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`CH_(3)NH_(2) (K_(b) = 5 xx 10^(-4))` के `0.1` मोल को HCl के 0.08 मोल में मिलाया जाता हैं तथा आयतन 1 लीटर कर लिया जाता हैं| विलयन में `[H^(+)]` का मान हैं:
A. `8 xx 10^(-2)` M
B. `8 xx 10^(-11)` M
C. `1.6 xx 10^(-11)` M
D. `8 xx 10^(-5)` M

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Correct Answer - b
`CH_(3)NH_(2) + HCl to CH_(3)NH_(3)^(+)Cl^(-)`
`{:(0,1,0.08,0),(0.02,0,0,08):}`
यह क्षारीय बफर विलयन हैं|
`[OH^(-)]=K_(b) xx [क्षार]/[लवण] = 5 xx 10^(-4) xx 0.02/0.08`
`therefore [H^(+)] = 10^(-14)/([OH^(-)]) = 10^(-14)/(1.25 xx 10^(-4)) = 8 xx 10^(-11)` M

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