PLEASE I'VE TRIED EVERYTHING

A solution is made by mixing 16.0g of Sr(OH)2 and 45.0mL of 0.250M HNO3. WHAT IS THE CONCENTRATION OF OH-. I was able to find all the concentrations EXCEPT THAT. i KEEP GETTING 5.84 nd it keeps telling me to check my calculations. ok the equation is
Sr(OH)2 + 2HNO3 -> Sr(NO3)2 + 2H2O

MUCH HELP IS APPRECIATED...SERIOUSLY

45.0ml * 0.250M *1L/1000ml=0.18moles HNO3
16g Sr(OH)2/121.63g/m=0.13m Sr(OH)2

0.18m HNO3* Sr(OH)2/2HNO3= 0.09m Sr(OH)2

so 0.18m HNO3 uses 0.09MSr(OH)2 leaving 0.13m- 0.09m=0.04m Sr(OH)2

0.04m Sr(OH)2 = 0.08m of OH-

0.08m /0.045L OH- =1.78M OH-

Sr(OH)2 + 2HNO3 -> Sr(NO3)2 + 2H2O
1 2

16g .045 * .25

Moles of HNO3: .045 * .25 = .1125 mol
Moles of Sr(OH)2 : .5 * .1125 = .05625 mol
Concentration of Sr(OH)2: .05625/ .045 = 1.25 M
Concentration of OH- : 2 * 1.25 = 2.5 M

I believe the answer is zero as OH- has all been used up.