M1V1/n1 = M2V2/n2 M1 = molarity of EDTA = 0.01 M M2 = molarity of bottled water = ? V1 = volume of titre used = 3.46 mL V2 = volume of bottled water used = 40 mL N1 = mole equivalence = 1 N2 = mole equivalence = 1 Calculate the hardness of water. Express the result in the form of Ca2+ ppm and CaCO3 ppm. Hardness is expressed as the equivalent amount of Ca2+ in molarity from titration. This is how I did it, and I don’t know if it’s right or wrong: Express the result in the form of Ca2+ ppm and CaCO3 ppm: 0.01 x 0.0034 = M x 0.04 M = (0.01 x 0.0034) / 0.04 M = 8.5 x 10^-4 Ca2+ Moles per liter: 40 x 8.5 x 10^-4 = 0.034 g per liter Ppm = mg per liter 34 mg per 1 liter 34 ppm Ca2+ Weight CaCO3 = 100 grams Weight Ca2+ = 40 34 x (100/40) = 85 ppm CaCO3
The Correct Answer and Explanation is:
To calculate the hardness of water in terms of Ca²⁺ ppm and CaCO₃ ppm, you have the right approach but there are a few minor adjustments to make. Let’s break it down step by step:
Given:
- M₁ = 0.01 M (molarity of EDTA)
- V₁ = 3.46 mL (volume of EDTA used)
- V₂ = 40 mL (volume of bottled water used)
- N₁ = 1 (mole equivalence of EDTA)
- N₂ = 1 (mole equivalence of Ca²⁺)
Step 1: Use the equation to find the molarity of Ca²⁺ in the sample
From the equation M₁V₁ / N₁ = M₂V₂ / N₂, you can solve for M₂ (the molarity of Ca²⁺ in the water). M2=M1V1V2=0.01×3.4640=8.65×10−4 MM₂ = \frac{M₁V₁}{V₂} = \frac{0.01 \times 3.46}{40} = 8.65 \times 10^{-4} \, MM2=V2M1V1=400.01×3.46=8.65×10−4M
So, the molarity of Ca²⁺ in the water is 8.65 × 10⁻⁴ M.
Step 2: Calculate the mass of Ca²⁺ in 1 liter of water (in grams)
Now, to convert the moles of Ca²⁺ to grams, you multiply by the molar mass of Ca²⁺. The molar mass of Ca²⁺ is 40.08 g/mol. Mass of Ca²⁺=8.65×10−4 M×40.08 g/mol=0.0347 g/L\text{Mass of Ca²⁺} = 8.65 \times 10^{-4} \, M \times 40.08 \, \text{g/mol} = 0.0347 \, \text{g/L}Mass of Ca²⁺=8.65×10−4M×40.08g/mol=0.0347g/L
This is the mass of Ca²⁺ in 1 liter of water.
Step 3: Convert to ppm (Ca²⁺)
Since 1 ppm = 1 mg/L, you have: 0.0347 g/L=34.7 mg/L=34.7 ppm of Ca²⁺0.0347 \, \text{g/L} = 34.7 \, \text{mg/L} = 34.7 \, \text{ppm of Ca²⁺}0.0347g/L=34.7mg/L=34.7ppm of Ca²⁺
Step 4: Convert to CaCO₃ ppm
To convert the Ca²⁺ concentration to the equivalent amount of CaCO₃, use the molar mass ratio. The molar mass of CaCO₃ is 100.09 g/mol, while the molar mass of Ca²⁺ is 40.08 g/mol. So: ppm of CaCO₃=34.7×100.0940.08=86.6 ppm of CaCO₃\text{ppm of CaCO₃} = 34.7 \times \frac{100.09}{40.08} = 86.6 \, \text{ppm of CaCO₃}ppm of CaCO₃=34.7×40.08100.09=86.6ppm of CaCO₃
Final Answer:
- Hardness in Ca²⁺ ppm = 34.7 ppm
- Hardness in CaCO₃ ppm = 86.6 ppm
Your calculation is very close but a little off due to rounding. The final results for Ca²⁺ are around 34.7 ppm and for CaCO₃ around 86.6 ppm.
