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

5

Explanation

Pyridinium chloride = 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`