A solution consists of 0.2 M NH4OH and 0.2M NH4Cl. if Kb of NH4OH is 1.8×10-5,theOH- of the resulting solution is

pOH=pKb+logSB-logOH-=-log1.8×10-5+log0.20.2OH-=1.8×10-5

A solution consists of 0.2 M NH4OH and 0.2M NH4Cl. if Kb of NH4OH is 1.8×10-5,theOH- of the resulting solution is

pOH=pKb+logSB-logOH-=-log1.8×10-5+log0.20.2OH-=1.8×10-5

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A solution consists of 0.2 M NH₄OH and 0.2M ${NH}_{4} Cl . if K_{b} of {NH}_{4} OH is 1.8 \times 10^{- 5}, the [{OH}^{-}]$ of the resulting solution is

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$3.6 \times 10^{- 5} M$

$3.2 \times 10^{- 5} M$

$0.9 \times 10^{- 5} M$

$1.8 \times 10^{- 5} M$

answer is B.

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Detailed Solution

$pOH = {pK}_{b} + \log \frac{[S]}{[B]}$

$- \log [{OH}^{-}] = - \log (1.8 \times 10^{- 5}) + \log (\frac{0.2}{0.2})$

$[{OH}^{-}] = 1.8 \times 10^{- 5}$

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A solution consists of 0.2 M NH4OH and 0.2M NH4Cl. if Kb of NH4OH is 1.8×10-5,theOH- of the resulting solution is

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A solution consists of 0.2 M NH₄OH and 0.2M ${NH}_{4} Cl . if K_{b} of {NH}_{4} OH is 1.8 \times 10^{- 5}, the [{OH}^{-}]$ of the resulting solution is