A water sample was alkaline to both phenolphthalein and methyl orange. 50 mL of the watersample required 18 mL of 0.02 N H2SO4 for phenolphthalein end point and another 6 mLfor complete neutralisation. Describe the type and amount of alkalinity present.
Question
A water sample was alkaline to both phenolphthalein and methyl orange. 50 mL of the watersample required 18 mL of 0.02 N H2SO4 for phenolphthalein end point and another 6 mLfor complete neutralisation. Describe the type and amount of alkalinity present.
Solution
The water sample is alkaline to both phenolphthalein and methyl orange, which indicates the presence of both bicarbonate (HCO3-) and carbonate (CO3 2-) ions.
The volume of H2SO4 required for the phenolphthalein end point (18 mL) neutralizes the CO3 2- ions, while the additional volume required for the methyl orange end point (6 mL) neutralizes the HCO3- ions.
To calculate the amount of alkalinity, we use the formula:
Alkalinity (mg/L as CaCO3) = (Volume of acid * Normality of acid * 50,000) / Sample volume
For the phenolphthalein end point (carbonate alkalinity):
Carbonate alkalinity = (18 mL * 0.02 N * 50,000) / 50 mL = 360 mg/L as CaCO3
For the methyl orange end point (bicarbonate alkalinity):
Bicarbonate alkalinity = (6 mL * 0.02 N * 50,000) / 50 mL = 120 mg/L as CaCO3
So, the water sample contains 360 mg/L of carbonate alkalinity and 120 mg/L of bicarbonate alkalinity.
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