Which ion is present in the highest concentration in a 2 mol/dm³ aqueous solution of sodium sulphate?
Answer:
It would have to be the sodium ion.
Na2SO4 ===> 2 Na+ + SO4-2
So there will be twice as much sodium ion as sulfate. And some of the sulfate ion might be taken up by a weak reaction to form bisulfate ion:
SO4-2 + H+ <======> HSO4-
sodium
There will be twice as many sodium as sulphate.
The sodium ion. The formula for sodium sulfate is Na2SO4, meaning one formula unit is made up of two sodium ions and one sulfate ion. When this solid dissolves in water, the ions are separated from each other, and there will be twice as many sodium ions as sulfate ions.
Think of this simple analogy. A bike has one frame and two tires. Although that counts as one bike, if you take it apart, you will have twice as many tires as frames. The water is taking apart the sodium sulfate, resulting in twice as many sodium ions as sulfate ions.
First- write the formual of sodium sulfate = Na2SO4
Second- write the equation for the dissociation of sodium sulfate in water
Na2SO4(s) --> 2Na+(aq) + SO42-(aq)
Notice that for every 1 mol of Na2SO4 that dissolves, 2 mol of Na+ ions are produced and 1 mol of SO42- ions are produced.
So if you have 2 mol/dm3 of Na2SO4 (assuming all of the Na2SO4 dissolves), you would have 4 mol/dm3 of Na+ ions and 2 mol/dm3 of SO42- ions.
Therefore, Na+ ions have the highest concentration (double that of the sulfate ion)
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