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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» Correct Answer - C `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))`. |
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