Finding pH, pOH, H3O+, and OH- given 2 aq solutions. The problem I am having trouble with is here: https://imgur.com/a/pq0KN?

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

1) 150mL solution , pOH = 2.75
[OH-] = 10^-2.75
[OH-] = 1.78*10^-3M
Mol KOH in 150mL = 0.150L = 0.150*(1.78*10^-3) = 2.67*10^-4 moles KOH

2) 1.60L = 1600mL solutiuon pOH = 3.75
[OH-] = 10^-3.75
[OH-] = 1.78*10^-4M
Mol NaOH in 1.6L = 1.6* ( 1.78*10^-4) = 2.84*10^-4 moles
Total moles bases = 2.84*10^-4) + 2.67*10^-4) = 5.51*10^-4 moles bases in 1.750L solution
Molarity = (5.51*10^-4) / 1.750 = 0.000315 M

pOH = -log 0.000315 = 3.50
pH = 14.00 - pOH = 14.00 - 3.50 = 10.50
[OH-] = 3.15*10^-4M ( calculated above
[H3O+] = 10^-14 / ( 3.15*10^-4) = 3.17*10^-11M