Question

The calcium content of urine can be determined by the following procedure: Ca2+ is precipitated as calcium oxalate in basic solution: Ca2+(aq) + C2O4 2(aq)  Ca(C2O4)H2O(s) After the precipitate is washed with ice-cold water to remove free oxalate, the solid is dissolved in acid, which gives Ca2+ and H2C2O4 in solution. The dissolved oxalic acid is heated to 60C and titrated with standardized potassium permanganate, KMnO4, until the purple end point is reached. 5 C2O4 2(aq) + 2 MnO4 (aq) + 16 H+(aq)  10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l) (colorless) (purple) (colorless) Titration of 0.1978 g of Na2C2O4 (FW: 134.00 g/mol) required 38.65 mL of a standard permanganate, MnO4 , solution. The calcium in a 25.00 mL urine sample was precipitated by the procedure above, redissolved and required 41.35 mL of the standard MnO4  solution. What is the concentration of Ca2+ in the urine in units of mg /dL?

Answer 1

Answer:

25317 mg/dL

Explanation:

Based in the reaction:

5 C₂O₄²⁻(aq) + 2 MnO4⁻(aq) + 16 H⁺(aq) → 10 CO₂(g) + 2 Mn²⁺(aq) + 8H₂O(l)

5 moles of oxalate react with 2 moles of permanganate.

The concentration of the KMnO₄ is:

0,1978g Na₂C₂O₄ ₓ (1mol / 134,00g) × (2mol permangante/ 5 mol oxalate) / 0,03865L = 0,01528M KMnO₄

In the precipitation of Calcium, you can see moles of oxalate ≡ moles of Ca. Thus:

Moles of titrated oxalate:

0,04135L × (0,01528moles / L)× (5moles oxalate / 2 moles permanganate) = 1,579x10⁻³ moles of oxalate≡ moles of Ca²⁺.

1,579x10⁻³ moles Ca ₓ (40,078g / mol) = 0,06329g Ca / 25,00mL = 2,532x10⁻³g/mL.

In mg/dL =

2,532x10⁻³ g/mL ₓ (1000mg /g) ₓ (10000mL / 1dL) = 25317 mg/dL

I hope it helps!