I have no idea where to start with this one...

Calculate the concentration (M) of sodium ions in a solution made by diluting 40.0ml of a 0.474m solution of sodium sulfide to a total volume of 300ml

To calculate the concentration (M) of sodium ions in the solution, you need to consider the dilution factor and the initial concentration of sodium in the solution. Follow these steps:

Step 1: Calculate the initial number of moles of sodium ions in the original solution.
The initial volume (V1) is 40.0 mL and the initial concentration (C1) is 0.474 M. Thus, the initial number of moles (n1) can be calculated using the formula:

n1 = C1 * V1

n1 = 0.474 M * 40.0 mL
Note that the units of volume must be consistent with the units of concentration, so we use mL for both.

Step 2: Calculate the final volume of the diluted solution.
The final volume of the solution (V2) is given as 300 mL.

Step 3: Use the dilution formula to calculate the final concentration of sodium ions.
The dilution formula is given as:

C1 * V1 = C2 * V2

Substituting the known values:

0.474 M * 40.0 mL = C2 * 300 mL

Rearranging the equation to solve for C2:

C2 = (0.474 M * 40.0 mL) / 300 mL

C2 ≈ 0.0632 M

Therefore, the concentration (M) of sodium ions in the diluted solution is approximately 0.0632 M.

To calculate the concentration of sodium ions in the diluted solution, we need to use the concept of dilution. Dilution is a process of reducing the concentration of a solute in a solution by adding more solvent to it.

Here's how you can calculate the concentration (M) of sodium ions in the diluted solution:

Step 1: Determine the moles of sodium ions in the initial solution.
We can use the formula: moles = concentration (M) x volume (L).

Given:
- Volume of initial solution (V1) = 40.0 mL = 0.040 L
- Concentration of initial solution (C1) = 0.474 M

Moles of sodium ions (n1) = C1 x V1

Step 2: Calculate the final volume of the diluted solution.
The final volume of the diluted solution (V2) is given as 300 mL = 0.300 L.

Step 3: Apply the dilution formula to find the concentration of the diluted solution.
The formula for dilution is: C1V1 = C2V2, where C2 is the concentration of the diluted solution.

Rearranging the formula, we get:
C2 = (C1V1) / V2

Substituting the values:
C2 = (n1 / V1) / V2

Step 4: Calculate the concentration of sodium ions in the diluted solution.
Substitute the values of n1, V1, and V2 into the formula to calculate C2.

C2 = (n1 / V1) / V2

C2 = (0.474 M * 0.040 L) / 0.300 L

C2 = 0.0632 M

So, the concentration of sodium ions in the diluted solution is 0.0632 M.

Is this 0.474m or 0.474M? I assume you know there is a difference between the two.

M make it easier to do.

Start M Na2S = 0.474
New M Na2S after dilution = 0.474 x (40 mL/300 mL) = ?M Na2S
Since there are 2 Na ions for each molecule of Na2S, then (Na^+) will be just twice that of Na2S.