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.
The Correct Answer and Explanation is:
To calculate the hardness of water based on the given formula, let’s break it down:
We are given the following information:
- M1 = 0.01 M (molarity of EDTA)
- V1 = 3.46 mL (volume of EDTA used for titration)
- V2 = 40 mL (volume of bottled water used)
- N1 = 1 (mole equivalence of EDTA)
- N2 = 1 (mole equivalence of bottled water)
The formula for the calculation of molarity (M) is:M1×V1N1=M2×V2N2\frac{M1 \times V1}{N1} = \frac{M2 \times V2}{N2}N1M1×V1=N2M2×V2
We can rearrange the equation to solve for M2M2M2 (the molarity of bottled water):M2=M1×V1×N2V2×N1M2 = \frac{M1 \times V1 \times N2}{V2 \times N1}M2=V2×N1M1×V1×N2
Now, substitute the given values:M2=0.01 M×3.46 mL×140 mL×1M2 = \frac{0.01 \, \text{M} \times 3.46 \, \text{mL} \times 1}{40 \, \text{mL} \times 1}M2=40mL×10.01M×3.46mL×1M2=0.01×3.4640M2 = \frac{0.01 \times 3.46}{40}M2=400.01×3.46M2=0.034640M2 = \frac{0.0346}{40}M2=400.0346M2=0.000865 MM2 = 0.000865 \, \text{M}M2=0.000865M
So, the molarity of bottled water (M2) is 0.000865 M.
Next, to calculate the hardness of water in terms of Ca2+\text{Ca}^{2+}Ca2+ in ppm (parts per million) and CaCO3\text{CaCO}_3CaCO3 in ppm, we need to relate the molarity of Ca2+\text{Ca}^{2+}Ca2+ to the equivalent concentration of CaCO3\text{CaCO}_3CaCO3.
- Hardness as Ca²⁺ (ppm):
The molarity of Ca2+\text{Ca}^{2+}Ca2+ is equivalent to the molarity of EDTA used in the titration. Since 1 mole of EDTA reacts with 1 mole of Ca2+\text{Ca}^{2+}Ca2+, we can convert this molarity directly to ppm.
ppm Ca2+=M2×40,000\text{ppm Ca}^{2+} = M2 \times 40,000ppm Ca2+=M2×40,000ppm Ca2+=0.000865 M×40,000=34.6 ppm Ca2+\text{ppm Ca}^{2+} = 0.000865 \, \text{M} \times 40,000 = 34.6 \, \text{ppm Ca}^{2+}ppm Ca2+=0.000865M×40,000=34.6ppm Ca2+
- Hardness as CaCO₃ (ppm):
The molecular weight of CaCO3\text{CaCO}_3CaCO3 is approximately 100 g/mol, and the equivalent weight for calcium in CaCO3\text{CaCO}_3CaCO3 is 40 g/mol. Thus, we use a conversion factor of 2.5 to convert the Ca2+\text{Ca}^{2+}Ca2+ hardness to CaCO3\text{CaCO}_3CaCO3 hardness:
ppm CaCO3=ppm Ca2+×2.5\text{ppm CaCO}_3 = \text{ppm Ca}^{2+} \times 2.5ppm CaCO3=ppm Ca2+×2.5ppm CaCO3=34.6×2.5=86.5 ppm CaCO3\text{ppm CaCO}_3 = 34.6 \times 2.5 = 86.5 \, \text{ppm CaCO}_3ppm CaCO3=34.6×2.5=86.5ppm CaCO3
Final Result:
- Hardness of water in Ca²⁺: 34.6 ppm
- Hardness of water in CaCO₃: 86.5 ppm
This means that the hardness of the water sample is equivalent to 34.6 ppm of Ca2+\text{Ca}^{2+}Ca2+ or 86.5 ppm of CaCO3\text{CaCO}_3CaCO3.
