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A buffer solution is prepared in which the concentration of `NH_(3)` is `0.30 M` and the concentration of `NH_(4)^(+)` is `0.20 M`. If the equilibrium constant, `K_(b)` for `NH_(3)` equals `1.8xx10^(-5)`, what is the `pH` of this solution? (`log 2.7=0.43`)A. 9.08B. 9.43C. 11.72D. 8.73 |
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Answer» Correct Answer - B `pOH=pK_(b)+log.(|NH_(3)^(+)|)/(|NH_(3)|)` `=-log (1.8 xx 10^(-5))+log. (0.2)/(0.3)` `=4.74 +log .(0.2)/(0.2)=4.56` `pH ==14-4.56=9.44` |
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