How many mL of 0.0283 M Na2S2O3 do you need to dissolve 0.860 g of AgBr?

Answer 1

We need a balanced equation for this reaction. I'm assuming you are reacting, not dissolving the silver bromide.

This is the balanced reaction:

2Na₂S₂O₃ + AgBr → NaBr + Na₃(Ag(S₂O₃)₂)

This tells us that 1 moles of AgBr needs 2 moles of Na₂S₂O₃. This is a 2/1 molar ratio.

0.860g of AgBr, with a molar mass of 187.8g/mole, is (0.860g/187.8g/mole) = 0.00458 moles of NaBr. We need twice this number of moles of Na₂S₂O₃ : 2*(0.00458) = 0.00916 moles of Na₂S₂O₃ .

(0.00916 moles of Na₂S₂O₃ )*(158.1g/mole) = 1.448 g of Na₂S₂O₃ .

But we are given Na₂S₂O₃ as a 0.0283 M solution, which means 0.0283 moles/liter. We want 0.00916 moles, so find the liters, V, of 0.0283 M solution that will deliver that number of moles.

(0.0283 moles/liter)*V = 0.00916 moles

V = 0.324 liters, or 324 ml.