Calculate the ph of a solution that is made by mixing 50.0 ml of 0.14 m acetic acid and 50.0 ml of 0.21 m sodium acetate. express your answer to two decimal places.

After mixing 50.0 mL and 50.0 mL the molarities are cut in half to 
0.07 M acetic acid and 0.105 M sodium acetate 

the Ka for acetic acid = 1.8 x 10^-5

HA => H+ & A- 

1.8 x 10^-5 = [H+ ] [A-] / {HA] 

1.8 e-5 = [H+ ] [0.105] / [0.07] 

H+ = 1.2 x 10^-5 M 

pH = - log (1.2 x10^-5 M)

pH = 4.92

What are the options? i can you if you give them.