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