What is the OH- in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M BaOH2 ?

Number of milliequivalent of HCl = 20 x 0.050 x 1 = 1Number of milliequivalent of BaOH2= 2 x 30 x 0.1 = 6OH- of final solution = milliequivalents of Ba(OH)2-milliequivalent of HCltotal volume                                       =6-150 = 0.1 M