Question

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)`?
A. 0.12M
B. 0.10 M
C. 0.40 M
D. 0.0050 M

Answer 1

Correct Answer - B
20 mL of 0.50 M HCl `= 20xx0.050 m` mol
1.0 m mol = 1.0 meq. of HCl
30 mL of 0.10 M `Ba(OH)_(2)`
`=30xx0.1 m` mol
=3m mol `=3xx2` meq
= 6 meq `Ba(OH)_(2)`
1 meq of HCl will neutralize 1 meq of `Ba(OH)_(2)`
`Ba(OH)_(2)` left =4 meq.
Total volume = 50mL
`Ba(OH)_(2)` conc. in final solution
`=(5)/(50)N=0.1N=0.05M` ltbr. ` [OH^(-)]=2xx0.05 M= 0.10M`
Alternatively ,
`Ba(OH)_(2)+2HCl to BaCl_(2) +2H_(2)O`
2m mol of HCl neutralize 1 m mole of `Ba(OH)_(2)`
1m mol of HCl neutralize 0.5 m mol of `Ba(OH)_(2)`
`Ba(OH)_(2)` left =3-0.5 m mol =2.5 m mol
`[Ba(OH_(2))=(22.5)/(50)M=0.05M`
or ` [OH^(-)]=2 xx 0.05 =01M`