What volume of a .500 M HCl solution is needed to neutralize each of the following:

10 ml of a .300 M NaOH solution
10 mL of a .200 M Ba(OH)2 solution

1 answer

Please remember to include the leading zero when writing values.

Start with the equation for the neutralisation

HCl + NaOH -> NaCl + H2O

so one mole of HCl neutralises 1 mole of NaOH

we have

10 x 0.300 mmoles of NaOH so we need the same number of mmoles of HCl

so if V is the volume of HCl

10 x 0.300 = V x 0.500

solve for V for the volume of HCl

repeat for Ba(OH)2