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The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was diluted to 2.000 L. After the solution was buffered to pH 10, a 10.00 mL aliquot was titrated with 26.23 mL of 0.003474 M EDTA. The calcium in a second 10.00 mL aliquot was isolated as CaC2O4, redissolved in acid, and titrated with 20.49 mL of the EDTA solution. (Note: Normal levels for magnesium are 15 to 300 mg per day and for calcium are 50 to 400 mg per day.)
a. How many mg of Ca were in the original sample?
b. How many mg of Mg were in the original sample?

Sagot :

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Answer:

Explanation:

From the given information:

The concentration of metal ions are:

[tex][Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}[/tex]

[tex][Ca^{2+}]=0.007118 \ M[/tex]

[tex][Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}[/tex]

[tex]=0.001994 \ M[/tex]

Mass of Ca²⁺ in 2.00 L urine sample is:

[tex]= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}[/tex]

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

[tex]= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}[/tex]

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

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