What is the ph of a sodium acetate (nac2h3o2) solution prepared by adding 0.820 grams of sodium acetate to 100.0 ml of water at 25.0 °c? the ka at 25.0 °c for acetic acid is 1.8 ⋅ 10-5?

According to the reaction equation:
CH3COONa+ H2O ↔ CH3COOH + OH-
when we have 0.82 g of sodium acetate in 100 mL
So we have 8.2 g per liter & when we have the molar mass of CH3COONa=82 g/mol
we have to get the molarity of CH3COONa = weight/molar mass
                                                                        = 8.2 / 82 =0.1 M
So                   CH3COONa + H2O ↔ CH3COOH + OH-
initial c              0.1                                       0               0 
equilibrium C (X-0.1)                                     X               X
when Kb= Kw / Ka and we have Kw = 1x10^-14 & Ka = 1.8x10^-5
So Kw/Ka = [CH3COOH][OH-] / [ CH3COONa]
(1x10^-14)/(1.8x10^-5) = X^2 / (X-0.1)
5.6x10^-10 = X^2/(0.1-X)
5.6x10^-11  - 5.6x10^-10 X = X^2
∴X= 7.48x 10 ^-6 ∴[OH] = 7.48x10^-6 M
when POH = -㏒[OH]
                   = -㏒(7.48x10^-6) = 5.13
∴PH = 14 - POH = 14 - 5.13 = 8.87 

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Sorry i dont understand the question can you rephase it?