Calcium levels in blood can be determined by adding oxa-late ion to precipitate calcium oxalate, CaC2O4, followed by dissolving the precipitate in aqueous acid and titrating the resulting oxalic acid (H2C2O4) with KMnO4:
How many milligrams of Ca2+ are present in 10.0 mL of blood if 21.08 mL of 0.000 988 M KMnO4 solution is needed for the titration?

Question

Calcium levels in blood can be determined by adding oxa-late ion to precipitate calcium oxalate, CaC2O4, followed by dissolving the precipitate in aqueous acid and titrating the resulting oxalic acid (H2C2O4) with KMnO4: 
How many milligrams of Ca2+ are present in 10.0 mL of blood if 21.08 mL of 0.000 988 M KMnO4 solution is needed for the titration?

Accepted Answer

Hello everyone today we are being given the following problem. The balanced equation for the reaction of bro. Mean with Aquarius style sulfate in neutral solution is as follows how much br to our browning ingram's is in a solution of 37.20 mL of 0.175 moller of sodium So faint is needed to titrate the b. r. two solution study. Um So fight. The first thing that you wanna do is you wanna start with are given. So we want to start with our given And we were given 37.20 middle leaders. And so we want to convert Our leaders are middle leaders into leaders with the conversion factor that one leader is equal to middle leaders. Next we want to use the polarity of our sodium sulfite To get rid of our leaders and transform that into moles. And so we can do that with 0. moles of in a two S over one leader. And so now we want to do a multiple ratio. So we'll say one more cell fight, It's equal to one mole of sodium soul fight. And then the next step is to get rid of our most of sulfate into moles of browning. So we have one mole of browning Is equal to one mole of soul fight. And I'm getting these numbers from our chemical equation. So the one wall comes from the fact that the coefficient in front of our br two or bromine is one and for sulfite it is two but gets simplified to one and then last but not least. We multiplied by the molar mass of browning, Which is 159.808 g. And when our units cancel out, we end up with an answer of 0.5202g of brown. Mean and this is our final answer. Overall, I hope this helped until next time.

Verified Solution

Key Concepts

Precipitation Reactions

Titration

Stoichiometry