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`K_(b)` for `NH_(4)OH` is `1.8 xx 10^(-5)`. The `[overset(Theta)OH]` of `0.1 M NH_(4)OH` isA. `5.0 xx 10^(-2)`B. `4.20 xx 10^(-3)`C. `1.34 xx 10^(-3)`D. `1.8 xx 10^(-6)` |
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Answer» Correct Answer - C `K_(b) = 1.8 xx 10^(-5)` `pK_(b) = 4.7447` `pOH_(W_(B)) = (1)/(2) (pK_(b) - log 0.1)` `= (1)/(2) (4.7447 +1) = 2.87`. `[overset(Theta)OH] = "Antilog" (-2.87)` `= "Antilog" (-2-0.87 + 1-1)` `= "Antilog" (bar(3).13)` `= 1.34 xx 10^(-3)` |
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