If `100 mL` of `1.00 M HCl` and `100mL` of `0.80 M NaOH` solution are mixed, the molarity of `Cl^(-)` ions in the resulting solutions will be

Options

A. `0.10 M`
B. `0.50 M`
C. `0.40 M`
D. `0.30 M`

Answer

2

Explanation

Millimoles of `HCl=V_(mL)xxM=(100 mL)(1.00 M)`
`100` mmol
Millimoles of `NaOH=(100 mL)(0.80 M)=80` mmol
`" "HCl+NaOH rarr NaCl+H_(2)O`
`{:("Rxn ratio:", 1 mmol,1 mmol,1 mmol,1 mmol),("Start:",100 mmol,80 mmol,0 mmol,),("Change:",-80 mmol,-80 mmol,+80 mmol,))/(After rxn: 20 mmol,0 mmol, 80mmol)`
Molarity of unused `HCl=(20 mmol HCl)/(200 mL)`
`=0.10 M HCl`
Molarity of `NaCl` formed=`(80 mmol)/(200 mL) NaCl`
`=0.40 M NaCl`
Both `HCl` and `NaCl` are strong electrolytes, so the solution is `0.10 M` in `H^(+) (aq.), (0.10+0.40)M=0.50 M` in `Cl^(-)`, and `0.40 M` in `Na^(+)` ions.