If 25.0mL of sodium nitrate solution (0.100mol/L) is added to 10.0mL of sodium carbonate solution (0.150 mol/L), what is the concentration of sodium ions in the resulting mixture? Assume the volumes are additive.
moles + moles = total moles.
volume + volume = total volume.
M = total moles/total volume.
Those aren't mole, those are concentraiton
Hello, i am still needing help
Hello yourself. I know those are not moles. I told you how to work the problem. Do you know how to get moles? Convert each to moles. Find moles Na in each. Add everything together to get final moles etc etc. Let me know if you still need help but tell me what you don't understand.
To find the concentration of sodium ions in the resulting mixture, we need to calculate the total number of moles of sodium ions and then divide it by the total volume of the mixture.
Here's how you can solve the problem step by step:
Step 1: Calculate the number of moles of sodium nitrate.
We are given the volume of sodium nitrate solution as 25.0 mL and the concentration as 0.100 mol/L. To calculate the number of moles, we multiply the volume by the concentration:
moles of sodium nitrate = (25.0 mL) * (0.100 mol/L) = 2.50 mmol
Note: Since the volume was given in milliliters and the concentration in moles per liter, we need to convert milliliters to liters by dividing by 1000.
Step 2: Calculate the number of moles of sodium carbonate.
We are given the volume of sodium carbonate solution as 10.0 mL and the concentration as 0.150 mol/L. We can calculate the number of moles using the same formula as in step 1:
moles of sodium carbonate = (10.0 mL) * (0.150 mol/L) = 1.50 mmol
Step 3: Calculate the total number of moles of sodium ions.
Since sodium nitrate contains one mole of sodium ions for every mole of sodium nitrate, the total moles of sodium ions due to sodium nitrate is 2.50 mmol.
Similarly, sodium carbonate contains two moles of sodium ions for every mole of sodium carbonate. Therefore, the total moles of sodium ions due to sodium carbonate is 1.50 mmol * 2 = 3.00 mmol.
To calculate the total moles of sodium ions in the resulting mixture, we sum up the moles from sodium nitrate and sodium carbonate: 2.50 mmol + 3.00 mmol = 5.50 mmol.
Step 4: Calculate the total volume of the mixture.
Since the volumes of the solutions are additive, the total volume of the mixture is the sum of the volumes of sodium nitrate and sodium carbonate: 25.0 mL + 10.0 mL = 35.0 mL.
Step 5: Calculate the concentration of sodium ions in the resulting mixture.
To find the concentration, we divide the total moles of sodium ions by the total volume of the mixture:
concentration of sodium ions = (5.50 mmol) / (35.0 mL) = 0.157 mol/L
So, the concentration of sodium ions in the resulting mixture is 0.157 mol/L.