# What volume of water would you add to 15.00 mL of a 6.77 M solution of nitric acid in order to get a 1.50 M solution?

May 1, 2014

This dilution problem uses the equation

${M}_{a} {V}_{a} = {M}_{b} {V}_{b}$

${M}_{a}$ = 6.77M - the initial molarity (concentration)
${V}_{a}$ = 15.00 mL - the initial volume
${M}_{b}$ = 1.50 M - the desired molarity (concentration)
${V}_{b}$ = (15.00 + x mL) - the volume of the desired solution

(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x

59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.