Calculate the amount of `NH_(4)C1` required to dissolve in `500mL` of water to have a `pH = 4.5, K_(b) = 2.0 xx 10^(-5)`.

Answer

C

Explanation

`NH_(4)C1` is a salt of `W_(B)//S_(A). (pK_(b) =- (log 2xx 10^(-5)) =- 0.3 +5 = 4.7`
`:. pH = (1)/(2) (pK_(w) - pK_(b) - logC)`
`4.5 = (1)/(2) (14 - 4.7 - logC)`
`9 = 9.3 - logC`,
`log C = 0.3`
`C = "Antilog" (0.3) ~~ 2M~~2 moles//L ~~1 mol//500 mL` Weight of `NH_(4)C1 = 53.5g.//500mL (Mw NH_(4)C1 = 53.5 g mol^(-1))`.