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What is the [OH^-] in the final solution...

What is the `[OH^-]` in the final solution prepared by mixing 20.0mL of 0.50 M HCl with 3.0 mL of 0.10 M `Ba(OH)_2?`

A

0.40M

B

0.0050M

C

0.12M

D

0.10M

Text Solution

Verified by Experts

The correct Answer is:
D
Millimoles of `H^+` produced `=20 xx 0.05=1`
Millimoles of `OH^-` produced `=30 xx 0.1xx 2=6 ("Each "Ba(OH)_2" gives "2OH^(-))`
Millimoles of `OH^-` remaining in solution =6-1=5
Total volume of solution =20+30=50mL
`[OH^-]=(5)/(50)=0.1 M`
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