Volume-titration and pH

Question

What volume (in mL) of Ba(OH)2 0.23 M is necessary to titrate 5.0 mL of 1.3 M H2SO4

Answer 1

Ba(OH)2 + H2SO4 ==> BaSO4 + 2H2O ... balanced equation

moles H2SO4 present = 5.0 ml x 1 L/1000 ml x 1.3 mol/L = 0.0065 moles H2SO4

moles Ba(OH)2 needed to neutralize this = 0.0065 moles Ba(OH)2 since mole ratio is 1:1

volume Ba(OH)2 needed to obtain 0.0065 moles: (x L)(0.023 mol/L) = 0.0065 moles

(x L)(0.35 mol/L) = 0.0065 moles

x = 0.02826 L x 1000 ml/L = 28.26 ml = 28 mls Ba(OH)₂ needed (to 2 significant figures)