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.a.Alkalinity due to OH– = 240 ppm, Alkalinity due to CO32– = 240 ppmb.Alkalinity due to OH– = 340 ppm, Alkalinity due to CO32– = 140 ppmc.Alkalinity due to OH– = 00 ppm, Alkalinity due to CO32– = 480 ppmd.Alkalinity due to OH– = 24 ppm, Alkalinity due to CO32– = 24 ppm
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.a.Alkalinity due to OH– = 240 ppm, Alkalinity due to CO32– = 240 ppmb.Alkalinity due to OH– = 340 ppm, Alkalinity due to CO32– = 140 ppmc.Alkalinity due to OH– = 00 ppm, Alkalinity due to CO32– = 480 ppmd.Alkalinity due to OH– = 24 ppm, Alkalinity due to CO32– = 24 ppm
Solution
The type and amount of alkalinity present in the water sample can be determined by the volume of H2SO4 required to neutralize the sample.
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The total volume of H2SO4 used is 18 mL for phenolphthalein end point and another 6 mL for complete neutralization, which is 24 mL in total.
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The normality of H2SO4 is 0.02 N.
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The alkalinity due to OH- and CO32- can be calculated using the formula: Alkalinity (ppm) = (Volume of H2SO4 (mL) * Normality of H2SO4 * 50,000) / Volume of water sample (mL)
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For phenolphthalein end point (OH- and CO32-): Alkalinity = (18 mL * 0.02 N * 50,000) / 50 mL = 360 ppm
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For complete neutralization (OH-, CO32- and HCO3-): Alkalinity = (24 mL * 0.02 N * 50,000) / 50 mL = 480 ppm
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The alkalinity due to HCO3- is the difference between the total alkalinity and the phenolphthalein alkalinity, which is 480 ppm - 360 ppm = 120 ppm.
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Therefore, the alkalinity due to OH- and CO32- is 360 ppm and the alkalinity due to HCO3- is 120 ppm.
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Since the water sample is alkaline to both phenolphthalein and methyl orange, it contains both OH- and CO32-.
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However, the question does not provide enough information to determine the specific amounts of OH- and CO32- alkalinity.
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Therefore, none of the options provided (a, b, c, d) are correct based on the given information and calculations.
Similar Questions
100 mL of a raw water sample required 25 mL of N/50 H2SO4 for neutralisation to phenolphthalein end point. After this methyl orange indicator was added to it and the end point was 28.0 mL. Calculate the alkalinity of water as CaCO3 equvivalents in parts per million.a.Alkalinity due to HCO32– = 60ppm Alkalinity due to CO32– = 220 ppmb.Alkalinity due to OH– = 220 ppm Alkalinity due to CO32– = 60 ppmc.Alkalinity due to HCO32– = 220 ppm Alkalinity due to CO32– = 60 ppmd.Alkalinity due to OH– =60 ppm Alkalinity due to CO32– =220 ppm
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.
A water sample is alkaline to both phenolphthalein as well as methyl orange. 200 mL of a this water sample on titration with N/50 HCl required 9.4 mL of the acid to phenolphthalein end point. When a few drops of methyl orange are added to the same solution and titration is further continued, the yellow color of the solution just turned red after addition of 30.4 mL of the acid solution. Calculate the alkalinity present in the water sample.a.P = 152 ppm., M = 47ppm and HCO3– alkalinity = 58 ppm, CO32– alkalinity = 94 ppmb.P = 1520 ppm., M = 470ppm and HCO3– alkalinity = 940 ppm, CO32– alkalinity = 580 ppmc.P = 47 ppm., M = 152 ppm and HCO3– alkalinity = 58 ppm, CO32– alkalinity = 94 ppmd.P = 470 ppm., M = 1520 ppm and HCO3– alkalinity = 580 ppm, CO32– alkalinity = 940 ppm
If a solution of pure water and sodium hydroxide is titrated with H2SO4 for the alkalinity measurement. Then the correct option would be...a.The Phenolphthalein alkalinity will be the total alkalinityb.The Phenolphthalein alkalinity will be the bicarbonate alkalinityc.The Methyl orange alkalinity will be the total alkalinityd.The Phenolphthalein alkalinity will not be the total alkalinity
100 ml of an alkaline water in presence of phenolphthalein, 50 ml of the N/100 HCl. The resulting mixture in presence of methyl orange indicator consumes 35 ml of HCl. Predict the alkalinities and calculate their amounts in ppm as CaCO3 equivalent.
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