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?
- 0.40 M
- 0.0050 M
- 0.12 M
- 0.10 M
Answer
(4) 0.10 M
Solution
Millimoles of H+ = 20.0 0.050 = 1.0
Millimoles of OH- = 30.0 0.10 2 = 6.0
Excess OH- = 6.0 - 1.0 = 5.0 millimol
Total volume = 20.0 + 30.0 = 50.0 mL
∴ [OH-] = 5.050.0 = 0.10 M
Answer: (4)