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

Moles of `H^+` ions in 20 mL of 0.05 M HCl
`=0.05/1000xx20=1xx10^(-3)` mol
Moles of `OH^-` ion in 30 mL of 0.10 M `Ba(OH)_2`
`=2xx0.10/1000xx30`
`=6xx10^(-3)` mol
Moles of `OH^-` ions left unneutralised
=`6xx10^(-3) -1xx10^(-3)`
`=5xx10^(-3)` mol
Total volume of soution ,
=20+30=50 mL
`therefore [OH^-]=(5xx10^(-3))/50xx1000`=0.1 M