What is the [OH⁻] in the final solution prepared by mixing 20 mL of 0.05 M Hcl with 30 mL of 0.1 M Ba(OH)₂?

Options

(a) 0.4 M
(b) 0.005 M
(c) 0.12 M
(d) 0.1 M

Correct Answer:

0.1 M

Explanation:

No of milliequivalent of HCl = 20 x 0.05 =1.0
No of milliequivalent of Br(OH)₂ = 30 x 0.1 x 2 = 60
after neutralization, no of milliequivalents in 50 ml of solution = 6-1=5
Total vol of solution = 20 + 30 = 50 ml
No of milliequivalent of OH⁻ is 5 in 50 ml
[OH⁻] = (5 x 100 / 10 ) x 10⁻³ = 0.1 M