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Solution of `0.1 N NH_(4)OH` and `0.1 N NH_(4)Cl` has `pH 9.25`, then find out `K_(b)` of `NH_(4)OH`.A. `9.25`B. `4.75`C. `3.75`D. `8.25` |
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Answer» Correct Answer - B It is a basic buffer. This, `pOH=pK_(b)+log.(C_("salt"))/(C_("base"))` For any aqueous solution at 298 K, we have `pH+pOH=pK_(w)=14` or `pOH=14-pH` `= 14-9.25` `=4.75` we are given `C_("salt")=C_("base")`. Thus, `log.(C_("salt"))/(C_("base"))=log 1= 0` Hence, `pOH= pK_(b)=4.75`. |
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