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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
No. of milli equivalent of HCl = `20 xx 0.05 = 1.0`
No. of milli equivalent of
`Ba(OH)_(2)= 30 xx 0.10 xx 2 = 6.0`
After neutralization, no. of milli equivalents in 50mL. Of solution No. of milli equivalent of solution = `(6-1) = 5`
No. of milli equivalent of `OH^(-)` is 5 in 50 mL .
`[OH^(-)] = (5xx1000)/50 xx 10^(-3) = 0.1 M `
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