Respuesta :
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.
[tex] Ca^{2+} (aq) + EDTA^{4-} (aq) \rightarrow [Ca-EDTA]^{2-} (aq) [/tex]
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 = [tex] 10.14 mL \times\frac{1L}{1000 mL} = 0.01014 L [/tex]
The moles of EDTA can be calculated using molarity formula which is given below.
[tex] Molarity =\frac{Moles}{L} [/tex]
[tex] 0.05 M =\frac{moles}{0.01014L} [/tex]
[tex] moles = 0.05 \times 0.01014 = 0.000507 [/tex]
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 = [tex] 0.000507mol EDTA \times\frac{1 mol Ca^{2+}}{1 mol EDTA} [/tex]
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.
[tex] mg (CaCO_{3}) = mol (CaCO_{3}) \times Molar Mass (CaCO_{3}) \times\frac{1000mg}{1g} [/tex]
Molar mass of CaCO₃ is 100.1 g/mol
Let us plug in the values in above formula.
[tex] mg (CaCO_{3}) = 0.000507 mol \times \frac{100.1g}{mol} \times\frac{1000mg}{1g} [/tex]
We have 50.75 mg CaCO₃
Step 5 : Find ppm using the formula
ppm CaCO₃ is calculated using following formula.
[tex] ppm(CaCO_{3})=\frac{mg (CaCO_{3})}{L (CaCl_{2})} [/tex]
The volume of CaCl₂ solution is 25 mL which is 0.025 L.
[tex] ppm (CaCO_{3}) =\frac{50.75 mg (CaCO_{3})}{0.025L} = 2030 ppm [/tex]
The concentration of Ca²⁺ ions is 2030 ppm CaCO₃ in the given solution.
