Question

I'm doing something wrong...
Determine the pOH (to two decimal places) of the solution that is produced by mixing 5.32 mL of 5.64×10-5 M Ca(OH)2 with 174 mL of 4.44×10-4 M MgH2.

(5.64x10-5) x (.00532L) x (2mol/1mol)=6.0x10-7
(4.44x10-4) x (.174L) x (1mol/1mol) = 7.73x10-5
[OH-]= (6.0x10-7) + (7.73x10-5)/.00532 + .174
[OH-]= 4.34x10-4
pOH= -log 4.34x10-4
pOH= 3.36
3.36 is wrong apparently and I really don't know what's wrong... could someone point me in the right direction?