Question
80.0 grams of KBr is mixed with sufficient water to prepare 1865 mL of solution. This solution is then mixed with 46.3 mL of 0.425 M KCl solution. 12 mL of this mixture is subsequently added to 61.0 mL of distilled water. Calculate the molarity of K+ ions in the final mixture? KBr and KCl dissolve in water according to the following processes.
KCl(s)+H2O(l) yields K+(aq)+Cl-(aq)
KBr(s)+H2O(l) yields K+(aq)+Br-(aq)
KCl(s)+H2O(l) yields K+(aq)+Cl-(aq)
KBr(s)+H2O(l) yields K+(aq)+Br-(aq)
Answers
80g KBr/molar mass = moles KBr in 1865 mL.
Add moles KCl = M x L = 0.0463*0.425=??
Add mols KBr + moles KCl for total moles.
Add volume 1.865L + 0.0463L = total volume
Next step is to find how much is pulled from the solution. We took 12 mL; therefore, we took
moles in 12 mL = total moles x 12/total volume.
Then M = final mols/final volume in L.
Add moles KCl = M x L = 0.0463*0.425=??
Add mols KBr + moles KCl for total moles.
Add volume 1.865L + 0.0463L = total volume
Next step is to find how much is pulled from the solution. We took 12 mL; therefore, we took
moles in 12 mL = total moles x 12/total volume.
Then M = final mols/final volume in L.
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