1.

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 ?

Answer»

Some `KMnO_4`is LEFT unreacted
0.02 M `MN^(2+)` ions are present in the solutions
`K_2SO_3` is PARTIALLY left unreacted
Newly formed `SO_4^(2-)` has 0.05 M cocentration

Solution :`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`


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