A25.00-ml sample of 0.723 m hclo4 is titrated with a 0.273 m koh solution. the h3o + concentration after the addition of 10.0 ml of koh is m.

  Moles HClO4 = 0.02500 L x 0.723 M=0.0181 
moles KOH = 0.0100 L x 0.273 M= 0.00273 

total volume = 25.00 + 10.0 = 35.0 mL = 0.0350 L 

moles H+ in excess = 0.0181 - 0.00273 =0.0154 
[H+]= 0.0154 mol/ 0.0350 L=0.440 M

It will fall off and tip over.