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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. `8.73`B. `9.08`C. `9.43`D. `11.72` |
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Answer» Correct Answer - C `pOH=pK_(b)+log``([Conjugate acid])/([base])` `= -log1.8xx10^(-5)+log``(0.2)/(0.3)` `= -log 1.8xx10^(-5)+log 0.66` `= 4.744-0.176=4.567` `:. pH=14-4.567=9.423` |
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