Hoping for chem help: pH and oxidation?

Question

Theres six questions im stuck on again, all help appreciated :)

13. The [OH-] = 5.0 x 10^(-12) for a certain carbonated drink. Determine the [H+], the pH, and the pOH.

41. How many millileters of a 2.5 M solution of sodium hydroxide are needed to neutralize 125 ml of 1.0 M solution of nitric acid?

73. Will a precipitate of PbSO4 form when 400 ml of 0.0050 M MgSO4 is mixed with 600 ml of 0.0020 M pb(NO3)2? Ksp = 6.3 x 10^(-7)

109. Calculate the oxidation number of for each of the following:

1. Cr2O3

2. H2Cr2O7

3. CrSO4

d. CrO4^-2

110. Calculate the oxidation numer for sulfur in each of the following:

1. H2S

2. SO2

3. H2SO3

4. S2O3^-2

147. Use the change in oxidation number to determine which elements are oxidized and which are reduced in these reactions

1. C + H2SO4 --> CO2 + SO2 + H2O

2. HNO3 + HI --> NO + I2 + H2O

Answer 1

13. [H+] = 10^-14 / 5.0 x 10^-12 =0.0020 M

pH = - log [H+] = 2.7

pOH = - log [OH-] = 11.3

41. Moles HNO3 = 0.125 L x 1.0 M = 0.125

V of HCl = 0.125 mol / 2.5 = 0.050 mL => 50.0 mL

73. Moles SO42- = 0.400 L x 0.0050 M = 0.0020

moles Pb2+ = 0.600 L x 0.0020 M = 0.0012

Total Volume = 1.0 L

[SO42-] = 0.0020 / 1.0 L = 0.0020 M

[Pb2+] = 0.0012 / 1.0 L = 0.0012 M

Qsp = 0.0020 x 0.0012 =2.4 x 10^-6 >> Ksp

Precipitation should occur

109. Cr = +3 ; O = -2

H = +1 ; Cr = +6 ; O = -2

Cr = +2 ; S = +6 ; O = -2

Cr = +6 ; O = -2

110 . H = +1 ; S = -2

S = +4 ; O = -2

H = +1 ; S = +4 ; O = -2

S = +2 ; O = -2

147 . C goes from 0 to + 4 : it is oxidized and S goes from +6 to + 4 : it is reduced

N goes from +5 to +2 : it is reduced and I goes from -1 to 0 and it is oxidized