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What is the [OH^(-)] in the final soluti...

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

0.10 M

B

0.40 M

C

0.0050 M

D

0.12 M

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(a)V_(a)=0.05xx20=1`
`N_(b)V_(b)=0.2xx30=6`
`N_(b)V_(b)gt N_(a)V_(a)`
`N=(N_(b)V_(b)-N_(a)V_(a))/("Final volume")=(6-1)/(50)=(5)/(50)=10^(-)=10^(-1)=0.1M=[OH^(-)]`
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