difficulty with these chem questions
 

hey people i am new to this site and require some help with some uni style chem questions i missed the lecture so am having a bit of trouble getting consistant answers. can anyone help buy getting an answer to the following questions...


1)

redox balncing:
Which of the following represents a balanced, overall equation for the oxidation of iron (II) in acid solution?
Cr2O72-(aq) + Fe2+(aq) 2Cr3+(aq) + Fe3+(aq)

a. Cr2O72-(aq) + 6Fe2+(aq) + 14H+(aq) 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
b. Cr2O72-(aq) + 6Fe2+(aq) + 7H+(aq) 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
c. Cr2O72-(aq) + 6Fe2+(aq) + 14H+(aq) Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
d. Cr2O72-(aq) + Fe2+(aq) + 14H+(aq) 3/42Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)


2)
What is the pH of a solution made by mixing 1600 mL of 0.05 M HNO3 and 400 mL of water?

a. -1.4
b. 1.4
c. -1.3
d. 2.0
e. 2.1


3)

3.0 L of 0.2 M K2CrO4 and 6.0 L of 0.4 M BaCl2 were mixed. Assuming that the barium chromate is insoluble the molarity of Ba2+ in the final solution is :

a. 1.8 M
b. 0.2 M
c. 0.6 M
d. 0.4 M
e. none of these


cheers guys any help would be great.

answer for 2 and 3

2
you got 1,6 liter of a 0,05 mol/ liter solution ---> 0,05 * 1,6 = 0,08 mol
you add 0,4 liter which makes 2 liter 0,08 mol in 2 liter = 0,04 mol/l
that makes 0,04 mol of H+
the pH is the negative decadic log of the conz of H+ pH = -log[H+] = -log [0,04] = 1,4

pH can never be negative!!!!

3
you got 3l*0,2mol/l = 0,6 mol K2CrO4
and 6l*0,4mol/l = 2,4 mol BaCl2
you got 9l of solution, you generate 0,6 mol BaCrO4
1,8 mol Ba2+ stay in solution
1,8mol/9l = 0,2 mol/l

question 1:

for each of the possibilities -->

just count up the numbers of particles on both sides....

then count up the numbers of charges on both sides...

then see which one is balanced for both!

if this is a problem take your shoes off and use your toes as well :D

or count in binarys (0-1-0-1) gets you up to more than a million :wink:

"pH can never be negative!!!! "

pH can be neagtive, i.e. if you have 2 moles per litre of HCl, then the pH would be:

pH = -log(2) = -0.301.