1.) A solution is prepared by dissolving 5.00 g of stannic nitrate in enough water to make 250.0 mL of stock solution. A 15.0 mL aliquot (portion) of this stock solution is then removed and added to 75.0 mL of water. Calculate the concentrations of all ions in the final solution.

This is my first time learning this so please do the above as an example. I have more to do so this can be an example.

Step 1 is to calculate the concn of the stock solution so you will know what you're starting with.

mols Sn(NO3)4 = grams/molar mass
Step 2. The M of the stock soln is mols/L soln or mol/0.250. Let's call this y molar.

Then dilutions are handled with the dilution formula which is
c1v1 = c2v2 where
c = concn
v = volume.
In this problem you will have
15.0mL x yM = ?M x 90 mL and you solve for ?M. This is all true IF volumes are additive. Technically they are not (and they aren't here) but the difference is so small it won't make any difference. I'm sure the asker means for you to add the 75 mL to 15 mL to end up with a solution volume of 90 mL.

Now you have M of the diluted solution of Sn(NO3)4.
Since there is 1 mol Sn^4+ per 1 mol Sn(NO3)4 the M Sn^4+ is the same as M of the solution.
M of the NO3^- is 4x M of Sn(NO3)4 because there are 4 mols NO3^- for each mol Sn(NO3)4.

Sure! Let's break down the problem step by step.

First, we need to calculate the concentration of the stock solution. We are given that 5.00 g of stannic nitrate is dissolved in enough water to make 250.0 mL of solution.

To find the concentration, we need to convert the mass of stannic nitrate to moles and then divide it by the volume of the solution.

Stannic nitrate (Sn(NO3)4) has a molar mass of 388.75 g/mol. So, we can calculate the moles of stannic nitrate as follows:

moles of Sn(NO3)4 = mass / molar mass
moles of Sn(NO3)4 = 5.00 g / 388.75 g/mol

Next, we need to calculate the molarity (concentration) of the stock solution:

Molarity = moles / volume (in liters)
Molarity = moles of Sn(NO3)4 / volume of solution (in liters)
Molarity = moles of Sn(NO3)4 / (250.0 mL / 1000)

Now, we can calculate the concentration of the stock solution by plugging in the values:

Molarity = (5.00 g / 388.75 g/mol) / (250.0 mL / 1000)
Molarity ≈ 0.0129 mol/L

So, the concentration of stannic nitrate in the stock solution is approximately 0.0129 mol/L.

Now, let's move on to calculating the concentrations of all ions in the final solution.

We start with a 15.0 mL aliquot (portion) of the stock solution and dilute it by adding 75.0 mL of water. This means the final volume of the solution is 15.0 mL + 75.0 mL = 90.0 mL.

Given that, we can calculate the final concentrations of ions using the concept of dilution. The number of moles of ions remains the same before and after dilution, but the volume changes.

The moles of Sn(NO3)4 in the final solution can be calculated as:
moles of Sn(NO3)4 = (Molarity) x (volume of final solution in L)
moles of Sn(NO3)4 = (0.0129 mol/L) x (90.0 mL / 1000)

Next, we need to calculate the concentration of each ion in the final solution. Since stannic nitrate dissociates into components, we will consider the ions separately.

The concentration of stannic ions (Sn4+) will be the same as the moles of Sn(NO3)4 divided by the final volume of the solution:
Concentration of Sn4+ = moles of Sn4+ / volume of final solution (in L)

Similarly, the concentration of nitrate ions (NO3-) will be:
Concentration of NO3- = moles of NO3- / volume of final solution (in L)

Now, substitute the known values to calculate the concentrations. Remember that the moles of each ion will be the same as the moles of Sn(NO3)4:

Concentration of Sn4+ = (moles of Sn(NO3)4) / (volume of final solution (in L))
Concentration of Sn4+ = (0.0129 mol/L) x (90.0 mL / 1000)

Concentration of NO3- = (moles of Sn(NO3)4) / (volume of final solution (in L))
Concentration of NO3- = (0.0129 mol/L) x (90.0 mL / 1000)

Evaluate the above expressions to find the concentrations of each ion in the final solution.

I hope this example helps you understand the process of calculating concentrations in a solution. Let me know if you have any more questions or if there's anything else I can assist you with!