chemistry: Ion concentrations

In an aqueous solution of ions of a species is related to the number of males of that species per concentration of the substance in the aqueous solution.

Molarity is the number of moles of a solute divided by the total volume of the solution. M=n/v

here is an example problem

Example 2

Determine the concentration of

K^{+}

in an aqueous solution of 0.238 M

K N O_{3}

SOLUTION

Since there is one mole of potassium in

K N O_{3}

, multiply the concentration of the species by the number of moles of the atom to obtain:

[K^{+}] = (0.238 M K N O_{3}) \times (1 m o l K^{+}) = 0.238 M

Although not asked, there is also one mole of nitrite ions in one mole of

K N O_{3}

, so its concentration is also 0.238 M:

[N O_{3}^{-}] = (0.238 M K N O_{3}) \times (1 m o l N O_{3}^{-}) = 0.238 M

FOLLOWUP

The stoichiometry always dictates the concentration, which was a simple 1:1 ratio for

K N O_{3}

. However, for more complex situations, different ratios will be encountered. For instance, if consider the dissolving of

A l_{2} (S O_{4})_{3}

A l_{2} (S O_{4})_{3} \to 2 A l_{(a q)}^{3 +} + 3 S O_{4 (a q)}^{2 -}

If the concentration of

A l_{2} (S O_{4})_{3}

is 0.019 M, what is the concentration of

A l_{(a q)}^{3 +}

? Simply multiply 0.019 M by the stoichiometric factor of

A l_{(a q)}^{3 +}

A l_{2} (S O_{4})_{3}

, which is 2:1. The concentration of

A l_{(a q)}^{3 +}

then becomes 0.038 M:

[A l^{3 +}] = (0.019 M A l_{2} (S O_{4})_{3}) \times (2 m o l A l^{3 +}) = 0.238 M

Although not asked, the concentration of

S O_{4}^{2 -}

is 0.057 M via the same argument;

http://chemwiki.ucdavis.edu/Core/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Unique_Features_of_Aqueous_Solutions

[S O 2 - 4] = (0.019 M) \times (3 m o l S O 2 - 4) = 0.057 M

chemistry

Thursday, March 10, 2016

Ion concentrations

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