0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution. Calculate pH of the resulting solution, assuming no change in volume. (K, for pyridine = 1.5 xx 10^(-9)M)

0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2

1.	0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution. Calculate pH of the resulting solution, assuming no change in volume. (K, for pyridine = 1.5 xx 10^(-9)M)
Answer» SOLUTION :Pyridiniumchloride = 0.15 mole Pyridine 500 ` cm ^(3)"of "0.2 M = 0.1 ` mole ` pOH = pK_b +log "" ([s]) /(["Base" ]) ` ` =- log (1.5 xx 10 ^(-9) ) +log"" (0.15)/(0.1) ` ` =9- log 1. 5+ log 1.5 =9` ` PH =14- 9 = 5`