A HCl is titrated with a NaOH solution of known concentration. How much volume of NaOH 0.120 M is necessary to neutralize 14.00 mL of HCl 0.500 M.

Answer 1

Balanced equation for the reaction taking place:

HCl + NaOH ===> NaCl + H2O

You can use V1M1 = V2M2 in which case you have...

(x ml NaOH)(0.120 M) = (14.00 ml HCl)(0.500 M HCl)

x = 58.33 mls of NaOH = 58.3 mls (to 3 sig. figs.)

You could also find moles of HCl present and equate that to volume of NaOH:

moles HCl = 0.01400 L x 0.500 mol/L = 0.007 moles HCl

moles NaOH needed = 0.007 moles HCl x 1 mol NaOH/mole HCl = 0.007 moles NaOH

Volume NaOH = (x L)(0.120 mol/L) = 0.007 moles and x = 0.583 L = 58.3 mls (to 3 sig. figs.)

Answer 2

This would be a M1V1 = M2V2 problem. You would simply find out M1 and plug in the values for V1, M2, V2.