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In the measurement of alkalinity if the volume of 0.02 N H2SO4 used is 25 ml then alkalinity of the sample
represented as CaCO3 is (volume of sample is 5l)
represented as CaCO3 is (volume of sample is 5l)
______________mg/L
Correct answer is '5'. Can you explain this answer?
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In the measurement of alkalinity if the volume of 0.02 N H2SO4 used is...
Calculation of alkalinity:
To calculate the alkalinity of the sample, we need to determine the amount of acid required to neutralize the alkaline components present in the sample. In this case, 0.02 N H2SO4 is used as the titrant to determine the alkalinity.
Given:
Volume of 0.02 N H2SO4 used = 25 ml
Volume of sample = 5 L
Conversion:
We need to convert the volume of the titrant used to the volume of the sample.
Calculation:
1. Convert the volume of 0.02 N H2SO4 used to liters:
Volume of 0.02 N H2SO4 (in L) = 25 ml ÷ 1000 = 0.025 L
2. Calculate the normality of the sample:
Normality of the sample = Normality of H2SO4 × Volume of H2SO4 (in L)
Normality of the sample = 0.02 N × 0.025 L = 0.0005 N
3. Calculate the alkalinity in mg/L:
Alkalinity (in mg/L) = Equivalent weight of CaCO3 × Normality of the sample × 1000
The equivalent weight of CaCO3 is 50 g/mol.
Alkalinity (in mg/L) = 50 × 0.0005 × 1000 = 25 mg/L
Final Answer:
The alkalinity of the sample, represented as CaCO3, is 25 mg/L.
To calculate the alkalinity of the sample, we need to determine the amount of acid required to neutralize the alkaline components present in the sample. In this case, 0.02 N H2SO4 is used as the titrant to determine the alkalinity.
Given:
Volume of 0.02 N H2SO4 used = 25 ml
Volume of sample = 5 L
Conversion:
We need to convert the volume of the titrant used to the volume of the sample.
Calculation:
1. Convert the volume of 0.02 N H2SO4 used to liters:
Volume of 0.02 N H2SO4 (in L) = 25 ml ÷ 1000 = 0.025 L
2. Calculate the normality of the sample:
Normality of the sample = Normality of H2SO4 × Volume of H2SO4 (in L)
Normality of the sample = 0.02 N × 0.025 L = 0.0005 N
3. Calculate the alkalinity in mg/L:
Alkalinity (in mg/L) = Equivalent weight of CaCO3 × Normality of the sample × 1000
The equivalent weight of CaCO3 is 50 g/mol.
Alkalinity (in mg/L) = 50 × 0.0005 × 1000 = 25 mg/L
Final Answer:
The alkalinity of the sample, represented as CaCO3, is 25 mg/L.
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In the measurement of alkalinity if the volume of 0.02 N H2SO4 used is 25 ml then alkalinity of the samplerepresented as CaCO3 is (volume of sample is 5l)______________mg/LCorrect answer is '5'. Can you explain this answer?
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