posted by Monty on .
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
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
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