How many milliliters of 0.225 M NH4C2H3O2 are needed to make 750.0 mL of 0.1667 M NH4C2H3O2?

dilution formula

ok so did I do it right??

0.225(750.0)=0.1667(v)
168.75/0.1667=1012.3??

Isn't this the same as the above?

0.225 x mL = 0.1667 x 750 mL.
mL = ?
The problem asks for "how many mL of 0.225M must ......."
so you have the v (unknown) in the wrong place.

To solve this problem, we can use the concept of dilution. The equation for dilution is often written as C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

In this case, the initial concentration (C1) is given as 0.225 M, the initial volume (V1) is unknown, the final concentration (C2) is given as 0.1667 M, and the final volume (V2) is given as 750.0 mL.

Let's substitute these values into the dilution equation and solve for V1:

(C1)(V1) = (C2)(V2)
(0.225 M)(V1) = (0.1667 M)(750.0 mL)

First, let's convert the final volume from milliliters (mL) to liters (L) because the concentrations are given in moles per liter:

V2 = 750.0 mL = 750.0 mL / 1000 mL/L = 0.750 L

Let's plug in the values again:

(0.225 M)(V1) = (0.1667 M)(0.750 L)

Next, solve for V1 by dividing both sides of the equation by (0.225 M):

V1 = (0.1667 M)(0.750 L) / (0.225 M)

Calculating this gives:

V1 = 0.557 L or 557 mL

Therefore, you would need 557 mL of 0.225 M NH4C2H3O2 to make 750.0 mL of 0.1667 M NH4C2H3O2.