Molandy ab Hc=0.10Mvolume taken =7.59mlm*mol=M_(xV)=0.10 xx7.59=0.759of KdadderlBuffer of CH_(3)Nth andIt is Basic Buffermmal of CH_(8)NH_(2)=MNV=173.38 xx0.50=86.69m. mal of CrsMH_(3)ar^(-)=231.71 xx10=281.71when Acid is added to Basic Buffer concentration of sat ancreare and that of Base decreale80.0.759m. mal of HCl veact with ... See the full answer