SHOW WORK What volume of concentrated nitric acid (15.8 M) is needed to prepare 4.0 L of a 2.0 M solution?

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

Also you kinda cute what’s up :)

Hno3(aq) → h+(aq) + no3−(aq). picture ii best represents the strong acid hno3. strong acids are strong electrolytes