^-2
pH = - log[1*10^-12]
pH = 12
pOH = 2
See if you can get these last two yourself.
Problem One
==========
This is so rective, that I have no idea how it would be done really.
2Na + H2SO4 ===> Na2SO4 + H2
Edit
=====
I do agree with the poster above me. I thought of putting the same thing. This section is not that hard. You need only 3 equations.
pH + pOH = 14
pH = - log[H+ concentration]
[H+ concetration] = - 10^-pH
The resent is what are you asked for and what are you given? If you guess, you have a 33.33 chance of getting it right.
[OH-] = 10^-pOH
pOH = - log(OH- concentration) are not included because they are almost the same as the three equations above.
-
If someone answered your questions that wouldnt help you on your exams.
If you want I can help you through msn -> Sharbel_91@hotmail.com ( Tutor )
Then after explaining the subject for you we can solve these.
If you want I can help you through msn -> Sharbel_91@hotmail.com ( Tutor )
Then after explaining the subject for you we can solve these.