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Chemistry 8 Online
OpenStudy (anonymous):

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?

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