Molar Concentration = Molarity(M) = mole of speices(mol)/Total volume(L)

steps:

1)Write a dissociation/ionization equation.
2)Use stoicheometry to set up the mole ratio.

steps with example

To begin, it will be important for you to be able to write balanced reactions that show how these substances break down into ions. If you are given the the formula of the compound (either ionic or acid) you will need to be able to determine:

which ions are produced and

in what mole ratio.

For example, sodium carbonate, Na2CO3 dissociates into ions as:
Na2CO3 (s) → 2 Na+(aq)+ CO32-(aq)
Notice that two mole of Na+ ions are produced for every one mole of Na2CO3 . The total volume of the solution, however, remains unchanged.
If we have a 0.20 M solution of Na2CO3 , what will be the concentration of our two ions, Na+ and CO32-? This can easily be determined from the coefficients in our balanced equation:

## Concentration of ions in Solution

## formula

Molar Concentration = Molarity(M) = mole of speices(mol)/Total volume(L)## steps:

1)Write a dissociation/ionization equation.

2)Use stoicheometry to set up the mole ratio.

steps with exampleTo begin, it will be important for you to be able to write balanced reactions that show how these substances break down into ions. If you are given the the formula of the compound (either ionic or acid) you will need to be able to determine:

- which ions are produced and
- in what mole ratio.

For example, sodium carbonate, Na2CO3 dissociates into ions as:Na2CO3 (s) → 2 Na+(aq)+ CO32-(aq)

Notice that two mole of Na+ ions are produced for every one mole of Na2CO3 . The total volume of the solution, however, remains unchanged.

If we have a 0.20 M solution of Na2CO3 , what will be the concentration of our two ions, Na+ and CO32-? This can easily be determined from the coefficients in our balanced equation:

[Na+] = 2 × [Na2CO3] = 2 × 0.20 M = 0.40 M

[CO32-] = 1 × [Na2CO3] = 1 × 0.20 M = 0.20 M

Reference:http://www.saskschools.ca/curr_content/chem30_05/4_solutions/solution2_5.htm

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