How, would you prepare 200.0 mL of 0.100 M nitric acid solution from 6.0M stock solution?

Answer 1

Hello, Isabella,

We'll find this answer by first looking into how many moles of nitric acid are actually required.

Molar is defined as moles/liter. To find the moles in 200ml of 0.100 M nitric acid we need to recognize that it can be written as 0.100 moles/liter. If we multiply that by the volume we have (200ml) the volumes will cancel, leaving us moles. But the units are different - one is liters and the other ml. Let's convert the 200ml into liters: 200ml*(1 liter/1000ml) = 0.200 liter nitric acid.

Now we can find the moles required: 0.200L*(0.100moles/L) = 0.0200 moles HNO₃.

The concentrated HNO₃ stock solution is 6.0 moles/Liter. We only need 0.0200 moles, so we want a volume that when multiplied times the concentration, will give that amount:

(6 moles/liter)*(Liters Required) = 0.0200 moles

Liters Required = 0.00333 liters, or 3.33 ml.

Add 3.33 ml of 6.0M HNO₃ to 196.7 (200 - 3.33) ml of water to produce 200 ml of a 0.100M HNO₃ solution.

I hope this helps,

Bob

Answer 2

Use V1M1 = V2M2

V1 = initial volume of HNO3 = ?

M1 = initial molarity of HNO3 = 6.0 M

V2 = final volume of HNO3 = 200.0 ml

M2 = final molarity of HNO3 = 0.100 M

(V1)(6.0 M) = (200.0 ml)(0.100 M)

V1 = 3.3 ml

Take 3.3 ml of 6.0 M HNO3 and dilute to a FINAL volume of 200.0 mls