what is the concentration of the ions in 0.20M Na2CrO4?
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To determine the concentration of ions in a solution of Na2CrO4, we need to consider the dissociation of the Na2CrO4 compound.
Na2CrO4 dissociates into two sodium ions (2 Na+) and one chromate ion (CrO4 2-).
Therefore, the concentration of sodium ions (Na+) in the solution is twice the molarity of Na2CrO4:
Concentration of Na+ = 2 × 0.20 M = 0.40 M
The concentration of chromate ions (CrO4 2-) in the solution is the same as the molarity of Na2CrO4:
Concentration of CrO4 2- = 0.20 M
So, the concentration of ions in a 0.20 M Na2CrO4 solution is:
Sodium ions (Na+): 0.40 M
Chromate ions (CrO4 2-): 0.20 M
To find the concentration of the ions in Na2CrO4, we need to consider the dissociation of the compound in water.
Na2CrO4 dissociates into two sodium ions (Na+) and one chromate ion (CrO4^2-).
The concentration of sodium ions is twice the molar concentration of Na2CrO4 because each formula unit of Na2CrO4 produces two sodium ions upon dissociation.
Therefore, the concentration of sodium ions is 0.20M × 2 = 0.40M.
The concentration of chromate ions is the same as the molar concentration of Na2CrO4 because each formula unit of Na2CrO4 produces one chromate ion upon dissociation.
Therefore, the concentration of chromate ions is 0.20M.
In summary:
- The concentration of sodium ions (Na+) is 0.40M.
- The concentration of chromate ions (CrO4^2-) is 0.20M.