1.

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. `0.40M`B. `0.050M`C. `0.12M`D. `0.10M`

Answer» Correct Answer - D
Normality = n-factor x Molarity. For HCl, n-factor is 1 and for `Ba(OH)_(2)`, n-factor is 2. Thus,
`0.05 0M HCl = 0.50N HCl`
`0.10 M HCI = 0.20 N Ba(OH)_(2)`
Milliequivalents `(H^(+))` = Milliequivalents (HCI)
`=(20 ml)(0.050 N)`
`=1.0`
Millequivalents `(OH^(-))` = Millequivalents `[(Ba(OH)_(2)]`
`=(30 ml)(0.20 N)`
`=6.0`
1 meq of `H^(+)` will neutralized 1 meq of `OH^(-)` . Thus, 5 meqs of `OH^(-)` are left. Since the total volume is `20+30=50 mL` , we have
`N(OH^(-))=(meqs)/(mL)=(5)/(50)"meq mL"^(-1)`
`=0.1` meq `mL^(-1)`
Since `OH^(-)` is univalent, its normality and molarity are the same. Thus,
`M(OH^(-))=0.1 mol L^(-1)`
Alternatively,
`Ba(OH)_(2)+2HCl rarr BaCl_(2)+2H_(2)O`
According to equation, 2 millimoles of HCl neutralize 1 millimole of `Ba(OH)_(2)`. Thus, 1 mmol of HCI will neutralized `0.5` mmol of `Ba(OH)_(2)`. As a result, `3-0.5=2.5` mmol of `Ba(OH)_(2)` are left unneutralized.
`M[Ba(OH)_(2)]= (mmol)/(mL)=(2.5)/(50)=0.05`
`M(OH^(-))=2 M [Ba(OH)_(2)]`
`= 2(0.05)=0.1M`
Note that 1 unit of `Ba(OH)_(2)` yields two units of `OH^(-)` ions.


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