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?

This dilution problem uses the equation

##M_aV_a = M_bV_b##

##M_a## = 6.77M - the initial (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.

I hope this was helpful. SMARTERTEACHER