100 ml of 0.1 M NaCl and 100 ml of 0.2 MNaOH are mixed. What is the change in pH of NaCI solution ?

Options

A. Some `KMnO_4` is left unreacted
B. 0.02 M `Mn^(2+)` ions are present in the solutions
C. `K_2SO_3` is partially left unreacted
D. Newly formed `SO_4^(2-)` has 0.05 M cocentration

Answer

A, B, D

Explanation

`100xx0.2 =20` meq `O_3^(2-)`
`100xx0.5 =50` meq `MnO_4^(-)`
`:. SO_3^(2-)` in limiting reagent
`MnO_4^(-)` left = 30 meq
`:.[Mn^(+2)]=(20//5)/200=0.05M`
`[SO_4^(-2)]=(20//2)/200 = 0.05M`