Question

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

(a) 0.40 M

(b) 0.0050 M

(c) 0.12 M

(d) 0.10 M

Answer 1

The correct option is: (d) 0.10 M

Explanation:

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

Comments

Thanks sir