What is the `[OH^(-)]` in the final solution prepared by mixing 20 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

Moles of `H^(+)` ions in 20 mL of 0.05 M HCl
`=(0.05)/(1000)xx20=1xx10^(-3)` mol
Moles of `OH^(-)` ion is 30 mL of 0.10 M `Ba(OH)_(2)`
`=2xx(0.10)/(1000)xx30`
`=6xx10^(-3)` mol
Moles of `OH^(-)` ions left unneutralised
`=6xx10^(-3)-1xx10^(-3)`
`=5xx10^(-3)` mol
Total volume of solution,
`=20+30=50 mL`
`therefore [OH^(-)]=(5xx10^(-3))/(50)xx1000=0.1 M`