Question

A group in your lab did three tree trials of edta titration to quantify the concentration of cacl2 solution (analyte) during the first week. the concentration of edta solution was 0.05 mol/l. the initial reading of edta level in the buret was 0.00 ml, and when the titration reached the end point, the final reading was 10.14 ml. the volume of cacl2 solution was 25 ml. what is the concentration of ca2+ ions (in ppmcaco3) in cacl2 solution?

Answer 1

Answer : The concentration of Ca²⁺ ions is 2030 ppm CaCO₃ in the given solution.

Explanation :

Step 1 : Write chemical equation

The reaction of Ca ions and EDTA solution can be represented as follows.

$Ca^{2+} (aq) + EDTA^{4-} (aq) \rightarrow [Ca-EDTA]^{2-} (aq)$

Step 2 : Find moles of EDTA

The molarity of EDTA solution is 0.05 mo/L

The volume of EDTA required for the reaction is 10.14 mL

The volume in liters = $10.14 mL \times\frac{1L}{1000 mL} = 0.01014 L$

The moles of EDTA can be calculated using molarity formula which is given below.

$Molarity =\frac{Moles}{L}$

$0.05 M =\frac{moles}{0.01014L}$

$moles = 0.05 \times 0.01014 = 0.000507$

We have 0.000507 moles of EDTA.

Step 3 : Find moles of Ca ions.

The mole ratio of Ca ions and EDTA is 1 : 1.

We have 0.000507 mol EDTA.

Moles of Ca²⁺ ions = $0.000507mol EDTA \times\frac{1 mol Ca^{2+}}{1 mol EDTA}$

We have 0.000507 moles of Ca²⁺ ions.

Step 4 : Find mg of CaCO₃

The concentration of Ca²⁺ ions needs to be expressed as ppm CaCO₃.

We assume that all the Ca²⁺ ions come from CaCO₃.

So moles of CaCO₃ are same as moles of Ca²⁺ ions which is 0.000507.

$mg (CaCO_{3}) = mol (CaCO_{3}) \times Molar Mass (CaCO_{3}) \times\frac{1000mg}{1g}$

Molar mass of CaCO₃ is 100.1 g/mol

Let us plug in the values in above formula.

$mg (CaCO_{3}) = 0.000507 mol \times \frac{100.1g}{mol} \times\frac{1000mg}{1g}$

We have 50.75 mg CaCO₃

Step 5 :  Find ppm using the formula

ppm CaCO₃ is calculated using following formula.

$ppm(CaCO_{3})=\frac{mg (CaCO_{3})}{L (CaCl_{2})}$

The volume of CaCl₂ solution is 25 mL which is 0.025 L.

$ppm (CaCO_{3}) =\frac{50.75 mg (CaCO_{3})}{0.025L} = 2030 ppm$

The concentration of Ca²⁺ ions is 2030 ppm CaCO₃ in the given solution.