# What volume of 12.0 N and 6.0 N HCl must be mixed to give a 2.5 L solution with a concentration of 10.0 N HCl?

Relevance
•  2.5 L solution with a concentration of 10.0 N HCl ===

2.5 X 10.0 =  25 mol HCl needed

.. let x - vol of 12 N .. and y = vol of 6 N

total vol 2.5  so  y = 2.5 - x

12.0x  +  6.0y = 25 mol

12.0x + 6.0(2.5 - x)  = 25  <<  solve for x

6x = 10

x = 1.666666L of 12.0N

y = 0.833333L of 6.0N

round for sig figs

• N₁V₁ + N₂V₂ = N₃V₃ .....(1)

we know V₁+V₂ =V₃, then V₂ = V₃-V₁

plug in (1)

N₁V₁ + N₂(V₃-V₁) = N₃V₃

N₁V₁ + N₂V₃ - N₂V₁ = N₃V₃

(N₁ - N₂)V₁ = (N₃ - N₂)V₃

V₁ = (N₃ - N₂)V₃ / (N₁ - N₂)

V₁ = (10.0 - 6.0)2.5 / (12.0 - 6.0)

V₁ = 1.7 L

V₂ = 2.5 - 1.7 = 0.8 L

We need 1.7L of 12.0N HCl solution and 0.8 L of 6.0N HCl solution

• Let the volume of 12 N HCl = V1

volume of 6 N HCl = V2

The resulting molarity = 10 N

wehave , ( N1 V1 + N1 V2 / V1 +V2 ) = 10N ----------- 1

V1 + V2 = 2.5 L

∴ V1 = 2.5 -V2

substituting this value in the equation 1 we get,

(12* ( 1-V2 ) + 6 V2 / 1 L) = 10 N

12 - 6 V2 = 10

∴ V2 = 0.33 L

V1 = 1 - (1/3) L

= (2/3) L