Finding pH, pOH, H3O+,&OH- given 4 aq solutions.(Assume volumes are additive) Here is the question and my work: https://imgur.com/a/7tCIh?

see this pic
http://i65.tinypic.com/73j5t3.png

the idea is for you to fill in that table and
.. (1) sum up volume
.. (2) sum up moles H+
.. (3) sum up moles OH-
.. (4) assume 1 mol H+ cancels out 1 mole OH-
.. (5) calculate final molarity of OH- or H+.. whichever is in XS
.. (6) convert that back to pH

from the pic I provided, .. final values after mixing are
.. volume = 2.2425 L
.. moles of OH- = 0.765 - 0.059 = 0.706
.. molarity of OH- (and NaOH) = 0.706 mol / 2.2425l = 0.2912M
.. pOH = -log(0.2912M) = 0.536
.. pH = 14.000 - 0.536 = 13.464
.. molarity of H+ = molarity H3O+ = 10^(-13.464) = 3.44x10^-14M

For the first solution, [OH⁻] = 150 mL = 0.150 L
For the first solution, [OH⁻] = 10⁻²˙⁷⁵ M = 10⁻²˙⁷⁵ mol/L
For the first solution, number of moles of OH⁻ = (10⁻²˙⁷⁵ mol/L) × (0.150 L)

For the second solution, [OH⁻] = 10⁻³˙⁷⁵ M = 10⁻³˙⁷⁵ mol/L
For the second solution, number of moles of OH⁻ = (10⁻³˙⁷⁵ mol/L) × (1.600 L)

Volume of the resulting solution = (0.150 + 1.600 ) L = 1.750 L

[OH⁻] of the resulting solution = [(10⁻²˙⁷⁵ mol/L) × (0.150 L) + (10⁻³˙⁷⁵ mol/L) × (1.600 L)] / (1.750 L) = 3.15 × 10⁻⁴ M
pOH = -log[OH⁻] = -log(3.15 × 10⁻⁴) = 3.5
[H₃O⁺] = Kw/[OH⁻] = (1.00 × 10⁻¹⁴) / (3.15 × 10⁻⁴) M = 3.17 × 10⁻¹¹ M
pH = pKw - pOH = 14.0 - 3.5 = 10.5

pH problems....

Just an aside. You don't have to write down every digit the calculator gives you. Simply go by the number of significant digits and add one extra "guard digit" for intermediate answers.

NaOH is a strong base and completely, ionized. The acids are strong acids and completely ionized. Therefore, there are no equilibria to worry about. Simply find the moles of OH- initially, and the moles of H+ initially and remember that H+ + OH- --> H2O. Then find whether there will be an excess of H+ or OH- and do the pH, etc. calculations based on the the new concentration. (Add up all the volumes to get the final volume.)
Some useful relationships:
pH + pOH = 14.00 ..... (at 25C)
[H+] x [OH-] = Kw = 1.00x10^-14 ...... (at 25C)
pH = -log[H+].... [H+] = 10^-pH .... pOH = -log[OH-] .... [OH-] = 10^-pOH

NaOH .... pH = 13.77... pOH = 0.230 .... [OH-] = 10^-pOH = 0.589 mol/L
0.589 mol / L x 1.300L = 0.765 mol OH-

HNO3 .... 0.090 mol / L x 0.125L = 0.01125 mol H+

HClO4 .... pOH = 11.3 .... pH = 2.70 .... [H+] = 0.001995 M
0.001995 mol H+ / L x 0.475L = 0.000948 mol H+

HBr ..... 0.090 mol/L x 0.525L = 0.04725 mol H+

Total volume = 1.300L + 0.125L + 0.475L + 0.525L = 2.425L

Total H+ ions = 0.01125 mol H+ + 0.000948 mol H+ + 0.04725 mol H+ = 0.0594 mol H+
The H+ ions will neutral 0.0594 OH- ions, leaving 0.706 mol OH-
The concentration of OH- = 0.706 mol / 2.425L = 0.291M OH-

pOH = 0.536
pH = 13.464
[H+] = 3.44x10^-14M
[OH-] = 0.291M