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)`?

Options

A. 0.10 M
B. 0.40 M
C. 0.0050 M
D. 0.12 M

Answer

A

Explanation

No. of milli equivalent of `HCl = 20 xx 0.05 = 1.0`
No. of milli equivalent of
`Ba(OH)_(2) = 30 xx 0.10 xx 2 = 6.0`
After neutralization, no. of milli equivalents in 50 ml. of solution = (6-1) = 5
No. of milli equivalent of `OH^(-)` is 5 in 50 ml.
`[OH^(-)] = (5 xx 100)/(50) xx 10^(-3) = 0.1M`