Answered

Get detailed and accurate responses to your questions with IDNLearn.com. Our experts are ready to provide prompt and detailed answers to any questions you may have.

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 :

Ajeigbeibraheemidn

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

Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.