What is the [OH-] in th final solution prepared by mixing 20.0 ml of 0.050 M HCl with 30.0 ml of 0.10M Ba(OH)2 ?

What is the [OH-] in th final solution prepared by mixing 20.0 ml of 0

1.	What is the [OH-] in th final solution prepared by mixing 20.0 ml of 0.050 M HCl with 30.0 ml of 0.10M Ba(OH)2 ?
Answer» Millimoles of H+ produced = 20 x 0.05 = 1 Millimoles of OH- produced = 30 x 0.1 x 2 = 6 (... Each Ba(OH)2 GIVES 2OH-.) . .. Millimoles of OH- remaining in solution = 6 - 1 = 5 TOTAL volume of solution = 20 + 30 = 50 mL . .. [OH-] = 5/50 = 0.1 M