How do you calculate concentration of ions in a solution?
Answer
593.4k+ views
Hint As we know that the concentration of ions in a solution depends on the mole ratio between the substance that is dissolved and the anions and cations it forms in the solution.
Complete Step by step solution:
- If we have a compound that dissociates into anions and cations, then it is found that the minimum concentration of each of the products that is cation or anion will be equal to the concentration of the original compound.
- Let us consider an example of NaCl:
$NaCl\left( aq \right)\to N{{a}^{+}}\left( aq \right)+C{{l}^{-}}\left( aq \right)$
- It is found that when sodium chloride dissociates into sodium cations and chloride anions.
- We can see here that the mole of NaCl will produce 1 mole of sodium and 1 mole of chloride.
We can see here that this means if we have a NaCl solution that is having a concentration of 1 M. Then the chloride ion concentration will be 1 M and the sodium ion concentration will be 1 M as well.
- As we know that if the volume of the solution remains constant, and the number of moles doubles, then it is found that the concentration will be two times greater for that ion.
Note:
- We can write the concentration of the compound mathematically as:
\[\begin{align}
& {{C}_{compound}}=\dfrac{{{n}_{compound}}}{V}\Rightarrow V=\dfrac{{{n}_{compound}}}{{{C}_{compound}}} \\
& {{C}_{ion}}=\dfrac{{{n}_{ion}}}{V}={{n}_{ion}}\dfrac{1}{V}={{n}_{ion}}\dfrac{{{C}_{compound}}}{{{n}_{compound}}} \\
& {{C}_{ion}}={{C}_{compound}}\dfrac{{{n}_{ion}}}{{{n}_{compound}}} \\
\end{align}\]
- We can see here that the mole ratio between the original compound and ion that it will form will basically determine the concentration of the respective ion in the solution.
Complete Step by step solution:
- If we have a compound that dissociates into anions and cations, then it is found that the minimum concentration of each of the products that is cation or anion will be equal to the concentration of the original compound.
- Let us consider an example of NaCl:
$NaCl\left( aq \right)\to N{{a}^{+}}\left( aq \right)+C{{l}^{-}}\left( aq \right)$
- It is found that when sodium chloride dissociates into sodium cations and chloride anions.
- We can see here that the mole of NaCl will produce 1 mole of sodium and 1 mole of chloride.
We can see here that this means if we have a NaCl solution that is having a concentration of 1 M. Then the chloride ion concentration will be 1 M and the sodium ion concentration will be 1 M as well.
- As we know that if the volume of the solution remains constant, and the number of moles doubles, then it is found that the concentration will be two times greater for that ion.
Note:
- We can write the concentration of the compound mathematically as:
\[\begin{align}
& {{C}_{compound}}=\dfrac{{{n}_{compound}}}{V}\Rightarrow V=\dfrac{{{n}_{compound}}}{{{C}_{compound}}} \\
& {{C}_{ion}}=\dfrac{{{n}_{ion}}}{V}={{n}_{ion}}\dfrac{1}{V}={{n}_{ion}}\dfrac{{{C}_{compound}}}{{{n}_{compound}}} \\
& {{C}_{ion}}={{C}_{compound}}\dfrac{{{n}_{ion}}}{{{n}_{compound}}} \\
\end{align}\]
- We can see here that the mole ratio between the original compound and ion that it will form will basically determine the concentration of the respective ion in the solution.
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