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 Ba ( OH )₂ ? -

Step-by-step solution

20 ~mL of 0.50 M HCl =20 0.050 ~m ~mol =1.0 ~m ~mol =1.0 meq. of HCl 30 ~mL of 0.10 M Ba ( OH )₂ aligned & =30 0.1 ~m ~mol & =3 ~m ~mol =3 2 meq & =6 meq Ba ( OH )₂ aligned 1 meq of HCl will neutralize 1 meq of Ba ( OH )₂ Ba ( OH )₂ left =5 meq. Tatal volume =50 ~mL Ba ( OH )₂ conc. in final solution = 5 50 ~N =0.1 ~N =0.05 M [ OH ⁻ ]=2 0.05 M =0.10 M Alternatively, Ba ( OH )₂+2 HCl BaCl ₂+2 H ₂ O 2 ~m ~mol of HCl neutralize 1 ~m mole of Ba ( OH )₂1 ~m ~mol of HCl neutralize 0.5 ~m ~mol of Ba ( OH )₂ Ba ( OH )₂ 1 e