In above question concentration of Triethyl ammonium ion `[C_(6)NH_(16)]^(+)` in resulting solution will be :
In above question concentration of Triethyl ammonium ion `[C_(6)NH_(16)]^(+)` in resulting solution will be :
A. `100K_(b)`
B. `200K_(b)`
C. `10K_(b)`
D. `K_(b)`
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`K_(b)=([OH^(-)][C_(6)NH_(16)^(+)])/([C_(6)NH_(15)])rArr ((10^(-2))/(2)[C_(6)NH_(16)^(+)])/((10^(-2))/(2))rArr[C_(6)NH_(16)^(+)]=K_(b)`
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