Answer :
Final answer:
When 36.6 ml of 0.0205 M AgNO3 reacts completely with 0.610 M Na2S2O3, there will be no Ag+ ions left. This is because the reaction completely consumes the AgNO3, resulting in a Ag+ ion concentration of 0 M.
Explanation:
The question is asking about finding the concentration of silver ions (Ag+) after a chemical reaction between 36.6 ml of 0.0205 M AgNO3 and 0.610 M Na2 ~S2 O3. To solve this, we should first calculate the moles of AgNO3 and Na2S2O3. The moles are found by multiplying the volume (in litres) and the molarity (M).
For AgNO3: Moles= 0.0366 L * 0.0205 M = 0.00075 mol
For Na2S2O3: Moles= 0.0366 L * 0.610 M = 0.022 mol
Since the reaction equation is AgNO3 + 2 Na2S2O3 -> Ag2S2O3 + 2 NaNO3, we can see that it takes two moles of Na2S2O3 to react with one mole of AgNO3. So, all of the AgNO3 will react fully with the Na2S2O3, and you will have no Ag+ ions left. Therefore, the concentration of Ag+ ions is effectively 0 M after the reaction.
Learn more about Chemical Reaction here:
https://brainly.com/question/34137415
#SPJ11