How many moles of `NH_(4)C1` should be added to `200mL` solution of `1.18 M NH_(4)OH` to have a `pH` of `9.60. K_(b)` of `NH_(4)OH= 2 xx 10^(-5)`

How many moles of `NH_(4)C1` should be added to `200mL` solution of `1

1.	How many moles of `NH_(4)C1` should be added to `200mL` solution of `1.18 M NH_(4)OH` to have a `pH` of `9.60. K_(b)` of `NH_(4)OH= 2 xx 10^(-5)`
Answer» Correct Answer - A It is basic buffer. `pH = 9.6, pOH = 14 - 9.6 = 4.4` `K_(b) = 2 xx 10^(-5), pK_(b) =- log (2xx10^(-5)) =- 0.3 +5 = 4.7` `pOH = pK_(b) + "log"(["Salt"])/(["Base"])` `4.4 = 4.7 + "log"(["Salt"])/((0.18))` `- 0.3 = log["Salt"] - log (0.18) = log ["Salt"] + 0.7447` `- 0.3 - 0.7447 = log ["Salt"]` `:. log ["Salt"] =- 1.0447` `["Salt"] = "Antilog" (-1-0.447+1-1)` `= "Antilog" (bar(2).9553) = 9.55 xx 10^(-2)M`. Moles of salt `= (9.55 xx 10^(-2) xx 200)/(1000)` `= 1.91 xx 10^(-2)= 0.019` mol.

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How many moles of `NH_(4)C1` should be added to `200mL` solution of `1.18 M NH_(4)OH` to have a `pH` of `9.60. K_(b)` of `NH_(4)OH= 2 xx 10^(-5)`

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