`0.15` mole of pyridinium chloride has been added into `500 cm^(3)` of `0.2M` pyridine solution. Calculate pH and hydroxyl ion contration in the resulting solution, assuming no change in volume. `(K_(b)"for pyridine" = 1.5xx10^(-9)M)`

`0.15` mole of pyridinium chloride has been added into `500 cm^(3)` of

1.	`0.15` mole of pyridinium chloride has been added into `500 cm^(3)` of `0.2M` pyridine solution. Calculate pH and hydroxyl ion contration in the resulting solution, assuming no change in volume. `(K_(b)"for pyridine" = 1.5xx10^(-9)M)`
Answer» `["Pyridinium chlodide"]= (0.15//500)xx1000=0.3M` `["Pyridine"]= 0.2M` :. A mixture of pyridine and its salt pyridinium chloride forms a basic a buffer and therefore, `pOH= -log K_(b)+log ["Salt"]//["Base"]` or `pOH= -log 1.5xx10^(-9)+log(0.30//0.20)` `= -log 1.5+9 log 10+log 1.5=9` `:. [OH^(-)]=10^(-9)` and `[H^(+)]=10^(-5)` So `pH=5`

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`0.15` mole of pyridinium chloride has been added into `500 cm^(3)` of `0.2M` pyridine solution. Calculate pH and hydroxyl ion contration in the resulting solution, assuming no change in volume. `(K_(b)"for pyridine" = 1.5xx10^(-9)M)`

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