Suppose 3.13g of potassium chloride KCl are dissolved in 500.mL of water. Find the composition of the resulting electrolyte solution.In particular, list the chemical symbols (including any charge) of each dissolved ion in the table below. List only one ion per row.Then, calculate the concentration of each ion in mEqL, and write the concentration in the second column of each row. Be sure you round your answers to the correct number of significant digits.
Question
Suppose 3.13g of potassium chloride KCl are dissolved in 500.mL of water. Find the composition of the resulting electrolyte solution.In particular, list the chemical symbols (including any charge) of each dissolved ion in the table below. List only one ion per row.Then, calculate the concentration of each ion in mEqL, and write the concentration in the second column of each row. Be sure you round your answers to the correct number of significant digits.
Solution
The potassium chloride (KCl) will dissociate into its ions when dissolved in water. The ions are:
- Potassium ion (K+)
- Chloride ion (Cl-)
To calculate the concentration of each ion in mEq/L, we first need to convert grams to moles. The molar mass of KCl is approximately 74.55 g/mol.
So, 3.13g of KCl is 3.13g / 74.55 g/mol = 0.042 mol.
Since KCl dissociates into one K+ ion and one Cl- ion, we have 0.042 mol of each ion.
The volume of the solution is 500 mL, which is 0.5 L.
So, the concentration of each ion in mol/L is 0.042 mol / 0.5 L = 0.084 mol/L.
To convert this to mEq/L, we multiply by the valence of the ion and by 1000 (since 1 Eq = 1 mol of charge, and 1 mEq = 0.001 Eq).
For K+, the valence is +1, so the concentration is 0.084 mol/L * 1 * 1000 = 84 mEq/L.
For Cl-, the valence is -1, so the concentration is 0.084 mol/L * 1 * 1000 = 84 mEq/L.
So, the composition of the electrolyte solution is:
| Ion | Concentration (mEq/L) |
|---|---|
| K+ | 84 |
| Cl- | 84 |
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