[G-funk]

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.

[Grumples]

For 1) write out the oxidation and reduction half reactions:
Fe2+ ---> Fe3+ +e-
Cr2O72- + 14 H+ +6e- ---> 2Cr3+ +7H2O

since you need conservation of mass with the electrons, you multiply the oxidation half-rxn by 6, so you have 6 electrons on either side of the equation.  the correct answer is A.

For 2) remember that molarity1*volume1 = molarity2 * volume2 if the moles are kept constant.  therefore 1.600L * .05M = 2L * X (where x is new molarity).  X = .04.
Since pH = -log [H+] and since you can assume that HNO3 will dissociate completely, you know that -log (.04) = 1.40 is the pH.

For 3) BaCrO4 is in a ration of one mole cation to one mole anion.  what you need to do is find out how many moles of each are in the original solutions (M*V=n).  They tell you that you have a limiting and excess equation, and that Ba2+ is in excess, so all you have to do is set up a box:

             Ba2+ +  CrO42- ---->  BaCrO4
initial       x             y                  0               (i didn't do the calculations, but x=
change    -y            -y                  y                moles of Ba2+ originally, and y =
final        x-y           0                 y                 moles of CrO42- originally)

solve for the final conc. of Ba2+ by dividing x-y by the total volume.