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Calculate the pH of `0.5L` of a`0.2 M NH_(4)Cl-0.2 M NH_(3)` buffer before and after addition of (a)` 0.05` mole of `NaOH` and (b) `0.05` mole of `HCl`. Assume that the volume remains constant.[Given:`pK_(a)`of `NH_(3)=4.74]` |
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Answer» `pOH=4.74+log((0.1)/(0.1))=4.74` `pH=9.26` (a) `{:(,NH_(4)^(+),+,H^(+),rarr,NH_(4)^(+)),(t=0,0.1"mole",,0.05"mole",,0.1"mole"),(,0.05"mole",,-,,0.15"mole"):}` `pOH=4.74+log((pOH=4.74+log((0.15)/(0.05))=4.26` `:. pH=9.74` (b)` {:(,NH_(3)^(+),+,H^(+),rarr,NH_(4)^(+)),(t=0,0.1"mole",,0.05"mole",,0.1"mole"),(,0.05"mole",,-,,0.15"mole"):}` `pOH=4.74+log((0.15)/(0.05))=5.22` `:.pH=8.78`. |
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