an aqueous solution contains 1.2mM of total ions. if the solution is NaCl (aq), what is the concentration of chloride ions?
1 answer:
Explanation:
As the total concentration is given as 1.2 mM. And, it is also given that salt present in the solution is NaCl.
As sodium chloride is an ionic compound so, when it is added to water then it will dissociate into sodium and chlorine ions as follows.
So, it means in total there will be formation of 2 ions when one molecules of NaCl dissociates.
Therefore, concentration of chlorine ions will be calculated as follows.
Concentration of ions =
= 0.6 mM
Thus, we can conclude that the concentration of chloride ions is 0.6 mM.
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Answer:
0.0714 M for the given variables
Explanation:
The question is missing some data, but one of the original questions regarding this problem provides the following data:
Mass of copper(II) acetate:
Volume of the sodium chromate solution:
Molarity of the sodium chromate solution:
Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:
Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:
Moles of the sodium chromate solution would be found by multiplying its volume by molarity:
Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.
Write the net ionic equation for this reaction:
Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:
According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:
The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:
Explanation:
The given data is as follows.
= 10 mM =
M
= 750 ml,
= 5 ml
= ?
Therefore, calculate the molarity of given NaCl stock as follows.
= 1.5 M
Thus, we can conclude that molarity of given NaCl stock is 1.5 M.