InterviewSolution
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InterviewSolution
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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. 0.40 MB. 0.0050 MC. 0.12 MD. 0.10 M |
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Answer» Correct Answer - D Millimoles of `H^(+)` produced `=20 xx 0.05 =1` Millimoles of `OH^(-)` produced `=30 xx 0.1 xx 2` , (Each `Ba(OH)_(2)` given two `OH^(-)` ions) =6 Millimoles of `OH^(-)` ions in the remaining solution =6 -1 =5 Total volume of solution =20 + 30 =50 mL `[OH^(-)] = (5)/(50) =0.1 M` |
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