A sample of water was alkaline both to phenolphthalein and methyl orange. 100 ml of this water sample required 20 ml of N/50 H2SO4 for phenolphthalein end point and another 15 ml for complete neutralization. Calculate the type and amount of alkalinity in ppm.
Question
A sample of water was alkaline both to phenolphthalein and methyl orange. 100 ml of this water sample required 20 ml of N/50 H2SO4 for phenolphthalein end point and another 15 ml for complete neutralization. Calculate the type and amount of alkalinity in ppm.
Solution
To solve this problem, we need to understand that the alkalinity of water is due to the presence of bicarbonates, carbonates, and hydroxides. The reaction with phenolphthalein indicates the presence of carbonates and hydroxides, while the reaction with methyl orange indicates the presence of bicarbonates.
Step 1: Calculate the alkalinity due to hydroxides and carbonates (phenolphthalein alkalinity) The volume of H2SO4 used is 20 ml and the normality is N/50. The equivalent weight of CaCO3 (which we use as a standard for expressing alkalinity) is 50.
Alkalinity (as CaCO3) = Volume of acid (ml) * Normality of acid * Equivalent weight of CaCO3 * 1000 / Volume of water sample (ml)
= 20 * (1/50) * 50 * 1000 / 100 = 200 ppm
Step 2: Calculate the total alkalinity (methyl orange alkalinity) The total volume of H2SO4 used is 20 ml + 15 ml = 35 ml.
Total alkalinity (as CaCO3) = 35 * (1/50) * 50 * 1000 / 100 = 350 ppm
Step 3: Determine the type and amount of alkalinity The difference between the total alkalinity and the phenolphthalein alkalinity gives the bicarbonate alkalinity.
Bicarbonate alkalinity = Total alkalinity - Phenolphthalein alkalinity = 350 ppm - 200 ppm = 150 ppm
So, the water sample contains 200 ppm of carbonates and hydroxides (phenolphthalein alkalinity) and 150 ppm of bicarbonates (methyl orange alkalinity).
Similar Questions
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