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# 0.1 Mole Of CH3NH2 (Kb=5x10-4) is mixed with 0.08 Mole of HCl and diluted to one Litre. What will be the H+ Concentration In the solution?

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## The reaction occurring is :${\mathrm{CH}}_{3}{\mathrm{NH}}_{2}+\mathrm{HCl}\to {\mathrm{CH}}_{3}{\mathrm{NH}}_{2}^{+}{\mathrm{Cl}}^{-}\phantom{\rule{0ex}{0ex}}0.10.080\phantom{\rule{0ex}{0ex}}0.0200.08\phantom{\rule{0ex}{0ex}}\left(0.1-0.08\right)$The reaction mixture contains unreacted methylamine(CH3NH2) and hydrochloride salt(HCl).Thus, it is a basic buffer solution.The expression for the hydroxide ion concentration is[OH-] = ${\mathrm{k}}_{\mathrm{b}}×\left[\frac{\mathrm{Base}}{\mathrm{Conjugate}\mathrm{acid}}\right]$Substitute values in the above expression= $5×{10}^{-4}×\left[\frac{0.02}{0.08}\right]=1.25×{10}^{-4}$But $\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{k}}_{\mathrm{W}}={10}^{-14}$$\left[{\mathrm{H}}^{+}\right]\left[1.25×{10}^{-4}\right]={10}^{-14}\phantom{\rule{0ex}{0ex}}\left[{\mathrm{H}}^{+}\right]=\frac{{10}^{-14}}{\left[1.25×{10}^{-4}\right]}=8×{10}^{-11}\phantom{\rule{0ex}{0ex}}$0.1 Mole Of CH3NH2 (Kb=5x10-4) is mixed with 0.08 Mole of HCl and diluted to one Litre. The H+ Concentration In the solution is 8x10-11

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