Answer:
- 0.0144 M
- 1440 ppm
Explanation:
Ca(In)²⁺ + EDTA → Ca(EDTA)²⁺ + In
We use the volume of EDTA consumed in the titration to <u>calculate the moles of Ca⁺² ions</u>:
- 0.012 L * 0.0600 M *
= 7.20x10⁻⁴ mol Ca⁺²
Now we <u>calculate the molarity</u>:
- 7.20x10⁻⁴ mol Ca⁺² / 0.050 L = 0.0144 M
To calculate in ppm, we <u>use the moles of Ca⁺² and convert to mg of CaCO₃:</u>
- 7.20x10⁻⁴ mol Ca⁺² = 7.20x10⁻⁴ mol CaCO₃
- 7.20x10⁻⁴ mol CaCO₃ * 100g/mol *
= 72 mg CaCO₃
Finally, <u>the concentration in ppm</u> is:
- 72 mg CaCO₃ / 0.050L = 1440 ppm