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

Answer

D

Explanation

Millimoles of `H^(+)` produced `=20 xx 0.05 =1`
Millimoles of `OH^(-)` produced `=30 xx 0.1 xx 2` ,
(Each `Ba(OH)_(2)` given two `OH^(-)` ions) =6
Millimoles of `OH^(-)` ions in the remaining solution =6 -1 =5
Total volume of solution =20 + 30 =50 mL
`[OH^(-)] = (5)/(50) =0.1 M`